Assigned problems for MAA134 Differential Equations at the course occasion in period 3 Academic Year 2014/15 Version 2015-06-16 Block 1 – ODE of first order 1. Introduction to differential equations .............................. ZC 1.1–1.3 Determine the order of the differential equation and whether it is linear or not (with respect to y). √ 2 2 1) (5x2 − 2)y 000 − 2xy 0 + y = x4 8) 3y 00 = 7yy 0 15) y (4) e− x +y + 2y 0 = y INL 1.1 2) y (4) y = x3 y 3) x4 y (3) + x7 y = 0 2 00 0 4) x y + 5xyy + y = sin(x) 0 x 3 5) yy = e y 6) cos(x)y 00 + x2 y 0 + 3y = 0 7) x2 y (5) − ln(x)y 000 + xy = 0 2. 9) ex y (4) + tan(x)y 00 + x3 y = 1 p 10) xy 0 + y 00 /x = x2 + 1 √ 11) xy (3) = x5 y 0 + x2 y 0 12) y − 4yy 000 =0 √ 13) y 00 /(1 + x2 ) = y x cot(x) p 14) 7y 0 + 2 1 + y 2 = 0 17) 5y = y (5) y 00 √ 18) x7 xy (4) + cos(x)y 00 = y 19) e9x y 00 + sin(x)y 0 + xy = 0 20) y 000 y 0 − xy 00 y = 0 First-order ordinary differential equations INL 2.1 .................... ZC 2.1–2.5 Sketch the phase portrait of the autonomous differential equation and classify the stationary points. Alse, state the equilibrium solutions and sketch the typical solution curves. 1) dy = −(y + 2)y 2 (y − 1)3 dx 8) dy = y 3 (y − 2)(y − 7) dx 2) dy = (y + 1)y(y − 1)2 dx 9) dy = −(y + 3)2 (y + 1)3 y dx 3) dy dy = −(y + 3)2 (y + 1)(y − 2)3 10) = (y + 4)y 2 (y − 2)3 dx dx 4) √ √ 16) ln(x)y 00 + y/ x = tan( x)y 0 dy = (y + 2)y 3 (y − 3)3 dx dy 5) = −y 2 (y − 2)(y − 4) dx 11) 15) dy = (y + 4)y 3 (y − 1)2 dx 16) dy = (y + 2)2 (y + 1)y 2 dx 17) dy = (y + 3)3 y(y − 5)2 dx 18) dy = y 4 (y − 2)3 (y − 4)2 dx 19) dy = (y + 1)4 y(y − 2)3 dx 20) dy = (y + 3)2 y 3 (y − 6)3 dx dy = −(y + 1)y(y − 3)2 dx dy 12) = (y + 1)2 y 2 (y − 1) dx 6) dy = (y + 1)y(y − 4)3 dx 13) dy = (y + 2)3 y 2 (y − 4) dx 7) dy = −(y + 5)y 2 (y − 1) dx 14) dy = (y + 1)2 y 4 (y − 2)3 dx 2 2.. FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS INL 2.2 3 Solve the initial-value problem. Also, determine the interval of existence for the solution. 1) yy 0 = −x , 6) y 0 = 5y + y 2 , y(2) = −3 . 2) y 0 = y 2 − 9 , 16) y 0 = 2x3 e−2y , y(0) = −4 . √ 12) y 0 = y 2 cos(x) , 17) 2xyy 0 = 1 + y 2 , 7) y 0 = y(0) = −1 . 2 11) yy 0 = xe−y , p y(0) = ln(2) . xy , y(1) = 1 . y(0) = 2 . y(π/6) = 2/3 . y(5/3) = −2 . √ 2 3) y 0 = xex −2y , √ y( 2) = ln(2e) . 8) y 0 = 2xey , y(2) = − ln(2) . 9) xy 0 + 2y = y 2 , 4) y 0 = (y − 1)2 , y(1) = 3 . y(−1) = 0 . 5) y 0 = y 2 + 4 , √ y(0) = 2 3 . INL 2.3 10) y 0 ln(y) = xy , y(0) = 1/e2 . 13) y 0 = (x + 1) y , 18) y 0 = x3 y 4 , y(1) = 4 . 14) y 0 = y−2 x+1 y(1) = 2 . 2 , √ 19) y 0 = y 2 x , y(4) = 3 . y(−2) = 3 . p 15) y 0 = x 1 − y 2 , √ y( π) = 1/2 . 3 Solve the initial-value problem. Also, determine the interval of existence for the solution. 1) xy 0 + 3y = 3 , 6) xy 0 − 5y = 2x , y(1) = 8 . y(1) = 5 . 2) y 0 + 3xy = x , 7) x2 y 0 + 2xy = 1 , y(0) = 0 . y(2) = 1 . 3) y 0 − 9x2 y = 27x2 , 8) y 0 − 4xy = 2x , y(0) = 9 . y(0) = 1 . 4) x2 y 0 − xy = 3 , 9) y 0 − 8x3 y = 16x3 , y(1) = −1 . 5) y 0 + 6x2 y = x2 , y(0) = 3 . 10) x2 y 0 + 5xy = −2 , y(0) = 1/3 . y(1) = 1 . √ 11) xy 0 − 2y = x x , √ 16) xy 0 + 4y = 2 x , y(4) = 80 . y(1) = −1 . 2 12) y 0 + 6xy = e−3x , 17) y 0 − 12x3 y = x3 , y(0) = −2 . y(0) = 1/12 . 13) y 0 − 7xy = 14x , 18) x2 y 0 + 9xy = 8/x4 , y(1) = 3 . y(0) = 5 . 14) y 0 + 15x4 y = −3x4 , 19) y(0) = 4/5 . √ xy 0 − y = 1 , y(4) = 1 . √ 15) x2 y 0 − 6xy = 16/x , 20) xy 0 + 8y = x2 x , y(−1) = 3 . y(1) = −1 . INL 2.4 2 20) yy 0 + x2 e2x +y = 0 , p y(0) = ln(3) . Solve the initial-value problem where the function U is defined by U (x−ξ) = 1) y 0 + 8y = 16 U (x − 2) , y(0) = 1 . 2) y 0 + 3y = −9 U (x − 3) , y(0) = 1 . 3) y 0 + 9y = 27 U (x − 4) , y(0) = 1 . 4) y 0 + 6y = −12 U (x − 5) , y(0) = 1 . 5) y 0 + 2y = 6 U (x − 6) , y(0) = 1 . 6) y 0 + 5y = −5 U (x − 7) , y(0) = 1 . 7) y 0 + 7y = 14 U (x − 8) , y(0) = 1 . 8) y 0 + 3y = −6 U (x − 9) , y(0) = 1 . 9) y 0 + 10y = 30 U (x − 6) , y(0) = 1 . 10) y 0 + πy = 2 U (x − 5) , y(0) = 1 . 11) y 0 + 2y = −4 U (x − 4) , y(0) = 1 . 12) y 0 + 12y = 24 U (x − 3) , y(0) = 1 . 0, 1, x < ξ, . x ≥ ξ. 13) y 0 + 4y = −20 U (x − 7) , y(0) = 1 . 14) y 0 + 8y = 24 U (x − 5) , y(0) = 1 . 15) y 0 + 5y = −15 U (x − 8) , y(0) = 1 . 16) y 0 + 7y = 28 U (x − 9) , y(0) = 1 . 17) y 0 + 9y = −18 U (x − 5) , y(0) = 1 . 18) y 0 + 6y = 42 U (x − 4) , y(0) = 1 . 4 19) y 0 + 13y = −26 U (x − 6) , y(0) = 1 . INL 2.5 20) y 0 + 4y = 12 U (x − 3) , y(0) = 1 . Prove that the differential equation is exact and determine (at least on implicit form) the general solution. 1) (10xy 3 + 35x4 y 3 ) dx + (15x2 y 2 + 21x5 y 2 ) dy = 0 11) (1 + x)ex y − y 3 dx + (xex − 3xy 2 ) dy = 0 2) (2x/y 3 + y 2 ) dx + (2xy − 3x2 /y 4 ) dy = 0 12) (6x2 y + 10xy 3 ) dx + (2x3 + 15x2 y 2 ) dy = 0 3) (y 5 + 6x2 ln(y)) dx + (5xy 4 + 2x3 /y) dy = 0 13) (6x/y − 2y 3 /x3 ) dx + (3y 2 /x2 − 3x2 /y 2 ) dy = 0 4) (16x3 y 2 − 2xy 3 ) dx + (8x4 y − 3x2 y 2 ) dy = 0 √ √ 14) (6y 2 /x4 +12x y) dx+(3x2 / y −4y/x3 ) dy = 0 5) (2xe3y − 9x2 y 2 ) dx + (3x2 e3y − 6x3 y) dy = 0 15) (5x4 y 3 + 5xy 4 ) dx + (3x5 y 2 + 10x2 y 3 ) dy = 0 6) (3x2 y 2 + 6xy 5 ) dx + (2x3 y + 15x2 y 4 ) dy = 0 √ √ 16) (6x2 y y−2y 3 /x3 ) dx+(3y 2 /x2 +3x3 y) dy = 0 7) (y 2 /x + 4xy) dx + (2 ln(x)y + 2x2 ) dy = 0 17) (8xy 2 − 15x4 ln(y)) dx + (8x2 y − 3x5 /y) dy = 0 8) (3y 3 − 4x3 y 2 ) dx + (9xy 2 − 2x4 y) dy = 0 18) (y 2 /x − 3y 3 /x2 ) dx + (9y 2 /x + 2y ln(x)) dy = 0 √ √ 9) (6x3 + y 3 + 2x ln(y)) dx + (3xy 2 + x2 /y) dy = 0 19) (6x2 y − y 5 ) dx + (x3 / y − 5xy 4 ) dy = 0 20) (4x3 /y 2 − 2xy 3 ) dx − (2x4 /y 3 + 3x2 y 2 ) dy = 0 10) (2xy 3 − 2y 2 ) dx + (3x2 y 2 − 4xy) dy = 0 INL 2.6 Prove that the differential equation is homogeneous and determine (at least on implicit form) all solutions. 1) (x + y)y 0 = x − y 8) (y + xey/x )dx = xdy √ 2) (y + 2 xy)dx = xdy p 3) xyy 0 = y 2 + x 4x2 + y 2 9) xyy 0 = x2 + 3y 2 p 10) (y 2 + x x2 + y 2 )dx = xydy 2 15) x2 + xyy 0 = 2y 2 16) x2 dy = (2xy + y 2 )dx 17) 4x2 + 2xyy 0 = 3y 2 0 2 4) (x3 + y 3 )dx = xy 2 dy 11) 3x + 2y + 4xyy = 0 5) x(x + y)y 0 = y(x − y) 12) y(3x + y)dx + x(x + y)dy = 0 6) (x + 3y)dx = (y − 3x)dy 13) x2 = y 2 + xyy 0 19) y 2 − 3xy − 2x2 = (x2 − xy)y 0 7) (x − y)y 0 = x + y 14) 4xydx + x2 dy = y 2 dx 20) (3xy+y 2 )dx+(x2 +xy)dy = 0 INL 2.7 18) (x2 + 2y 2 )dx = xydy Solve the initial-value problem. Also, determine the interval of existence for the solution. 1) 2xy 0 + y(1 + xy) = 0 , 6) 2xy 0 = 5(1/y − y) , y(1) = 1/2 . 2) y 0 = (xy − 1)y , y(0) = 1 . y(−2) = −3 . 7) y 0 = y + xy 1/3 , y(0) = −8 . 11) xy 0 = y(y − 1) , y(−1) = 1/3 . 12) y 0 = (ex y − 1)y , y(−1) = −1 . 3) (x + x2 )y 0 = y(1 − y) , 8) 3y 0 = −4x3 y(1 + 2y 3 )13) xy 0 = 2y(1 + 2y 2 ) , , y(1) = 2 . y(1) = 1/4 . y(0) = 1/2 . 4) y 0 = y + 1/y , y(0) = −1 . 9) y 0 = 2xy(1 − y) , y(0) = 3 . 5) y 0 = y(2 − y) , 10) xy 0 + y = x2 y −2 , y(0) = 1 . y(1) = 1 . 16) xy 0 = y + 2xy −3 , y(3) = −1 . 17) xy 0 + x3 y 2 = 2y , y(−1) = 2 . 18) 2xy 0 + x2 y 3 + 2y = 0 , y(−e) = −1/(2e) . 14) y 0 + xy(2 + y 3 ) = 0 , 19) xy 0 + 4y = x4 y 2 , y(0) = −1 . y(1) = 1 . 15) xy 0 = (x4 y − 1)y , y(−1) = 6 . 20) xy 0 = 2y + y −1 , y(1) = −2 . 3.. MODELING WITH FIRST-ORDER DE INL 2.8 5 The differential equation has an integrating factor which only depends on x (odd-numbered problems) or y (even-numbered problems). Find the equation for the solution curve which contains the point (1, 2) . 1) (12x + y 3 /x) dx + 3y 2 dy = 0 11) (x + 3y 2 ) dx + 2xy dy = 0 2) 10x/y dx + (5x2 /y 2 − 6/y) dy = 0 12) 2xy dx + (3x2 + 2/y) dy = 0 3) (4y 3 /x2 + 20x) dx + 12y 2 /x dy = 0 13) (6xy + 14y 3 ) dx + (2x2 + 21xy 2 ) dy = 0 4) 9x2 y dx + (6x3 + 16y 2 ) dy = 0 14) (8x3 y + 3x2 y 3 ) dx + (6x4 + 5x3 y 2 ) dy = 0 5) (16y 3 /x − 6/x) dx + 24y 2 dy = 0 15) (3y 4 + 15x2 y 2 ) dx + (4xy 3 + 6x3 y) dy = 0 6) 7y 3 dx + (28xy 2 − 10) dy = 0 16) (6xy 2 − 4x3 y) dx + (9x2 y − 2x4 ) dy = 0 7) (45x3 y 2 + 16x2 ) dx + 18x4 y dy = 0 17) (28x2 y + 24xy 2 ) dx + (7x3 + 16x2 y) dy = 0 8) 18x2 y dx + (18x3 − 12y 3 ) dy = 0 18) (18x5 y − 10xy 2 ) dx + (9x6 − 20x2 y) dy = 0 9) (4y 7 − 30x4 ) dx + 14xy 6 dy = 0 19) (27y 2 + 20x2 y 3 ) dx + (18xy + 12x3 y 2 ) dy = 0 10) 10xy 3 dx + (20x2 y 2 + 16y 2 ) dy = 0 3. 20) (8xy 4 − 5y 2 ) dx + (20x2 y 3 − 15xy) dy = 0 Modeling with first-order DE INL 3.1 .............................................. ZC 3.1–3.3 Modeling with first-order linear differential equations 1) A rapidly falled ill person with a high fever requires immediate diagnosis. One has access to only a relatively slow thermometer and therefore need to be able to calculate the fever long before the thermometer reads the temperature. It is assumed that the rate at which the temperature read by the thermometer changes per unit time, at any time is proportional to the difference between temperature read by the thermometer and the temperature of the medium which surrounds the measuring point. A reasonable assumption is also that the smallness of the thermometer and the corresponding negligible heat content not appreciably disturb the temperature in the body of the ill person. The thermometer is normally kept at room temperature (20 degrees). Three (3) seconds after the thermometer has been inserted into the mouth of the patient it reads 30 degrees. Another three seconds later, the thermocouple meter reads 35 degrees. Using the collected data, make a quick calculation of the person’s temperature. What is the conclusion, i.e. how high fever has the ill person? 2) A radioactive material decays at a rate proportional to the amount present at time t. Assume that the half-life is 50 years, and that at a given time there is 100 mg of the material. How long ago was it 10 g of the material? 3) In a simplified model (No. 1) by how much a student learns from a particular course material it is assumed that the learning per unit time at a given time is proportional (α) to the difference of how much it is possible to learn (the total amount of course material M ) and how much has been learned up to that time. In a somewhat more realistic model (No. 2), it is also taken into account that a student has time to forget parts of the previously learned, and then at a rate which is proportional (β) to the student’s knowledge at the time. Formulate and solve the differential equations for the two models. In so doing, assume that the student’s knowledge is zero at the time 0. Compare the solutions for very large t. Furthermore, describe the meaning of the special cases α β (α much greater than β) and α β. 4) Temperature changes in a thermometer are assumed to be described by Newton’s cooling/warming law. A feverish person has with the help of such a thermometer measured its own body temperature to be 38o C. What does the thermometer read three minutes after the fever measurement if it shows 22o C two minutes after the fever measurement and if the room temperature is 20o C? It is considered reasonable to suppose that the thermometer’s small extent and the corresponding negligible heat content not substantially disturb the temperature of the surrounding medium (the person’s body and 6 the air respectively). 9) In an isolated population where nobody dies, individuals fall ill with a mild disease at a rate which in each moment is proportional to the number of healthy individuals. Let the constant of proportionality be equal to 10−8 s−1 , the number of individuals in the entire population equal to 50 000, and the number of diseased at the time t0 equal to 30. Find the rate at which people become ill as a function of time t. Also, determine the number of diseased individuals for large times. 5) In an isolated population where nobody dies, individuals fall ill with a mild disease at a rate which in each moment is proportional to the number of healthy individuals. Let the constant of proportionality be equal to 10−7 s−1 , the number of individuals in the entire population equal to 100 000, and the number of diseased at the time t0 equal to 20. Find the rate at which people become ill as a function of time t. Also, determine the number of diseased individuals for large times. 10) In a simplified model (No. 1) by how much satiation a rabbit gets by a certain amount of 6) A rapidly falled ill person with a high fever food, it is assumed that the change of the satiety requires immediate diagnosis. One has acper unit time at a given time is proportional cess to only a relatively slow thermometer (α) to the difference of how great satiation and therefore need to be able to calculate that is possible to get (maximum value S) the fever long before the thermometer reads and the amount of satiation that has been the temperature. It is assumed that the achieved up to that time. In a somewhat more rate at which the temperature read by the realistic model (No. 2), it is also taken into thermometer changes per unit time, at any account that the rabbit’s hunger all the time time is proportional to the difference between ”depletes” the current satiety, and then at a temperature read by the thermometer and the rate that is proportional (β) to the rabbit’s temperature of the medium which surrounds satiety at the time. Formulate and solve the the measuring point. A reasonable assumption differential equations for the two models. In so is also that the smallness of the thermometer doing, assume the rabbit’s satiety is zero at the and the corresponding negligible heat content time 0. Compare the solutions for very large not appreciably disturb the temperature in the t. Furthermore, describe the meaning of the body of the ill person. special cases α β (α much greater than β) and α β. The thermometer is normally kept in a heated state of 30 degrees. Five (5) seconds after the thermometer has been inserted into the mouth 11) The water in a small lake is constant in volume (Vlake ), but is translated by two streams with a of the patient it reads 35 degrees. Another flow on each u0 /2 (volume per unit time) into five seconds later, the thermocouple meter reads the lake, and by a third stream with a flow u0 37,6 degrees. Using the collected data, make a bringing water from the lake. One of the two quick calculation of the person’s temperature. streams which represent the in-flow is from time What is the conclusion, i.e. how high fever has 0 polluted with the impurity content η (amount the ill person? of impurity per unit volume). Assume that the 7) A radioactive material decays at a rate proporimpurity diffusion speed in the lake is much tional to the amount present at time t. Assume greater than the average rate at which water that the half-life is 15 years, and that at a given flows in (through) the lake. How much pollution time there is 10 g of the material. How long will is there in the lake at the time t, and how large it be before there is only 0.7 g remaining of the is the amount pollution after a long period of material? time? 8) Temperature changes in a thermometer are 12) The number of bacteria in a bacterial culture is assumed to be described by Newton’s coolincreasing at a rate that is proportional to the ing/warming law. A feverish person has with number at the time t. At a certain moment, the help of such a thermometer measured its there are 5 000 bacteria while an hour before own body temperature to be 39o C. What does there was 2 000. How many will there be two the thermometer read four minutes after the hours after the first time? fever measurement if it shows 25o C one minute after the fever measurement and if the room 13) Temperature changes in a thermometer are temperature is 19o C? It is considered reasonable assumed to be described by Newton’s coolto suppose that the thermometer’s small extent ing/warming law. A feverish person has with and the corresponding negligible heat content the help of such a thermometer measured not substantially disturb the temperature of the its own body temperature to be 40o C. What surrounding medium (the person’s body and does the thermometer read five minutes after the air respectively). the fever measurement if it shows 25o C three 3.. MODELING WITH FIRST-ORDER DE 7 minutes after the fever measurement and if the person’s body and the air respectively). room temperature is 21o C? It is considered reasonable to suppose that the thermometer’s 14) A radioactive material decays at a rate proportional to the amount present at time t. Assume small extent and the corresponding negligible that the half-life is 30 years, and that at a given heat content not substantially disturb the time there is 0.3 mg of the material. How long temperature of the surrounding medium (the ago was it 5 mg of the material? INL 3.2 Modeling with first-order non-linear differential equations 1) At time 0, there are 50 grams of chemical A and 80 grams of chemical B. The chemicals are combined whereby they react together in the proportions 1 : 2 forming the chemical C, i.e. for each 3 grams of the final product C, it will be used 1 gram of substance A and 2 grams of substance B. The rate of the reaction is assumed to be proportional to the product of the remaining amounts of the chemicals A and B not converted to C. It is also observed that 60 grams of chemical C is formed in 1/30 minutes. How much is formed in 1/15 minutes? 2) At time 0, there are 100 grams of substance A. The substance disintegrates into two substances B and C, formed in a ratio of 5:3, i.e. for each decayed gram of original substance A it is created 5/8 gram of subject B and 3/8 gram of substance C. The rate of the disintegration is assumed to be proportional to the square of the remaining amount of substance A, and the time until the amount of original substance has been halved equals 10 minutes. How many grams are there of substance B at time t (in minutes counted) if the grams at time 0 equals zero. 3) At time 0, there are 75 grams of chemical A and 50 grams of chemical B. The chemicals are combined whereby they react together in the proportions 3 : 1 forming the chemical C, i.e. for each 4 grams of the final product C, it will be used 3 grams of substance A and 1 gram of substance B. The rate of the reaction is assumed to be proportional to the product of the remaining amounts of the chemicals A and B not converted to C. It is also observed that 25 grams of chemical C is formed in 1/10 minutes. How much is formed in 2/5 minutes? 4) At time 0, there are 50 grams of substance A. The substance disintegrates into two substances B and C, formed in a ratio of 2:5, i.e. for each decayed grams of original substance A it is created 2/7 gram of subject B and 5/7 gram of substance C. The rate of the disintegration is assumed to be proportional to the square of the remaining amount of substance A, and the time until the amount of original substance has been halved equals 20 minutes. How many grams are there of substance C at time t (in minutes counted) if the grams at time 0 equals zero. 5) At time 0, there are 60 grams of chemical A and 100 grams of chemical B. The chemicals are combined whereby they react together in the proportions 2 : 3 forming the chemical C, i.e. for each 5 grams of the final product C, it will be used 2 grams of substance A and 3 grams of substance B. The rate of the reaction is assumed to be proportional to the product of the remaining amounts of the chemicals A and B not converted to C. It is also observed that 30 grams of chemical C is formed in 1/20 minutes. How much is formed in 1/10 minutes? 6) At time 0, there are 90 gram of substance A. The substance disintegrates into two substances B and C, formed in a ratio of 4:3, i.e. for each decayed grams of original substance A it is created 4/7 grams of subject B and 3/7 grams of substance C. The rate of the disintegration is assumed to be proportional to the square of the remaining amount of substance A, and the time until the amount of original substance has been halved equals 15 minutes. How many grams are there of substance B at time t (in minutes counted) if the grams at time 0 equals zero. 7) At time 0, there are 60 grams of chemical A and 25 grams of chemical B. The chemicals are combined whereby they react together in the proportions 4 : 1 forming the chemical C, i.e. for each 5 grams of the final product C, it will be used 4 grams of substance A and 1 gram of substance B. The rate of the reaction is assumed to be proportional to the product of the remaining amounts of the chemicals A and B not converted to C. It is also observed that 20 grams of chemical C is formed in 1/15 minutes. How much is formed in 1/5 minutes? 8) At time 0, there are 60 gram of substance A. The substance disintegrates into two substances B and C, formed in a ratio of 2:3, i.e. for each decayed grams of original substance A it is created 2/5 grams of subject B and 3/5 grams of substance C. The rate of the disintegration is assumed to be proportional to the square of the remaining amount of substance A, and the time until the amount of original substance has been 8 halved equals 12 minutes. How many grams are there of substance C at time t (in minutes counted) if the grams at time 0 equals zero. it will be used 1 gram of substance A and 3 grams of substance B. The rate of the reaction is assumed to be proportional to the product of the remaining amounts of the chemicals A and B not converted to C. It is also observed that 40 grams of chemical C is formed in 1/50 minutes. How much is formed in 1/10 minutes? 9) At time 0, there are 70 grams of chemical A and 80 grams of chemical B. The chemicals are combined whereby they react together in the proportions 3 : 4 forming the chemical C, i.e. for each 7 grams of the final product C, 14) At time 0, there are 40 gram of substance A. The substance disintegrates into two substances it will be used 3 grams of substance A and 4 B and C, formed in a ratio of 7:2, i.e. for grams of substance B. The rate of the reaction each decayed grams of original substance A it is is assumed to be proportional to the product of created 7/9 grams of subject B and 2/9 grams the remaining amounts of the chemicals A and of substance C. The rate of the disintegration is B not converted to C. It is also observed that 10 assumed to be proportional to the square of the grams of chemical C is formed in 1/40 minutes. remaining amount of substance A, and the time How much is formed in 1/20 minutes? until the amount of original substance has been halved equals 16 minutes. How many grams 10) At time 0, there are 80 gram of substance A. are there of substance B at time t (in minutes The substance disintegrates into two substances counted) if the grams at time 0 equals zero. B and C, formed in a ratio of 2:1, i.e. for each decayed grams of original substance A it is created 2/3 grams of subject B and 1/3 grams 15) At time 0, there are 70 grams of chemical A and 40 grams of chemical B. The chemicals of substance C. The rate of the disintegration are combined whereby they react together in is assumed to be proportional to the square of the proportions 5 : 4 forming the chemical C, the remaining amount of substance A, and the i.e. for each 9 grams of the final product C, time until the amount of original substance has it will be used 5 grams of substance A and 4 been halved equals 8 minutes. How many grams grams of substance B. The rate of the reaction are there of substance B at time t (in minutes is assumed to be proportional to the product of counted) if the grams at time 0 equals zero. the remaining amounts of the chemicals A and B not converted to C. It is also observed that 50 11) At time 0, there are 90 grams of chemical A grams of chemical C is formed in 1/35 minutes. and 45 grams of chemical B. The chemicals How much is formed in 1/7 minutes? are combined whereby they react together in the proportions 5 : 2 forming the chemical C, 16) At time 0, there are 55 gram of substance A. i.e. for each 7 grams of the final product C, The substance disintegrates into two substances it will be used 5 grams of substance A and 2 B and C, formed in a ratio of 1:4, i.e. for grams of substance B. The rate of the reaction each decayed grams of original substance A it is is assumed to be proportional to the product of created 1/5 grams of subject B and 4/5 grams the remaining amounts of the chemicals A and of substance C. The rate of the disintegration B not converted to C. It is also observed that 55 is assumed to be proportional to the square of grams of chemical C is formed in 1/25 minutes. the remaining amount of substance A, and the How much is formed in 1/5 minutes? time until the amount of original substance has been halved equals 7 minutes. How many grams 12) At time 0, there are 75 gram of substance A. are there of substance C at time t (in minutes The substance disintegrates into two substances counted) if the grams at time 0 equals zero. B and C, formed in a ratio of 1:3, i.e. for each decayed grams of original substance A it is 17) At time 0, there are 65 grams of chemical A created 1/4 grams of subject B and 3/4 grams and 80 grams of chemical B. The chemicals of substance C. The rate of the disintegration is are combined whereby they react together in assumed to be proportional to the square of the the proportions 3 : 5 forming the chemical C, remaining amount of substance A, and the time i.e. for each 8 grams of the final product C, until the amount of original substance has been it will be used 3 grams of substance A and 5 halved equals 25 minutes. How many grams grams of substance B. The rate of the reaction are there of substance C at time t (in minutes is assumed to be proportional to the product of counted) if the grams at time 0 equals zero. the remaining amounts of the chemicals A and B not converted to C. It is also observed that 35 13) At time 0, there are 50 grams of chemical A grams of chemical C is formed in 1/20 minutes. and 70 grams of chemical B. The chemicals How much is formed in 1/4 minutes? are combined whereby they react together in the proportions 1 : 3 forming the chemical C, 18) At time 0, there are 60 gram of substance A. i.e. for each 4 grams of the final product C, The substance disintegrates into two substances 3.. MODELING WITH FIRST-ORDER DE 9 B and C, formed in a ratio of 8:5, i.e. for is assumed to be proportional to the product of each decayed grams of original substance A it is the remaining amounts of the chemicals A and created 8/13 grams of subject B and 5/13 grams B not converted to C. It is also observed that 60 of substance C. The rate of the disintegration is grams of chemical C is formed in 1/30 minutes. assumed to be proportional to the square of the How much is formed in 1/6 minutes? remaining amount of substance A, and the time until the amount of original substance has been 20) At time 0, there are 45 gram of substance A. The substance disintegrates into two substances halved equals 14 minutes. How many grams B and C, formed in a ratio of 4:7, i.e. for are there of substance B at time t (in minutes each decayed grams of original substance A it is counted) if the grams at time 0 equals zero. created 4/11 grams of subject B and 7/11 grams 19) At time 0, there are 75 grams of chemical A of substance C. The rate of the disintegration and 50 grams of chemical B. The chemicals is assumed to be proportional to the square of are combined whereby they react together in the remaining amount of substance A, and the the proportions 6 : 5 forming the chemical C, time until the amount of original substance has i.e. for each 11 grams of the final product C, been halved equals 5 minutes. How many grams it will be used 6 grams of substance A and 5 are there of substance C at time t (in minutes grams of substance B. The rate of the reaction counted) if the grams at time 0 equals zero. INL 3.3 Modeling with systems of first-order differential equations Write a system of differential equations which at the time t describes how the amount of salt in a liquid varies in three containers A, B and C respectively. Also, state the corresponding initial values. It is assumed that the rates by which salt and liquid become well-mixed within the different containers are much greater than the flow-rates by which liquids are streaming through the system of containers. 1) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows from A to B, from B to C and from C to A. It also flows liquid from the environment into A, and from B to the environment. At the time 0 there are 200 liters of liquid in each of the containers, of which that in B is evenly mixed with 2 grams of salt. The liquid flowing into A has the flow rate φin = 4 liter/min and salinity ρin = 1 gram/liter. The other flow rates are φA→B = 2 liter/min, φB→C = 1 liter/min, φC→A = 3 liter/min and φut = 2 liter/min. 2) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows from A to B, from B to C and from C to B. It also flows liquid from the environment into A, and from C to the environment. At the time 0 there are 200 liters of liquid in each of the containers, of which that in A is evenly mixed with 3 grams of salt. The liquid flowing into A has the flow rate φin = 3 liter/min and salinity ρin = 2 gram/liter. The other flow rates are φA→B = 2 liter/min, φB→C = 3 liter/min, φC→B = 1 liter/min and φut = 4 liter/min. 3) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows from A to B, from B to C and from B to A. It also flows liquid from the environment into B, and from C to the environment. At the time 0 there are 200 liters of liquid in each of the containers, of which that in B is evenly mixed with 3 grams of salt. The liquid flowing into B has the flow rate φin = 4 liter/min and salinity ρin = 1 gram/liter. The other flow rates are φA→B = 1 liter/min, φB→C = 2 liter/min, φB→A = 2 liter/min and φut = 3 liter/min. 4) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows from A to B, from B to C and from C to A. It also flows liquid from the environment into A, and from B to the environment. At the time 0 there are 200 liters of liquid in each of the containers, of which that in A is evenly mixed with 2 grams of salt. The liquid flowing into A has the flow rate φin = 2 liter/min and salinity ρin = 3 gram/liter. The other flow rates are φA→B = 3 liter/min, φB→C = 1 liter/min, φC→A = 2 liter/min and φut = 1 liter/min. 5) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows from A to B, from B to C and from B to A. It also flows liquid from the environment into A, and from C to the environment. At the time 0 there are 200 liters of liquid in each of the containers, of which that in B is evenly mixed with 3 grams of salt. The liquid flowing into A has the flow rate φin = 1 liter/min and salinity ρin = 2 gram/liter. The other flow rates are φA→B = 2 liter/min, φB→C = 3 liter/min, φB→A = 1 liter/min and φut = 2 liter/min. 6) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows from A to B, from B to C and from C to B. 10 7) 8) 9) 10) It also flows liquid from the environment into φC→A = 2 liter/min and φut = 2 liter/min. B, and from C to the environment. At the time 0 there are 200 liters of liquid in each 11) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows of the containers, of which that in A is evenly from B to A, from B to C and from A to C. mixed with 2 grams of salt. The liquid flowing It also flows liquid from the environment into into B has the flow rate φin = 1 liter/min and B, and from C to the environment. At the salinity ρin = 3 gram/liter. The other flow rates time 0 there are 200 liters of liquid in each are φA→B = 2 liter/min, φB→C = 3 liter/min, of the containers, of which that in B is evenly φC→B = 1 liter/min and φut = 3 liter/min. mixed with 2 grams of salt. The liquid flowing Three different tanks A, B and C are with tubes into B has the flow rate φin = 5 liter/min and interconnected so that liquid in the system flows salinity ρin = 2 gram/liter. The other flow rates from A to B, from B to C and from C to A. are φB→A = 3 liter/min, φB→C = 1 liter/min, It also flows liquid from the environment into φA→C = 2 liter/min and φut = 4 liter/min. A, and from B to the environment. At the time 0 there are 200 liters of liquid in each 12) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows of the containers, of which that in B is evenly from A to B, from B to C and from B to A. mixed with 2 grams of salt. The liquid flowing It also flows liquid from the environment into into A has the flow rate φin = 1 liter/min and A, and from C to the environment. At the salinity ρin = 3 gram/liter. The other flow rates time 0 there are 200 liters of liquid in each are φA→B = 2 liter/min, φB→C = 1 liter/min, of the containers, of which that in C is evenly φC→A = 2 liter/min and φut = 3 liter/min. mixed with 5 grams of salt. The liquid flowing Three different tanks A, B and C are with tubes into A has the flow rate φin = 1 liter/min and interconnected so that liquid in the system flows salinity ρin = 2 gram/liter. The other flow rates from A to B, from B to A, from B to C and from are φA→B = 2 liter/min, φB→C = 1 liter/min, C to B. It also flows liquid from the environment φB→A = 3 liter/min and φut = 2 liter/min. into C, and from A to the environment. At the time 0 there are 200 liters of liquid in each 13) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows of the containers, of which that in C is evenly from C to B, from B to A and from A to C. mixed with 3 grams of salt. The liquid flowing It also flows liquid from the environment into into C has the flow rate φin = 3 liter/min and C, and from A to the environment. At the salinity ρin = 2 gram/liter. The other flow rates time 0 there are 200 liters of liquid in each are φA→B = 2 liter/min, φB→A = 1 liter/min, of the containers, of which that in A is evenly φB→C = 3 liter/min, φC→B = 1 liter/min and mixed with 3 grams of salt. The liquid flowing φut = 1 liter/min. into C has the flow rate φin = 5 liter/min and Three different tanks A, B and C are with tubes salinity ρin = 1 gram/liter. The other flow rates interconnected so that liquid in the system flows are φC→B = 2 liter/min, φB→A = 3 liter/min, from A to B, from B to C and from C to B. φA→C = 1 liter/min and φut = 1 liter/min. It also flows liquid from the environment into A, and from C to the environment. At the 14) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows time 0 there are 200 liters of liquid in each from A to B, from B to C, from B to A and from of the containers, of which that in B is evenly C to A. It also flows liquid from the environment mixed with 2 grams of salt. The liquid flowing into C, and from A to the environment. At into A has the flow rate φin = 3 liter/min and the time 0 there are 200 liters of liquid in each salinity ρin = 1 gram/liter. The other flow rates of the containers, of which that in C is evenly are φA→B = 2 liter/min, φB→C = 3 liter/min, mixed with 2 grams of salt. The liquid flowing φC→B = 1 liter/min and φut = 4 liter/min. into C has the flow rate φin = 2 liter/min and Three different tanks A, B and C are with tubes salinity ρin = 3 gram/liter. The other flow rates interconnected so that liquid in the system flows are φA→B = 1 liter/min, φB→C = 2 liter/min, from A to B, from B to C and from C to A. φB→A = 3 liter/min, φC→A = 1 liter/min and It also flows liquid from the environment into φut = 3 liter/min. B, and from C to the environment. At the time 0 there are 200 liters of liquid in each 15) Three different tanks A, B and C are with tubes of the containers, of which that in A is evenly interconnected so that liquid in the system flows mixed with 3 grams of salt. The liquid flowing from A to B, from B to C and from A to C. into B has the flow rate φin = 2 liter/min and It also flows liquid from the environment into salinity ρin = 3 gram/liter. The other flow rates B, and from C to the environment. At the are φA→B = 1 liter/min, φB→C = 3 liter/min, time 0 there are 200 liters of liquid in each 3.. MODELING WITH FIRST-ORDER DE of the containers, of which that in A is evenly mixed with 3 grams of salt. The liquid flowing into B has the flow rate φin = 2 liter/min and salinity ρin = 4 gram/liter. The other flow rates are φA→B = 1 liter/min, φB→C = 2 liter/min, φA→C = 2 liter/min and φut = 5 liter/min. 11 C to B. It also flows liquid from the environment into A, and from C to the environment. At the time 0 there are 200 liters of liquid in each of the containers, of which that in C is evenly mixed with 4 grams of salt. The liquid flowing into A has the flow rate φin = 2 liter/min and salinity ρin = 3 gram/liter. The other flow rates are φA→B = 4 liter/min, φB→A = 2 liter/min, φB→C = 3 liter/min, φC→B = 2 liter/min and φut = 1 liter/min. 16) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows from A to C, from C to B, from B to C and from B to A. It also flows liquid from the environment into A, and from C to the environment. At 19) Three different tanks A, B and C are with tubes the time 0 there are 200 liters of liquid in each interconnected so that liquid in the system flows of the containers, of which that in C is evenly from A to C, from C to B and from B to A. mixed with 3 grams of salt. The liquid flowing It also flows liquid from the environment into into A has the flow rate φin = 5 liter/min and C, and from B to the environment. At the salinity ρin = 4 gram/liter. The other flow rates time 0 there are 200 liters of liquid in each are φA→C = 3 liter/min, φC→B = 2 liter/min, of the containers, of which that in B is evenly φB→C = 2 liter/min, φB→A = 1 liter/min and mixed with 7 grams of salt. The liquid flowing φut = 1 liter/min. into C has the flow rate φin = 3 liter/min and salinity ρin = 2 gram/liter. The other flow rates 17) Three different tanks A, B and C are with tubes are φ = 5 liter/min, φC→B = 2 liter/min, A→C interconnected so that liquid in the system flows φ = 3 liter/min and φut = 7 liter/min. B→A from A to B, from B to C and from C to B. It also flows liquid from the environment into A, and from B to the environment. At the 20) Three different tanks A, B and C are with tubes interconnected so that liquid in the system flows time 0 there are 200 liters of liquid in each from A to B, from B to C and from C to A. of the containers, of which that in A is evenly It also flows liquid from the environment into mixed with 4 grams of salt. The liquid flowing A, and from C to the environment. At the into A has the flow rate φin = 2 liter/min and time 0 there are 200 liters of liquid in each salinity ρin = 3 gram/liter. The other flow rates of the containers, of which that in B is evenly are φA→B = 4 liter/min, φB→C = 3 liter/min, mixed with 4 grams of salt. The liquid flowing φC→B = 1 liter/min and φut = 1 liter/min. into A has the flow rate φin = 4 liter/min and 18) Three different tanks A, B and C are with tubes salinity ρin = 3 gram/liter. The other flow rates interconnected so that liquid in the system flows are φA→B = 1 liter/min, φB→C = 3 liter/min, from A to B, from B to A, from B to C and from φC→A = 2 liter/min and φut = 2 liter/min. Block 2 – ODE of higher orders, Difference equations 4. Linear ODE of higher orders 1) 2) 3) 4) 5) 6) 7) ZC 4.1–4.4, 4.7, 4.10 Determine whether the functions f1 , f2 , f3 are linear independent on R (on R+ in versions 10 and 13) or not. 3 f1 (x) = 3 + x f1 (x) = 7x − x f1 (x) = |x| + 5 2 f2 (x) = 2x + 5x3 f2 (x) = x + 5 f2 (x) = x − 2x 15) 8) f3 (x) = 9x + 4x3 f3 (x) = |x| − 5 f3 (x) = 3x2 − 1 f1 (x) = e2x f1 (x) = x −x x −2x f1 (x) = 2e + 3e f2 (x) = 2x + 3x2 f2 (x) = e 9) f2 (x) = e−x − 2ex 16) f3 (x) = 4x − x3 f3 (x) = cosh(2x) f3 (x) = e−x + ex f1 (x) = x f1 (x) = ln(x) 2 f2 (x) = ln(x2 ) f2 (x) = 3x − x 10) f1 (x) = x + |x| f3 (x) = x ln(x) f3 (x) = 2x + x2 f2 (x) = 2|x| − 3x 17) 2 f3 (x) = x − |x| f1 (x) = |x| + 5 f1 (x) = x − x 2 f2 (x) = 2 + 5x f2 (x) = x + 5 11) 2 f3 (x) = x − 3 f3 (x) = x f1 (x) = cos(x) − 2 sin(x) f2 (x) = sin(x) 18) 2x f1 (x) = 1 + sin2 (x) f1 (x) = e − 2 f3 (x) = sin(2x) f2 (x) = e2x cosh(2x) f2 (x) = cos2 (x) − 3 12) f3 (x) = 3 + e2x f3 (x) = 5 2 4 f1 (x) = 2x + 3x 2 f1 (x) = x f1 (x) = x ln(x) f2 (x) = 7x2 − 8x4 19) 2 3 f2 (x) = x + ln(x) f2 (x) = 4x + x 13) f3 (x) = x2 f3 (x) = x f3 (x) = 2x3 − 6x2 3x x 2 f1 (x) = sinh(3x) f1 (x) = e + e f1 (x) = |x| − 2x f2 (x) = |x| f2 (x) = e2x f2 (x) = e3x 20) 14) 2 2x f3 (x) = ex − 3e2x f3 (x) = |x| + 3x f3 (x) = e INL 4.1 .................................. 12 4.. LINEAR ODE OF HIGHER ORDERS INL 4.2 13 Find the general solution of the differential equation. Hint: Begin by testing whether the DE has a solution on any of the forms eβx , x + β, xβ or (ln(x))β . 1) x(x + 1)y 00 + (2 − x2 )y 0 − (x + 2)y = 0 11) (2x + 1)y 00 + 4xy 0 − 4y = 0 2) x2 (1 − x)y 00 + x(x − 4)y 0 + 6y = 0 12) (x + 1)y 00 + xy 0 − y = 0 3) x2 (x + 1)y 00 + (2x + 1)(y − xy 0 ) = 0 13) x2 y 00 − (2x + 3)xy 0 + (2x + 3)y = 0 4) xy 00 − (2x + 1)y 0 − 3(x − 1)y = 0 14) xy 00 − 2(x + 1)y 0 + (x + 2)y = 0 5) x(x + 1)y 00 + (x + 2)y 0 − y = 0 15) x2 y 00 − x(x + 2)y 0 + (x + 2)y = 0 6) x2 y 00 − (4x + 3)(xy 0 − y) = 0 7) xy 00 − (2x + 1)y 0 + 2y = 0 00 0 16) xy 00 − (3x + 1)y 0 + 3y = 0 17) x2 y 00 − 2xy 0 + (x2 + 2)y = 0 Hint: The DE has a solution on the form x sin(x) 8) xy − (x + 1)y + y = 0 18) x2 y 00 + (2x2 − x)y 0 − 2xy = 0 9) x3 y 00 + xy 0 − y = 0 19) xy 00 + (x − 1)y 0 + 3(1 − 4x)y = 0 10) (x ln x)2 y 00 + x(ln x)2 y 0 − 6y = 0 INL 4.3 20) xy 00 − (x + 2)y 0 + 2y = 0 Solve the initial-value problem. 1) y 00 + y 0 − 6y = 0, y(0) = 3, y 0 (0) = −4 11) y 00 − 2y 0 − 8y = 0, y(0) = 1, y 0 (0) = −14 2) y 00 − 4y 0 + 4y = 0, y(0) = 3, y 0 (0) = 2 12) y 00 + 6y 0 + 9y = 0, y(0) = 2, y 0 (0) = −3 3) y 00 − 2y 0 − 3y = 0, y(0) = −1, y 0 (0) = 2 13) y 00 − 4y 0 − 12y = 0, y(0) = 2, y 0 (0) = 28 4) y 00 + 4y 0 + 4y = 0, y(0) = 2, y 0 (0) = 4 14) y 00 − 12y 0 + 36y = 0, y(0) = −3, y 0 (0) = −11 5) y 00 + 3y 0 − 10y = 0, y(0) = 8, y 0 (0) = −5 15) y 00 − 5y 0 + 6y = 0, y(0) = 3, y 0 (0) = 1 6) y 00 − 10y 0 + 25y = 0, y(0) = −2, y 0 (0) = −7 16) y 00 + 10y 0 + 25y = 0, y(0) = −4, y 0 (0) = 19 7) y 00 − y 0 − 12y = 0, y(0) = 1, y 0 (0) = 11 17) y 00 − 3y 0 − 28y = 0, y(0) = 7, y 0 (0) = −6 8) y 00 + 8y 0 + 16y = 0, y(0) = 3, y 0 (0) = 10 18) y 00 − 8y 0 + 16y = 0, y(0) = −1, y 0 (0) = 0 9) y 00 + y 0 − 2y = 0, y(0) = −2, y 0 (0) = −5 19) y 00 + 2y 0 − 15y = 0, y(0) = 10, y 0 (0) = −2 10) y 00 − 6y 0 + 9y = 0, y(0) = 2, y 0 (0) = 1 INL 4.4 20) y 00 + 12y 0 + 36y = 0, y(0) = 2, y 0 (0) = −10 Find the general solution of the differential equation. 1) y 00 − 10y 0 + 25y = xe−x + 3e5x 9) y 00 − 4y 0 + 4y = 3e2x − 3xe−x 2) y 00 − y 0 − 12y = e2x − 4xe4x 10) y 00 − 2y 0 − 3y = 2xe−x + 5e2x 3) y 00 + 8y 0 + 16y = xe3x − 2e−4x 11) y 00 + 4y 0 + 4y = xe3x − 4e−2x 4) y 00 + y 0 − 2y = 3xex + 5e−3x 12) y 00 + 3y 0 − 10y = 4xe−5x + 2e−x 5) y 00 − 6y 0 + 9y = 6e3x + xex 13) y 00 + 12y 0 + 36y = 2e−6x − 3xex 6) y 00 − 2y 0 − 8y = 2xe−2x − 5e3x 14) y 00 + 2y 0 − 15y = 2xe−5x + 6e−3x 7) y 00 + 6y 0 + 9y = 2xe2x + 5e−3x 15) y 00 − 8y 0 + 16y = −3e4x + 4xe2x 8) y 00 + y 0 − 6y = 7ex + xe2x 16) y 00 − 3y 0 − 28y = 5e4x − 2xe7x 14 17) y 00 + 10y 0 + 25y = e−5x + 2xe−3x 19) y 00 − 12y 0 + 36y = −4e6x + xe3x 18) y 00 − 5y 0 + 6y = −3xe2x − e−3x 20) y 00 − 4y 0 − 12y = 3e2x − 2xe6x INL 4.5 Find the general solution of the differential equation (either for x > 0 or x < 0). 1) x2 y 00 + 9xy 0 + 16y = 0 8) x2 y 00 − xy 0 − 3y = 0 15) x2 y 00 − 11xy 0 + 36y = 0 2) x2 y 00 + 2xy 0 − 2y = 0 9) x2 y 00 + 5xy 0 + 4y = 0 16) x2 y 00 − 3xy 0 − 12y = 0 3) x2 y 00 − 5xy 0 + 9y = 0 10) x2 y 00 + 4xy 0 − 10y = 0 2 00 0 2 00 17) x2 y 00 + 13xy 0 + 36y = 0 0 4) x y − xy − 8y = 0 11) x y − 9xy + 25y = 0 5) x2 y 00 + 7xy 0 + 9y = 0 12) x2 y 00 − 2xy 0 − 18y = 0 6) x2 y 00 + 2xy 0 − 6y = 0 13) x2 y 00 + 11xy 0 + 25y = 0 19) x2 y 00 − 7xy 0 + 16y = 0 7) x2 y 00 − 3xy 0 + 4y = 0 14) x2 y 00 − 4xy 0 + 6y = 0 20) x2 y 00 − 2xy 0 − 28y = 0 4. Non-linear ODE of higher orders INL 4.6 18) x2 y 00 + 3xy 0 − 15y = 0 ........................................... ZC 4.10 Find all solutions of the differential equation. 1) (y − 2)y 00 = (y 0 )2 6) (y + 1)y 00 = 2(y 0 )2 11) y 2 y 00 = y 0 2) yy 00 = 3(y 0 )2 7) yy 00 = (y − 1)(y 0 )2 12) (2y − 1)(y 0 )2 = y 2 y 00 17) y 00 = 12y 2 (y 0 )3 3) yy 00 + (y 0 )2 = yy 0 8) yy 00 + (y 0 )2 = 0 13) y 0 + (y 0 )2 = yy 00 18) 2yy 00 = 1 + (y 0 )2 4) y 00 = (y 0 )2 9) y 2 y 00 = (y 0 )3 14) yy 00 + (y 0 )3 = 0 19) yy 00 +2(y 0 )3 = 2(y 0 )2 5) y 00 = 2y(y 0 )3 10) (y 2 + 1)y 00 = 2y(y 0 )2 15) (y 2 + 1)y 0 = y 2 y 00 16) y(y−1)y 00 +(y 0 )2 = 0 20) y 00 = y 0 + (y 0 )3 7.. LAPLACE TRANSFORM 7. Laplace transform 2) 3) 4) 5) 6) 7) 8) 9) 10) ....................................................................... ZC 7.1–7.6 Solve the differential equation problem for t ≥ 0 and with y(0) = 2. 0, 0 ≤ t < 4 0, 0 ≤ t < 2 0 0 6 − 2t , 4 ≤ t < 7 5, 2 ≤ t < 5 11) y (t) + 7y(t) = y (t) + 7y(t) = 5, t ≥ 7 2t + 1 , t ≥ 5 2t + 9 , 0 ≤ t < 2 5 − 3t , 0 ≤ t < 5 0 0 7, 2 ≤ t < 4 0, 5 ≤ t < 7 12) y (t) + 9y(t) = y (t) + 4y(t) = 0, t ≥ 4 9, t ≥ 7 7 − t, 0 ≤ t < 6 6, 0 ≤ t < 3 0, 6 ≤ t < 8 7t − 4 , 3 ≤ t < 6 . 13) y 0 (t) + 5y(t) = y 0 (t) + 8y(t) = 3, t ≥ 8 0, t ≥ 6 6, 0 ≤ t < 4 0, 0 ≤ t < 6 4t + 3 , 4 ≤ t < 9 3t + 2 , 6 ≤ t < 9 14) y 0 (t) + 2y(t) = y 0 (t) + 3y(t) = 0, t ≥ 9 4, t ≥ 9 0, 0 ≤ t < 3 8 − 5t , 0 ≤ t < 4 0 0 8 − 6t , 3 ≤ t < 5 7 , 4 ≤ t < 8 15) y (t) + 3y(t) = y (t) + 2y(t) = 7, t ≥ 5 0, t ≥ 8 9t − 2 , 0 ≤ t < 2 9, 0 ≤ t < 3 0 0 4, 2 ≤ t < 8 0, 3 ≤ t < 7 16) y (t) + 8y(t) = y (t) + 5y(t) = 0, t ≥ 8 6t − 1 , t ≥ 7 0, 0 ≤ t < 5 4t + 7 , 0 ≤ t < 2 0 0 2 − 3t , 5 ≤ t < 9 5, 2 ≤ t < 9 17) y (t) + 5y(t) = y (t) + 9y(t) = 8, t ≥ 9 0, t ≥ 9 3, 0 ≤ t < 2 2, 0 ≤ t < 5 0 0 0, 2 ≤ t < 7 0, 5 ≤ t < 8 18) y (t) + 7y(t) = y (t) + 6y(t) = 2t + 6 , t ≥ 7 9t − 7 , t ≥ 8 4 − 5t , 0 ≤ t < 3 0, 0 ≤ t < 2 0 0 2, 3 ≤ t < 8 1 − 4t , 2 ≤ t < 6 19) y (t) + 3y(t) = y (t) + 8y(t) = 0, t ≥ 8 3, t ≥ 6 5, 0 ≤ t < 4 8, 0 ≤ t < 3 0 0 0, 4 ≤ t < 6 0, 3 ≤ t < 9 20) y (t) + 6y(t) = y (t) + 4y(t) = 7t − 5 , t ≥ 6 3t + 8 , t ≥ 9 INL 7.1 1) 15 16 Solve the integral equation for t ≥ 0. ( Z t 0, 0 ≤ t < 2 y(t − w) cos(w) dw = 1) y(t) + 2 e2−t , t ≥ 2 0 INL 7.2 t Z e−2τ cos(2τ ) y(t − τ ) dτ 2) y(t) = cos(3t) + 4 0 0, 0 ≤ t < 3 1, 3 ≤ t < 5 y(ξ) dξ = t + 0, t ≥ 5 t Z 3) y(t) + 2 0 4) 2e2t = 2y(t) + 5 t Z y(ω) sin 2(t − ω) dω 0 ( t Z y(ω) 5(t − ω) + 23 (t − ω)3 dω = 5) y(t) + 0 6) y(t) + e−t ( t Z eξ y(ξ) dξ = 0 0, 0 ≤ t < 5 e10−2t , t ≥ 5 ( t Z 0, 0 ≤ t < 2 4t − 8 , t ≥ 2 3 (t − ω) y(ω) dω + 7) 2y(t) = 27 0 Z ( t 8) y(t) + 2 e −3ξ y(t − ξ) dξ = 0 0, 0 ≤ t < 6 e2t−12 , t ≥ 6 t Z 0 0, 0 ≤ t < 3 t − 3, t ≥ 3 y(t − τ ) cos(3τ ) dτ 9) y (t) = 4 + 5 with y(0) = 0 0 10) e−t = y(t) + 2 t Z y(ξ) cos(t − ξ) dξ 0 t Z 11) Z y(λ) dλ + 3 0 ( t 0, 0 ≤ t < 5 y(t − ν) e−4ν dν = (60 − 12t)e20−4t , t ≥ 5 0 t Z (τ − t) y(τ ) dτ = y(t) 12) 3 sin(2t) + 0 Z 13) 3 ( t y(ξ) dξ + y(t) = 0 Z 0, 0 ≤ t < π 13 sin(2t) , t ≥ π t y(t − λ) cos(5λ) dλ 14) 1 = y(t) + 8 0 ( t Z 2 y(τ ) (t − τ ) 25 + 24(t − τ ) 15) 2y(t) + 2 0 16) y 00 (t) + 5 Z 0 ( t y 0 (t − ξ) cos(2ξ) dξ = dτ = 0, 0 ≤ t < 5 7(t − 5)2 , t ≥ 5 0, 0 ≤ t < 4 9 sin(2(t − 4)) , t ≥ 4 ( with y(0) = 0 y 0 (0) = 9 Z.. LINEAR DIFFERENCE EQUATIONS INL 7.3 17 Solve the differential equation problem for t ≥ 0. 1) y 00 + 8y 0 + 16y = 2δ(t − 3) − 5U (t − 8), y(0) = 3, y 0 (0) = 10 11) y 00 − 8y 0 + 16y = U (t − 3) − 5δ(t − 7), y(0) = −1, y 0 (0) = 0 2) y 00 + y 0 − 2y = 4U (t − 3) + δ(t − 5), y(0) = −2, y 0 (0) = −5 12) y 00 + 2y 0 − 15y = δ(t − 2) + U (t − 4), y(0) = 10, y 0 (0) = −2 3) y 00 − 6y 0 + 9y = U (t − 2) − 2δ(t − 7), y(0) = 2, y 0 (0) = 1 13) y 00 + 12y 0 + 36y = 3δ(t − 2) + 2U (t − 3), y(0) = 2, y 0 (0) = −10 4) y 00 − 2y 0 − 8y = δ(t − 6) + 5U (t − 9), y(0) = 1, y 0 (0) = −14 14) y 00 + y 0 − 6y = U (t − 5) − 3δ(t − 9), y(0) = 3, y 0 (0) = −4 5) y 00 + 6y 0 + 9y = 5U (t − 6) + δ(t − 4), y(0) = 2, y 0 (0) = −3 15) y 00 − 4y 0 + 4y = 4δ(t − 3) + 7U (t − 4), y(0) = 3, y 0 (0) = 2 6) y 00 − 4y 0 − 12y = 5δ(t − 3) − 2U (t − 4), y(0) = 2, y 0 (0) = 28 16) y 00 − 2y 0 − 3y = 6U (t − 2) + 2δ(t − 5), y(0) = −1, y 0 (0) = 2 7) y 00 − 12y 0 + 36y = δ(t − 2) + 3U (t − 9), y(0) = −3, y 0 (0) = −11 17) y 00 + 4y 0 + 4y = δ(t − 3) − U (t − 4), y(0) = 2, y 0 (0) = 4 8) y 00 − 5y 0 + 6y = U (t − 4) + 13δ(t − 5), y(0) = 3, y 0 (0) = 1 18) y 00 + 3y 0 − 10y = U (t − 3) + δ(t − 4), y(0) = 8, y 0 (0) = −5 9) y 00 + 10y 0 + 25y = δ(t − 3) − 5U (t − 6), y(0) = −4, y 0 (0) = 19 19) y 00 − 10y 0 + 25y = δ(t − 2) + U (t − 4), y(0) = −2, y 0 (0) = −7 10) y 00 − 3y 0 − 28y = 4δ(t − 2) + 9U (t − 8), y(0) = 7, y 0 (0) = −6 20) y 00 − y 0 − 12y = δ(t − 2) − 5U (t − 5), y(0) = 1, y 0 (0) = 11 Z. Linear difference equations INL z.1 .......................................................... Notes ∞ Find the number sequence {yn }n=0 which satisfies the recurrence relation and the initial conditions. 1) 6yn + yn−1 − 2yn−2 = 0 , y0 = 1 , y1 = 4 . 11) 12yn − 5yn−1 − 3yn−2 = 0 , y0 = 10 , y1 = 1 . 2) 4yn + 4yn−1 + yn−2 = 0 , y0 = 3 , y1 = −2 . 12) 25yn + 20yn−1 + 4yn−2 = 0 , y0 = −3 , y1 = −2 . 3) 12yn + yn−1 − yn−2 = 0 , y0 = 11 , y1 = 1 . 13) 6yn + yn−1 − yn−2 = 0 , y0 = 7 , y1 = 4 . 4) 10yn + 3yn−1 − yn−2 = 0 , y0 = 21 , y1 = 0 . 14) 12yn − yn−1 − yn−2 = 0 , y0 = 10 , y1 = 1 . 5) 9yn − 12yn−1 + 4yn−2 = 0 , y0 = −1 , y1 = 2 . 15) 25yn − 30yn−1 + 9yn−2 = 0 , y0 = 2 , y1 = −3 . 6) 20yn + yn−1 − yn−2 = 0 , y0 = 14 , y1 = 1 . 16) 8yn − 2yn−1 − 3yn−2 = 0 , y0 = 12 , y1 = 0 . 7) 9yn + 6yn−1 + yn−2 = 0 , y0 = −2 , y1 = 2 . 17) 16yn + 24yn−1 + 9yn−2 = 0 , y0 = −1 , y1 = 3 . 8) 16yn − 8yn−1 − 3yn−2 = 0 , y0 = 16 , y1 = −2 . 18) 6yn − yn−1 − yn−2 = 0 , y0 = −1 , y1 = 2 . 9) 10yn − yn−1 − 2yn−2 = 0 , y0 = 1 , y1 = 5 . 19) 9yn − 3yn−1 − 2yn−2 = 0 , y0 = 3 , y1 = −4 . 10) 16yn − 8yn−1 + yn−2 = 0 , y0 = 4 , y1 = 0 . 20) 25yn − 10yn−1 + yn−2 = 0 , y0 = 2 , y1 = 1 . 18 ∞ Find the number sequence {yn }n=0 which satisfies the difference equation problem. 0 , n 6= 5 , 0 , n 6= 6 , yn−1 − 3yn = n ≥ 1, 11) yn−1 − 2yn = n ≥ 1, 3, n = 5, 3, n = 6, y0 = 2 . y0 = −1 . 0 , n 6= 3 , 0 , n 6= 4 , yn−1 + 5yn = n ≥ 1, 12) yn−1 + 7yn = n ≥ 1, −5 , n = 3 , −5 , n = 4 , y0 = 4 . y0 = 4 . 0 , n 6= 4 , 0 , n 6= 7 , yn−1 − 4yn = n ≥ 1, 13) yn−1 + 8yn = n ≥ 1, 2, n = 4, 4, n = 7, y0 = 5 . y0 = −3 . 0 , n 6= 7 , 0 , n 6= 9 , yn−1 + 6yn = n ≥ 1, 14) yn−1 + 5yn = n ≥ 1, 4, n = 7, 10 , n = 9 , y0 = −2 . y0 = 4 . 0 , n 6= 6 , 0 , n 6= 8 , yn−1 − 7yn = n ≥ 1, 15) yn−1 + 9yn = n ≥ 1, −3 , n = 6 , −3 , n = 8 , y0 = 3 . y0 = 5 . 0 , n 6= 9 , 0 , n 6= 5 , yn−1 + 4yn = n ≥ 1, 16) yn−1 − 4yn = n ≥ 1, −3 , n = 9 , −6 , n = 5 , y0 = −5 . y0 = −3 . 0 , n 6= 6 , 0 , n 6= 4 , yn−1 + 2yn = n ≥ 1, 17) yn−1 + 8yn = n ≥ 1, −5 , n = 6 , 2, n = 4, y0 = −3 . y0 = −1 . 0 , n 6= 5 , 0 , n 6= 9 , yn−1 − 5yn = 18) yn−1 − 7yn = n ≥ 1, n ≥ 1, −4 , n = 9 , 2, n = 5, y0 = 3 . y0 = 4 . 0 , n 6= 8 , 0 , n 6= 7 , 19) yn−1 − 9yn = n ≥ 1, yn−1 + 3yn = n ≥ 1, 5, n = 7, 4, n = 8, y0 = 2 . y0 = 2 . 0 , n 6= 3 , 0 , n 6= 8 , 20) yn−1 − 6yn = n ≥ 1, yn−1 − 6yn = n ≥ 1, −3 , n = 8 , −3 , n = 3 , y0 = −5 . y0 = −5 . INL z.2 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) INL z.3 Solve the difference equation with the initial value y0 = 0 . 1) 4yn−1 + yn = (−4)n 8) 3yn−1 − yn = 3n 15) 8yn−1 + yn = (−8)n 2) 2yn−1 + yn = (−2)n 9) 5yn−1 + yn = (−5)n 16) 5yn−1 + yn = −(−5)n 3) 5yn−1 − yn = 5n 10) 4yn−1 − yn = 4n 4) 3yn−1 + yn = (−3)n 11) 6yn−1 + yn = (−6)n 5) 6yn−1 − yn = 6n 12) 7yn−1 − yn = 7n 6) 2yn−1 − yn = 2n 13) 9yn−1 − yn = 9n 19) 8yn−1 − yn = 8n 7) 7yn−1 + yn = (−7)n 14) 4yn−1 − yn = −4n 20) 3yn−1 + yn = −(−3)n 17) 9yn−1 + yn = (−9)n 18) 6yn−1 − yn = −6n Block 3 – Systems of ODE 8. System of first order linear DE ........................................... ZC 8.1–8.3 Find the general solution of the linear system and sketch a representative part of the phase portrait. dx/dt −2x + y dx/dt x + 4y dx/dt 2x − 2y 8) = 1) = 15) = dy/dt x − 2y dy/dt 2x − y dy/dt −2x − y ! ! − 4 x + 53 y dx/dt 4x − 2y dx/dt − 38 x − 13 y 2) = 9) = 103 dx/dt 1 dy/dt 3x − y dy/dt 16) = 3 x + 3y dy/dt − 23 x − 73 y ! 1 dx/dt −x dx/dt 2 x + 9y 3) = 10) = dy/dt x − 2y 1 dx/dt −3x + 2y dy/dt x + 2y 17) = 2 dy/dt −2x + 2y dx/dt 2x − y dx/dt 7x + 6y 4) = 11) = ! dy/dt 3x − 2y 12 6 dy/dt 2x + 6y x − y dx/dt 5 5 18) = dy/dt − 15 x + 13 dx/dt −3x − 2y dx/dt x + 2y 5 y 5) = 12) = dy/dt −x − 4y dy/dt 3x + 2y ! dx/dt −4x + 6y 11 9 19) = x + 5y dx/dt 2y dx/dt dy/dt −3x + 5y 6) = 13) = 65 14 dy/dt −x + 3y dy/dt x + y 5 5 ! 2 − 10 5 dx/dt 3 x − 3y dx/dt dx/dt x + 3y −4x − 2 y 20) = 7) = 14) = dy/dt − 43 x − 83 y dy/dt 2x + 2y dy/dt 3x + y INL 8.1 Sketch for t ≥ 0 the solution curve which satisfies the initial-value problem. ! dx/dt 3x + y x(0) 0 − 65 x + 32 y dx/dt x(0) −3 8) = , = 1) = , = dy/dt −x + y y(0) 1 2 4 dy/dt y(0) 0 −3x + 5y dx/dt 4x + 5y x(0) −1 dx/dt 2x + y x(0) −3 9) = , = 2) = , = dy/dt −5x − 6y y(0) 3 dy/dt −x y(0) 10 dx/dt 3x − 4y x(0) 0 dx/dt x + 4y x(0) 1 10) = , = , = 3) = dy/dt x−y y(0) 1 dy/dt −x + 5y y(0) 0 INL 8.2 dx/dt 2x + 3y x(0) −2 = , = dy/dt −3x − 4y y(0) −2 dx/dt 4y x(0) 0 5) = , = dy/dt −x + 4y y(0) 1 ! ! dx/dt − 32 x + 14 y x(0) 3 6) = , = y(0) 2 dy/dt −x − 12 y 4) dx/dt −4x − y 7) = , dy/dt x − 2y x(0) 2 = y(0) 1 dx/dt 5x + 4y = , dy/dt −x + y dx/dt x + 3y = , dy/dt −3x − 5y dx/dt 7x − 4y = , dy/dt 9x − 5y dx/dt 3x − y = , dy/dt 25x − 7y 11) 12) 13) 14) 19 x(0) −1 = y(0) 1 x(0) −2 = y(0) −2 x(0) 1 = y(0) 2 x(0) −1 = y(0) −6 20 15) 16) 17) dx/dt dy/dt dx/dt dy/dt = = dx/dt = dy/dt ! 2x − 5y , 5 4 x − 3y x(0) −2 = y(0) −3 dx/dt 5x − 3y 18) = , dy/dt 3x − y 5x − 3y , 12x − 7y ! −x − 13 y , 4 1 3x + 3y dx/dt 19) = dy/dt x(0) 1 = y(0) 1 x(0) 1 = y(0) 0 20) x(0) 0 = y(0) 1 ! −2x + 21 y , − 12 x − y x(0) −4 = y(0) 0 dx/dt −8x + 5y = , dy/dt −5x + 2y x(0) −1 = y(0) 1 INL 8.3 Sketch for t ≥ 0 the solution curve which satisfies the initial-value problem. dx/dt 3x + y x(0) 1 dx/dt x+y x(0) 1 1) = , = 11) = , = dy/dt −8x − y y(0) 1 dy/dt −2x − y y(0) 0 dx/dt −x − y x(0) 1 dx/dt 4x − 5y x(0) −3 12) = , = 2) = , = dy/dt 9x − y y(0) −3 dy/dt 5x − 4y y(0) −3 dx/dt −12x + 13y x(0) 0 dx/dt −x + 2y x(0) 1 13) = , = 3) = , = dy/dt −13x + 12y y(0) 5 dy/dt −x − 3y y(0) 0 dx/dt −x − 2y x(0) −2 dx/dt 2x − 4y x(0) 1 14) = , = 4) = , = dy/dt 8x − y y(0) −2 dy/dt 2x − 2y y(0) −2 dx/dt 2x − 3y x(0) 1 dx/dt 4x − 2y x(0) 4 15) = , = 5) = , = dy/dt 3x + 2y y(0) −2 dy/dt 5x + 2y y(0) −1 dx/dt −3x − 5y x(0) −2 dx/dt 3x + y x(0) −2 16) = , = 6) = , = dy/dt 5x + 3y y(0) 2 dy/dt −x + 3y y(0) 1 dx/dt 5x + 41y x(0) 5 dx/dt −11x + 2y x(0) 1 17) = , = 7) = , = dy/dt −x − 3y y(0) 0 dy/dt −4x − 7y y(0) 1 dx/dt −3x − y x(0) 1 dx/dt 3x − 2y x(0) 4 18) = , = 8) = , = dy/dt 2x − y y(0) 2 dy/dt 5x + y y(0) −1 ! 13 5 x(0) 2 dx/dt 2x + 2 y dx/dt 3x − 5y x(0) −5 , = 19) = 9) = , = 5 y(0) 3 dy/dt − 13 x + y dy/dt 5x − 3y y(0) 1 2 2 10) dx/dt 3x + 2y = , dy/dt −2x + 3y x(0) −1 = y(0) −1 dx/dt −2x + 2y 20) = , dy/dt −5x − 4y x(0) 3 = y(0) −2 10.. PLANE AUTONOMOUS SYSTEMS 10. Plane autonomous systems INL 10.1 1) 2) 3) 4) 5) 6) 7) 8) Classify the stationary point (origin) period) if periodic solutions occur. dx/dt ax − 2y = för a 6= −2. dy/dt 2x + 2y dx/dt 19x − 3ay = för a 6= 95 22 . dy/dt x + (a − 5)y dx/dt ax − 3y = för a 6= − 94 . dy/dt 3x + 4y dx/dt ax − 2y = för a 6= −4. dy/dt 2x + y dx/dt 3x − 2y = för a 6= − 43 . dy/dt 2x + ay dx/dt ax − 2y = för a 6= 2. dy/dt (1 − a)x + y dx/dt y = för alla a. dy/dt −x + ay dx/dt 2x + 7y = för a 6= − 49 8 . dy/dt −7x + 4ay 9) 10) dx/dt dy/dt dx/dt dy/dt = INL 10.2 1) 2) 3) 4) 5) 6) 21 −ax − 12 y för a 6= 94 . 9 x + y 2 = 3x + 54 y −5x + ay för a 6= − 25 12 . .............................................. ZC 10.1–10.3 of the system. Also, determine the orbital time (the dx/dt ax − 12y 11) = för a 6= −18. dy/dt 3x + 2y dx/dt −5x − 9y 12) = för a 6= − 95 , 0. dy/dt a2 x − ay dx/dt 2ax + 2y 13) = för a 6= 6. dy/dt −18x − 3y x + 25 y dx/dt för a 6= − 25 14) = 8 . dy/dt − 52 x + 2ay 15) 16) 17) 18) 19) 20) dx/dt ax − by = för |a| + |b| > 0. dy/dt bx + ay dx/dt 2ax + (a − 2)y = för a 6= − 56 . dy/dt 3x + 4y dx/dt 3x + ay = för a 6= 89 . dy/dt 5x + (3 − a)y dx/dt 5ax − 12y = för a 6= − 18 5 . dy/dt 3x + 2y dx/dt 3x − 5y = för a 6= − 100 3 . dy/dt 20x + ay dx/dt ax + 3y = för a 6= 56 . dy/dt (2 − a)x + 2y Find all stationary points of the system, and state for each of them whether it is asymptotically stable, stable or unstable. Also, state the classifications valid for the points regarded as stationary for the corresponding linearized system. 2 dx/dt x + x − 2xy dx/dt 14x − 2x2 − xy dx/dt 30x − 3x2 + xy = 7) = 13) = dy/dt xy + 5y dy/dt 16y − 2y 2 − xy dy/dt 60y − 3y 2 + 4xy dx/dt x + xy − 3x2 dx/dt 6x + x2 + 3xy dx/dt 2 − x2 + y = 8) = 14) = dy/dt 4y − 2xy − y 2 dy/dt 4xy − 12y dy/dt 2x2 − 2xy dx/dt xy − 2y − 4 dx/dt 20x − xy − 2x2 dx/dt y − xy + y 2 = 9) = 15) = dy/dt 4y 2 − x2 dy/dt 16y − y 2 − xy dy/dt 5x + xy − 2x2 dx/dt 4x − xy dx/dt 5x − x2 − xy dx/dt 4x − 2xy = 10) = 16) = dy/dt 5xy − y − y 2 dy/dt xy − 2y dy/dt 3y − xy − y 2 2 dx/dt 7x − xy − x2 dx/dt −3x + x2 − xy dx/dt x + 3y 2 − 4 = 11) = 17) = dy/dt xy − 5y dy/dt −5y + xy dy/dt 3xy − x2 dx/dt 3x − x2 − xy dx/dt x + x2 − 2xy dx/dt 2x − 2xy = 12) = 18) = dy/dt −2xy + 4y dy/dt xy − y dy/dt x2 − 2xy 2 − 3y Block 4 – Fourier series, Partial DE 11. Fourier series INL 11.1 ........................................................................... ZC 11.1–11.3 Find the Fourier series, the Fourier-cosinus series and the Fourier-sinus series of the function f , and sketch the corresponding graphs in the interval Iskiss . Finally, use the series to determine the sums of the two chosen of the following series: ∞ ∞ ∞ ∞ X X X X 1 (−1)n−1 1 (−1)n−1 , S = , S = . , S2 = S1 = 3 4 2n − 1 (2n − 1)2 n2 n2 n=1 n=1 n=1 n=1 1) f (x) = 2 − x2 /2 Df = (0, 2] Isketch = [−6, 6] S3 and S4 6) f (x) = x2 /3 − 1 Df = (0, 3) Isketch = [−9, 9] S3 and S4 11) f (x) = −(x + 2) Df = (−2, 0] Isketch = [−6, 6] S1 and S2 16) f (x) = −(x + 2)x Df = (−1, 0) Isketch = [−3, 3] S1 and S2 2) f (x) = 2x + 1 Df = [0, 1) Isketch = [−3, 3] S1 and S2 7) f (x) = −(2x + 1) Df = [−2, 0] Isketch = [−6, 6] S1 and S2 12) f (x) = 3x + 1 Df = [−1, 0] Isketch = [−3, 3] S1 and S2 17) f (x) = 3 − x Df = [0, 3] Isketch = [−9, 9] S1 and S2 3) f (x) = −x(4 + x) Df = (−2, 0) Isketch = [−6, 6] S3 and S4 8) f (x) = (4 − x)x Df = [0, 2) Isketch = [−6, 6] S2 and S3 13) f (x) = 1 + x2 /9 Df = (0, 3] Isketch = [−9, 9] S3 and S4 18) f (x) = x2 − 1 Df = (0, 1] Isketch = [−3, 3] S3 and S4 4) f (x) = x + 3 Df = [−3, 0] Isketch = [−9, 9] S1 and S2 9) f (x) = 3 − 2x Df = [0, 1) Isketch = [−3, 3] S1 and S2 14) f (x) = 2 − x/2 Df = [0, 2) Isketch = [−6, 6] S1 and S2 19) f (x) = x + 2 Df = [−2, 0) Isketch = [−6, 6] S1 and S2 5) f (x) = x − 1 Df = (0, 1] Isketch = [−3, 3] S1 and S2 10) f (x) = x(x + 6) Df = (−3, 0) Isketch = [−9, 9] S3 and S4 15) f (x) = (x + 3)2 Df = [−3, 0] Isketch = [−9, 9] S3 and S4 20) f (x) = (x − 3)2 + 3 Df = (0, 3) Isketch = [−9, 9] S2 and S3 12. Partial differential equations ........................................... ZC 12.1–12.5 INL 12.1 2) The temperature u of an ideal rod of length 3 is assumed to obey the partial differential equation u00xx = 71 u0t , 0 < x < 3, t > 0. The left endpoint of the rod is held at temperature 0 while the right is insulated from the environment. At time 0, the temperature is 4 degrees in the interior of the rod. Determine 1) The temperature u of an ideal rod of length 8 is assumed to obey the partial differential equation u00xx = 15 u0t , 0 < x < 8, t > 0. The endpoints of the rod are insulated from the environment. At time 0, the temperature is x degrees in the interior of the rod. Determine u for 0 < x < 8, t > 0. 22 12.. PARTIAL DIFFERENTIAL EQUATIONS u for 0 < x < 3, t > 0. 3) The temperature u of an ideal rod of length 5 is assumed to obey the partial differential equation u00xx = 2u0t , 0 < x < 5, t > 0. The endpoints of the rod are held at temperature 0. At time 0, the temperature is 3x degrees in the interior of the rod. Determine u for 0 < x < 5, t > 0. 23 10) The temperature u of an ideal rod of length 6 is assumed to obey the partial differential equation u00xx = 81 u0t , 0 < x < 6, t > 0. The left endpoint of the rod is insulated from the environment, while the right is held at temperature 0. At time 0, the temperature is 3 degrees in the interior of the rod. Determine u for 0 < x < 6, t > 0. 11) 4) The temperature u of an ideal rod of length 7 is assumed to obey the partial differential equation u00xx = 61 u0t , 0 < x < 7, t > 0. The left endpoint of the rod is insulated from the environment, while the right is held at temperature 0. At time 0, the temperature is 2 degrees in the interior of the rod. Determine u 12) for 0 < x < 7, t > 0. The temperature u of an ideal rod of length 9 is assumed to obey the partial differential equation u00xx = 14 u0t , 0 < x < 9, t > 0. The endpoints of the rod are insulated from the environment. At time 0, the temperature is 2x degrees in the interior of the rod. Determine u for 0 < x < 9, t > 0. The temperature u of an ideal rod of length 2 is assumed to obey the partial differential equation u00xx = 3u0t , 0 < x < 2, t > 0. The endpoints of the rod are held at temperature 0. At time 0, the temperature is x degrees in the interior of the rod. Determine u for 0 < x < 2, t > 0. 5) The temperature u of an ideal rod of length 6 is assumed to obey the partial differential equation u00xx = 41 u0t , 0 < x < 6, t > 0. The endpoints of the rod are held at temperature 0. At time 0, the temperature is 5 degrees in the interior of the rod. Determine u for 0 < x < 6, 13) The temperature u of an ideal rod of length t > 0. 8 is assumed to obey the partial differential equation u00xx = 12 u0t , 0 < x < 8, t > 0. 6) The temperature u of an ideal rod of length The left endpoint of the rod is insulated from 4 is assumed to obey the partial differential the environment, while the right is held at 00 0 equation uxx = 3ut , 0 < x < 4, t > 0. temperature 0. At time 0, the temperature is 5 The left endpoint of the rod is insulated from degrees in the interior of the rod. Determine u the environment, while the right is held at for 0 < x < 8, t > 0. temperature 0. At time 0, the temperature is 2x degrees in the interior of the rod. Determine 14) The temperature u of an ideal rod of length u for 0 < x < 4, t > 0. 3 is assumed to obey the partial differential equation u00xx = 19 u0t , 0 < x < 3, t > 0. The 7) The temperature u of an ideal rod of length endpoints of the rod are held at temperature 0. 9 is assumed to obey the partial differential At time 0, the temperature is 2 degrees in the equation u00xx = 12 u0t , 0 < x < 9, t > interior of the rod. Determine u for 0 < x < 3, 0. The left endpoint of the rod is held at t > 0. temperature 0 while the right is insulated from the environment. At time 0, the temperature is 15) The temperature u of an ideal rod of length x degrees in the interior of the rod. Determine 5 is assumed to obey the partial differential u for 0 < x < 9, t > 0. equation u00 = 4u0 , 0 < x < 5, t > 0. The xx t endpoints of the rod are insulated from the environment. At time 0, the temperature is 3x degrees in the interior of the rod. Determine u for 0 < x < 5, t > 0. 8) The temperature u of an ideal rod of length 2 is assumed to obey the partial differential equation u00xx = 19 u0t , 0 < x < 2, t > 0. The endpoints of the rod are insulated from the environment. At time 0, the temperature is x+3 16) The temperature u of an ideal rod of length degrees in the interior of the rod. Determine u 7 is assumed to obey the partial differential for 0 < x < 2, t > 0. equation u00xx = 18 u0t , 0 < x < 7, t > 0. The left endpoint of the rod is held at 9) The temperature u of an ideal rod of length temperature 0 while the right is insulated from 7 is assumed to obey the partial differential the environment. At time 0, the temperature is equation u00xx = 4u0t , 0 < x < 7, t > 2x degrees in the interior of the rod. Determine 0. The left endpoint of the rod is held at u for 0 < x < 7, t > 0. temperature 0 while the right is insulated from the environment. At time 0, the temperature is 17) The temperature u of an ideal rod of length 6 is assumed to obey the partial differential 9 degrees in the interior of the rod. Determine u for 0 < x < 7, t > 0. equation u00xx = 51 u0t , 0 < x < 6, t > 0. The 24 endpoints of the rod are insulated from the environment. At time 0, the temperature is 2−x degrees in the interior of the rod. Determine u for 0 < x < 6, t > 0. 5 is assumed to obey the partial differential equation u00xx = 61 u0t , 0 < x < 5, t > 0. The endpoints of the rod are held at temperature 0. At time 0, the temperature is 4x degrees in the interior of the rod. Determine u for 0 < x < 5, t > 0. 18) The temperature u of an ideal rod of length 3 is assumed to obey the partial differential equation u00xx = 2u0t , 0 < x < 3, t > 20) The temperature u of an ideal rod of length 4 is assumed to obey the partial differential 0. The left endpoint of the rod is held at equation u00xx = 17 u0t , 0 < x < 4, t > 0. temperature 0 while the right is insulated from The left endpoint of the rod is insulated from the environment. At time 0, the temperature is the environment, while the right is held at 7 degrees in the interior of the rod. Determine temperature 0. At time 0, the temperature is x u for 0 < x < 3, t > 0. degrees in the interior of the rod. Determine u 19) The temperature u of an ideal rod of length for 0 < x < 4, t > 0. INL 12.2 1) An ideal string of normal length 3 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 3) held fixed at displacement level 0. At time 0, the string is in motion such that no point of the string is displaced but has the velocity −3. Find the displacement u of the string for 0 < x < 3, t > 0, where it is assumed that the displacement obeys the wave equation 1 00 u00xx = 49 utt . 2) An ideal string of normal length 8 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 8) held fixed at displacement level 0. At time 0, the string is in rest such that interior points of the string are displaced according to x. Find the displacement u of the string for 0 < x < 8, t > 0, where it is assumed that the displacement obeys the wave equation 1 00 u00xx = 25 utt . 3) An ideal string of normal length 7 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 7) held fixed at displacement level 0. At time 0, the string is in rest such that interior points of the string are displaced 2. Find the displacement u of the string for 0 < x < 7, t > 0, where it is assumed that the displacement obeys the wave equation 1 00 u00xx = 36 utt . 4) An ideal string of normal length 5 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 5) held fixed at displacement level 0. At time 0, the string is in motion such that no point of the string is displaced but has the velocity x. Find the displacement u of the string for 0 < x < 5, t > 0, where it is assumed that the displacement obeys the wave equation 1 00 utt . u00xx = 64 5) An ideal string of normal length 4 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 4) held fixed at displacement level 0. At time 0, the string is in rest such that interior points of the string are displaced according to 2x. Find the displacement u of the string for 0 < x < 4, t > 0, where it is assumed that the displacement obeys the wave equation u00xx = 19 u00tt . 6) An ideal string of normal length 6 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 6) held fixed at displacement level 0. At time 0, the string is in motion such that no point of the string is displaced but has the velocity 2. Find the displacement u of the string for 0 < x < 6, t > 0, where it is assumed that the displacement obeys the wave equation 1 00 u00xx = 16 utt . 7) An ideal string of normal length 2 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 2) held fixed at displacement level 0. At time 0, the string is in rest such that interior points of the string are displaced 3. Find the displacement u of the string for 0 < x < 2, t > 0, where it is assumed that the displacement obeys the wave equation 1 00 u00xx = 81 utt . 8) An ideal string of normal length 9 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 9) held fixed at displacement level 0. At time 0, the string is in motion such that no point of the string is displaced but has the velocity −4. Find the displacement u of the string for 0 < x < 9, t > 0, where it is assumed that the displacement obeys the wave equation 1 00 u00xx = 36 utt . 9) An ideal string of normal length 6 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 6) held fixed at displacement level 0. At time 0, the string is in rest such 12.. PARTIAL DIFFERENTIAL EQUATIONS that interior points of the string are displaced according to −x. Find the displacement u of the string for 0 < x < 6, t > 0, where it is assumed that the displacement obeys the wave 1 00 equation u00xx = 64 utt . 25 positions 0 and 7) held fixed at displacement level 0. At time 0, the string is in motion such that no point of the string is displaced but has the velocity 4. Find the displacement u of the string for 0 < x < 7, t > 0, where it is assumed that the displacement obeys the wave equation u00xx = 91 u00tt . 10) An ideal string of normal length 7 can vibrate in a xu-plane. The string is in its both ends (the positions 0 and 7) held fixed at displacement 16) An ideal string of normal length 5 can vibrate in level 0. At time 0, the string is in motion such a xu-plane. The string is in its both ends (the that no point of the string is displaced but has positions 0 and 5) held fixed at displacement the velocity 2. Find the displacement u of the level 0. At time 0, the string is in rest such string for 0 < x < 7, t > 0, where it is assumed that interior points of the string are displaced that the displacement obeys the wave equation −2. Find the displacement u of the string 1 00 utt . u00xx = 25 for 0 < x < 5, t > 0, where it is assumed that the displacement obeys the wave equation 11) An ideal string of normal length 2 can vibrate in 1 00 utt . u00xx = 25 a xu-plane. The string is in its both ends (the positions 0 and 2) held fixed at displacement 17) An ideal string of normal length 3 can vibrate in level 0. At time 0, the string is in motion such a xu-plane. The string is in its both ends (the that no point of the string is displaced but has positions 0 and 3) held fixed at displacement the velocity −5. Find the displacement u of the level 0. At time 0, the string is in motion such string for 0 < x < 2, t > 0, where it is assumed that no point of the string is displaced but has that the displacement obeys the wave equation the velocity 5. Find the displacement u of the u00xx = 41 u00tt . string for 0 < x < 3, t > 0, where it is assumed that the displacement obeys the wave equation 12) An ideal string of normal length 9 can vibrate in 1 00 u00xx = 81 utt . a xu-plane. The string is in its both ends (the positions 0 and 9) held fixed at displacement 18) An ideal string of normal length 6 can vibrate in level 0. At time 0, the string is in rest such a xu-plane. The string is in its both ends (the that interior points of the string are displaced positions 0 and 6) held fixed at displacement 3. Find the displacement u of the string for level 0. At time 0, the string is in rest such 0 < x < 9, t > 0, where it is assumed that interior points of the string are displaced that the displacement obeys the wave equation according to −x. Find the displacement u of 1 00 u00xx = 81 utt . the string for 0 < x < 6, t > 0, where it is assumed that the displacement obeys the wave 13) An ideal string of normal length 3 can vibrate in equation u00xx = 14 u00tt . a xu-plane. The string is in its both ends (the positions 0 and 3) held fixed at displacement 19) An ideal string of normal length 5 can vibrate in level 0. At time 0, the string is in motion such a xu-plane. The string is in its both ends (the that no point of the string is displaced but has positions 0 and 5) held fixed at displacement the velocity x. Find the displacement u of the level 0. At time 0, the string is in motion such string for 0 < x < 3, t > 0, where it is assumed that no point of the string is displaced but has that the displacement obeys the wave equation the velocity 3. Find the displacement u of the 1 00 00 uxx = 49 utt . string for 0 < x < 5, t > 0, where it is assumed that the displacement obeys the wave equation 14) An ideal string of normal length 8 can vibrate in 1 00 utt . u00xx = 36 a xu-plane. The string is in its both ends (the positions 0 and 8) held fixed at displacement level 0. At time 0, the string is in rest such 20) An ideal string of normal length 4 can vibrate in a xu-plane. The string is in its both ends (the that interior points of the string are displaced positions 0 and 4) held fixed at displacement according to −3x. Find the displacement u of level 0. At time 0, the string is in rest such the string for 0 < x < 8, t > 0, where it is that interior points of the string are displaced assumed that the displacement obeys the wave 1 00 00 2. Find the displacement u of the string for equation uxx = 16 utt . 0 < x < 4, t > 0, where it is assumed 15) An ideal string of normal length 7 can vibrate in that the displacement obeys the wave equation 1 00 a xu-plane. The string is in its both ends (the u00xx = 49 utt . INL 12.3 26 1) En rektangulär, värmeledande platta med sidlängderna 5 och 2 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar de positiva koordinataxlarna. Den sida som tangerar y-axeln är av den kortare längden och hålls vid temperaturen 0 grader, medan dess motstående sida hålls isolerad från omgivningen. Den sida som tangerar x-axeln hålls vid temperaturen 0 grader, medan dess motstående sida hålls vid temperaturen 9 grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. 2) En rektangulär, värmeledande platta med sidlängderna 4 och 3 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar de positiva koordinataxlarna. Den sida som tangerar y-axeln är av den kortare längden och hålls isolerad från omgivningen, medan dess motstående sida hålls vid temperaturen 0 grader. Den sida som tangerar x-axeln hålls isolerad, medan dess motstående sida hålls vid temperaturen 7 grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. 3) En rektangulär, värmeledande platta med sidlängderna 9 och 5 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar de positiva koordinataxlarna. Den sida som tangerar y-axeln är av den kortare längden och hålls isolerad från omgivningen, medan dess motstående sida hålls vid temperaturen 0 grader. Den sida som tangerar x-axeln hålls vid temperaturen 0 grader, medan dess motstående sida hålls vid temperaturen 2 grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. 4) En rektangulär, värmeledande platta med sidlängderna 6 och 4 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar de positiva koordinataxlarna. Den sida som tangerar y-axeln är av den kortare längden och hålls vid temperaturen 0 grader, medan dess motstående sida hålls isolerad från omgivningen. Den sida som tangerar x-axeln hålls isolerad, medan dess motstående sida hålls vid temperaturen 8 grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. hålls isolerade från omgivningen. Den sida som tangerar x-axeln hålls isolerad, medan dess motstående sida hålls vid temperaturen 2 − x grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. 6) En rektangulär, värmeledande platta med sidlängderna 7 och 6 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar de positiva koordinataxlarna. De sidor som är parallella med y-axeln är de kortare och hålls vid temperaturen 0 grader. Den sida som tangerar x-axeln hålls vid temperaturen 0 grader, medan dess motstående sida hålls vid temperaturen 3 grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. 7) En rektangulär, värmeledande platta med sidlängderna 8 och 5 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar de positiva koordinataxlarna. De sidor som är parallella med y-axeln är de kortare och hålls isolerade från omgivningen. Den sida som tangerar x-axeln hålls vid temperaturen 0 grader, medan dess motstående sida hålls vid temperaturen 2x − 1 grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. 8) En rektangulär, värmeledande platta med sidlängderna 5 och 3 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar de positiva koordinataxlarna. De sidor som är parallella med y-axeln är de kortare och hålls vid temperaturen 0 grader. Den sida som tangerar x-axeln hålls isolerad, medan dess motstående sida hålls vid temperaturen 4 grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. 9) En rektangulär, värmeledande platta med sidlängderna 4 och 2 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar de positiva koordinataxlarna. De sidor som är parallella med y-axeln är de kortare och hålls isolerade från omgivningen. Den sida som tangerar x-axeln hålls vid temperaturen 0 grader, medan dess motstående sida hålls vid temperaturen 3 − x grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. 5) En rektangulär, värmeledande platta med sidlängderna 3 och 2 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar 10) En rektangulär, värmeledande platta med de positiva koordinataxlarna. De sidor som sidlängderna 9 och 4 är placerad i ett koordinatär parallella med y-axeln är de kortare och system på så vis att två av sidorna tangerar 12.. PARTIAL DIFFERENTIAL EQUATIONS 11) 12) 13) 14) 27 de positiva koordinataxlarna. De sidor som partiella differentialekvationen u00xx + u00yy = 0. är parallella med y-axeln är de kortare och hålls vid temperaturen 0 grader. Den sida 15) A rectangular, heat conductive plate with side lengths 7 and 5 is located in a coordinate system som tangerar x-axeln hålls vid temperaturen such that two of the sides are tangential to the 0 grader, medan dess motstående sida hålls positive coordinate axes. The side tangential vid temperaturen 2 grader. Bestäm den to the y-axis is held at the temperature of statiska temperaturfördelningen u i det inre av 0 degrees, while its opposite side is is kept plattan om fördelningen antas lyda den partiella 00 00 insulated from the environment. The side differentialekvationen uxx + uyy = 0. that is tangential to the x-axis is held at the En rektangulär, värmeledande platta med temperature of 0 degrees, while its opposite sidlängderna 6 och 5 är placerad i ett koordinatside is held at the temperature of 6 degrees. system på så vis att två av sidorna tangerar de Determine the static temperature distribution positiva koordinataxlarna. De sidor som är paru in the interior of the plate if the distribution allella med y-axeln är de kortare och hålls vid is assumed to obey the partial differential temperaturen 0 grader. Den sida som tangerar equation u00xx + u00yy = 0. x-axeln hålls isolerad, medan dess motstående sida hålls vid temperaturen 8 grader. Bestäm 16) En rektangulär, värmeledande platta med sidlängderna 9 och 6 är placerad i ett koordinatden statiska temperaturfördelningen u i det system på så vis att två av sidorna tangerar inre av plattan om fördelningen antas lyda den de positiva koordinataxlarna. Den sida som partiella differentialekvationen u00xx + u00yy = 0. tangerar y-axeln är av den kortare längden En rektangulär, värmeledande platta med och hålls isolerad från omgivningen, medan sidlängderna 5 och 4 är placerad i ett koordinatdess motstående sida hålls vid temperaturen system på så vis att två av sidorna tangerar 0 grader. Den sida som tangerar x-axeln de positiva koordinataxlarna. De sidor som hålls isolerad, medan dess motstående sida är parallella med y-axeln är de kortare och hålls vid temperaturen 4 grader. Bestäm den hålls isolerade från omgivningen. Den sida statiska temperaturfördelningen u i det inre av som tangerar x-axeln hålls isolerad, medan dess plattan om fördelningen antas lyda den partiella motstående sida hålls vid temperaturen 2x + 3 differentialekvationen u00xx + u00yy = 0. grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen 17) En rektangulär, värmeledande platta med sidlängderna 7 och 4 är placerad i ett koordinatantas lyda den partiella differentialekvationen system på så vis att två av sidorna tangerar de u00xx + u00yy = 0. positiva koordinataxlarna. De sidor som är parEn rektangulär, värmeledande platta med allella med y-axeln är de kortare och hålls vid sidlängderna 7 och 3 är placerad i ett koordinattemperaturen 0 grader. Den sida som tangerar system på så vis att två av sidorna tangerar x-axeln hålls isolerad, medan dess motstående de positiva koordinataxlarna. Den sida som sida hålls vid temperaturen 2 grader. Bestäm tangerar y-axeln är av den kortare längden den statiska temperaturfördelningen u i det och hålls vid temperaturen 0 grader, medan inre av plattan om fördelningen antas lyda den dess motstående sida hålls isolerad från ompartiella differentialekvationen u00xx + u00yy = 0. givningen. Den sida som tangerar x-axeln hålls isolerad, medan dess motstående sida 18) En rektangulär, värmeledande platta med sidlängderna 8 och 6 är placerad i ett koordinathålls vid temperaturen 5 grader. Bestäm den system på så vis att två av sidorna tangerar statiska temperaturfördelningen u i det inre av de positiva koordinataxlarna. De sidor som plattan om fördelningen antas lyda den partiella är parallella med y-axeln är de kortare och differentialekvationen u00xx + u00yy = 0. hålls isolerade från omgivningen. Den sida En rektangulär, värmeledande platta med som tangerar x-axeln hålls vid temperaturen sidlängderna 8 och 4 är placerad i ett koordinat0 grader, medan dess motstående sida hålls system på så vis att två av sidorna tangerar vid temperaturen 3x − 2 grader. Bestäm den de positiva koordinataxlarna. Den sida som statiska temperaturfördelningen u i det inre av tangerar y-axeln är av den kortare längden plattan om fördelningen antas lyda den partiella och hålls isolerad från omgivningen, medan differentialekvationen u00xx + u00yy = 0. dess motstående sida hålls vid temperaturen 0 grader. Den sida som tangerar x-axeln hålls vid 19) En rektangulär, värmeledande platta med temperaturen 0 grader, medan dess motstående sidlängderna 6 och 3 är placerad i ett koordinatsida hålls vid temperaturen 3 grader. Bestäm system på så vis att två av sidorna tangerar den statiska temperaturfördelningen u i det de positiva koordinataxlarna. Den sida som inre av plattan om fördelningen antas lyda den tangerar y-axeln är av den kortare längden 28 och hålls vid temperaturen 0 grader, medan dess motstående sida hålls isolerad från omgivningen. Den sida som tangerar x-axeln hålls isolerad, medan dess motstående sida hålls vid temperaturen 4 grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0. 20) En rektangulär, värmeledande platta med sidlängderna 8 och 7 är placerad i ett koordinat- system på så vis att två av sidorna tangerar de positiva koordinataxlarna. Den sida som tangerar y-axeln är av den kortare längden och hålls isolerad från omgivningen, medan dess motstående sida hålls vid temperaturen 0 grader. Den sida som tangerar x-axeln hålls vid temperaturen 0 grader, medan dess motstående sida hålls vid temperaturen 3 grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen antas lyda den partiella differentialekvationen u00xx + u00yy = 0.
© Copyright 2024