Sample Solutions of Assignment 3 for MAT3270A:2.8-3.2 Note: Any problems to the sample solutions, email Mr. Yao Xiao a (yxiaomath.cuhk.edu.hk) directly. September,2013 1. Determine if the following equation is exact. If it is exact, find out the solutions 0 0 (a). (x4 + 4y) + (4x − 3y 8 )y = 0 (b). (x + y) + (2x − y)y = 0 0 (c). (2xy 2 + 2y) + (2x2 y + 2x)y = 0 (d). (x log y + xy)dx + (y log x + xy)dy = 0 (e). (x2 +yx2 )3/2 dx + ydy 3 (x2 +y 2 ) 2 =0 Answer: (a). Since ∂y (x4 +4y) = 4 and ∂x (4x−3y 8 ) = 4, the equation 1 1 is exact. And x5 + 4xy − y 9 = C is the first integral. 5 3 (b). Since ∂y (x + y) = 1 and ∂x (2x − y) = 2, the equation is not exact. (c). Since ∂y (2xy 2 + 2y) = 4xy + 2 and ∂x (2x2 y + 2x) = 4xy + 2, the equation is exact. And x2 y 2 + 2xy = C is the first integral. y x (d). Since ∂y (x log y + xy) = + x and ∂x (y log x + xy) = + y, the y x equation is not exact. 3 2xy 3 2xy (e). Since ∂y ( (x2 +yx2 )3/2 ) = − and ∂x ( 2 y 2 3 ) = − , 2 2 5/2 2 2 (x +y ) 2 (x + y ) 2 (x + y 2 )5/2 1 = C is the first integral. the equation is exact. And − 2 (x + y 2 )1/2 2. Using the integrating factors given below to solve the corresponding ODEs. 0 1 xy 3 cos y+2e−x cos x dy y (a). x2 y 3 + x(1 + y 2 )y = 0, µ(x, y) = (b). ( siny y − 2e−x sin x)dx + 2 = 0, µ(x, y) = yex (c). (3x + y6 )dx + ( xy + 2 xy )dy = 0, µ(x, y) = xy 1 2 Answer: (a). Multiply µ(x, y) = 1 xy 3 on both sides of equation, one has 1 + y2 0 y =0 y3 2 1 d( x2 − 2 + ln y) = 0 2 y 2 1 2 x − 2 + ln y = C 2 y x+ ⇒ ⇒ (b). Multiply µ(x, y) = yex on both sides of equation, one has (ex sin y − 2ysin x)dx + (ex cos y + 2 cos x)dy = 0 ⇒ d(ex sin y + 2y cos x) = 0 ⇒ ex sin y + 2y cos x = C (c). Multiply µ(x, y) = xy on both sides of equation, one has ⇒ ⇒ (3x2 y + 6x)dx + (x3 + 2y 2 )dy = 0 2 d(x3 y + 3x2 + y 3 ) = 0 3 2 x3 y + 3x2 + y 3 = C 3 3. Find out the integrating factors of the following ODEs and solve the corresponding ODEs (a). (3x2 y+2xy+y 3 )dx+(x2 +y 2 )dy = 0 (c). ydx+(2xy−e−2y )dy = 0 (b). dx+( xy −sin y)dy = 0 (d). ex dx+(ex cot y+2y csc y)dy = 0 My − Nx Answer: (a). Since My −Nx = 3x2 +3y 2 = 3N , µx = µ = 3µ. N Hence we can choose µ = e3x . Multiple µ on both side of ODE, we 3 have e3x (3x2 y + 2xy + y 3 )dx + e3x (x2 + y 2 )dy = 0 1 ⇒ d(e3x (x2 y + y 3 )) = 0 3 1 ⇒ e3x (x2 y + y 3 ) = C 3 1 1 My − Nx 1 (b). Since My − Nx = − = − M , µy = − µ = µ. Hence y y M y we can choose µ = y. Multiple µ on both side of ODE, we have ydx + (x − y sin y)dy = 0 ⇒ d(xy − sin y + y cos y) = 0 ⇒ xy − sin y + y cos y = C 2y − 1 My − Nx M , µy = − µ= y M 2y − 1 e2y µ. Hence we can choose µ = . Multiple µ on both side of y y ODE, we have (c). Since My − Nx = 1 − 2y = − ⇒ e2y e2y )ydx + ( )(2xy − e−2y )dy = 0 y y 1 e2y dx + (2xe2y − )dy = 0 y 2y d(e x − ln y) = 0 ⇒ e2y x − ln y = C ( ⇒ My − Nx µ= M cot yµ. Hence we can choose µ = sin y. Multiple µ on both side of (d). Since My − Nx = −ex cot y = − cot yM , µy = − ODE, we have ex sin ydx + (ex cos y + 2y)dy = 0 ⇒ d(ex sin y + y 2 ) = 0 ⇒ ex sin y + y 2 = C 4 4. Solve the following ODEs 0 (a). y = 0 2x+y , y(0) 3+3y 2 −x 2 2x (c). xy + y − y e =0 0 (b). y = = 0, y(0) = 0 (e). xdy − ydx = 2x2 y 2 dy, y(1) = −2 0 (g). y = y3 , y(0) 1−2xy 2 =1 y t−y 0 (d). y = ex+y (f). dy dx (h). (2y + 1)dx + = − 2xy+1 x2 +2y x2 −y dy x =0 Answer: (a). Original ODE can be written as (2x + y)dx + (x − 3y 2 − 3)dy = 0 Since My − Nx = 0, the equation is exact. Then we have d(x2 + xy − y 3 − 3y) = 0 ⇒ x2 + xy − y 3 − 3y = y 3 (0) − 3y(0) = 0 (b). Original ODE can be written as ydt + (y − t)dy = 0 Since My − Nt = 2 = choose µ = 2 My − Nt 2 M ,µy = − µ = − µ. Hence we can y M y 1 . Multiple µ on both side of ODE, we have y2 ⇒ ⇒ (c). Multiply 1 t dt + (1 − 2 )dy = 0 y y t d(ln y + ) = 0 y t ln y + = C y 1 on both sides of original equation, we have x2 y − y 2 e2x 1 dx + dy = 0 x2 x 1 − 2ye2x 1 2 − 2ye2x Since My −Nx = + 2 = = x2 x x2 2 1 − µ. Hence we can choose µ = 2 . Multiple y y (1) 2 My − Nx M ,µy = − µ= y M µ on both side of ODE 5 (1), we have ( ⇒ ⇒ 1 e2x 1 − )dx + dy = 0 x2 y x2 xy 2 R e2x 1 d( + dx) = 0 xy x2 R e2x 1 + dx = C xy x2 Thus original ODE can’t have a nontrivial solution such that y(x = 0) = 0. Hence y ≡ 0 is the only solution. (d). Original ODE can be written as ex+y dx − dy = 0 My − Nx µ = −µ. Hence we can M choose µ = e−y . Multiple µ on both side of ODE, we have Since My − Nx = ex+y = M ,µy = − ex dx − e−y dy = 0 ⇒ d(ex + e−y ) = 0 ⇒ ex + e−y = C (e). Original ODE can be written as ydx + (2x2 y 2 − x)dy = 0 My − Nx 2 2 µ = − µ. Hence Since My − Nx = 2 − 4xy 2 = − N ,µx = x N x 1 we can choose µ = 2 . Multiple µ on both side of ODE, we have x y 1 dx + (2y 2 − )dy = 0 2 x x 2 3 y ⇒ d( y − ) = 0 3 x 2 3 y 2 3 10 ⇒ y − = y (1) − y(1) = − 3 x 3 3 (f). Original ODE can be written as (2xy + 1)dx + (x2 + 2y)dy = 0 6 Since My − Nx = 0, the equation is exact. Then we have (2xy + 1)dx + (x2 + 2y)dy = 0 ⇒ d(x2 y + x + y 2 ) = 0 ⇒ x2 y + x + y 2 = C (g). Original ODE can be written as y 3 dx + (2xy 2 − 1)dy = 0 Since My − Nx = y 2 = 1 My − Nx 1 M ,µy = − µ = − µ. Hence we can y M y 1 choose µ = . Multiple µ on both side of ODE, we have y 1 y 2 dx + (2xy − )dy = 0 y 2 ⇒ d(xy − ln y) = 0 ⇒ xy 2 − ln y = − ln y(0) = 0 (h). Original ODE can be written as y (2y + 1)dx + (x − )dy = 0 x 1 My − Nx 1 y µ = µ. Hence we can Since My − Nx = 1 − 2 = N ,µx = x x N x choose µ = x. Multiple µ on both side of ODE, we have ⇒ ⇒ (2xy + x)dx + (x2 − y)dy = 0 1 1 d(x2 y + x2 − y 2 ) = 0 2 2 1 2 1 2 2 x y+ x − y =C 2 2 5. Transform the given initial value problem into an equivalent problem with the initial point at the origin (a). dy = t2 + y 2 , y(1) = 2, dt (b). dy = 1 − y 3 , y(−1) = 3 dt 7 Answer: (a)Let t = s + 1, y = w + 2, then dy =1 dw dt =1 ds the original problem can be written as dw = (s + 1)2 + (w + 2), w(0) = 0. ds (b)Let t = s − 1, y = w + 3, then dy =1 dw dt =1 ds the original problem can be written as dw = 1 − (w + 3)2 , w(0) = 0. ds 6. Use the method of successive approximations to solve the given initial value problem: (1) Determine φn (t); (2) Find the limit of {φn }. 0 0 (a).y = 2(y + 1), y(0) = 0, (b)y = y + 1 − t, y(0) = 0. Answer: (a) If y = φ(t), then the corresponding integral equation is Z t φ(t) = 2(φ(s) + 1)ds 0 If the initial approximation is φ0 (t) = 0, then Z t φ1 (t) = 2ds = 2t 0 Z φ2 (t) = t 2(2s + 1)ds = 2t + 2t2 0 and Z φn (t) = t 2(φn−1 (s) + 1)ds = 0 n X 2k tk 1 k! 8 lim φn (t) = e2t − 1 n→∞ (b) If y = φ(t), then the corresponding integral equation is Z t φ(t) = (φ(s) + 1 − s)ds 0 If the initial approximation is φ0 (t) = 0, then Z t 1 (1 − s)ds = t − t2 φ1 (t) = 2! 0 Z t 1 1 2(1 − s2 )ds = t − t3 φ2 (t) = 2 3! 0 and Z t tn+1 2(φn−1 (s) + 1 − s)ds = t − φn (t) = (n + 1)! 0 lim φn (t) = t n→∞ Problem 2,4,5 on page 155 Find the Wronskian of the given pair of functions. (2). cos t, sin t (4). x, ee2x s (5). et sin t, e2t cos t Answer: The computation is easy, so we just give the final result. (2). W = 1 (4). W = 2x2 e2x (5). W = −e4t Problem 7,9,12 on page 155-156 In the following problems determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution. 9 7.ty 00 + 3y = t, y(2) = 1, y 0 (2) = 2. 9. t(t − 4)y 00 + 3ty 0 + 5y = 2, y(3) = 0, y 0 (3) = −1 12. (x − 2)y 00 + y 0 + (x − 2)(tan x)y = 0, y(4) = 1, y 0 (4) = 2 Answer: (7). The original solution can written as 3 y 00 + y = 1, t hence the longest interval is t > 0. Then the points of discontinuity of the coefficients are t = 0. Therefore, the longest open interval, containing the initial point t = 1, in which all the coefficients are continuous, is 0 < t < ∞. (9).The original solution can written as y 00 + 5 2 3t y0 + y= t(t − 4) t(t − 4) t(t − 4) 3t 5 2 , q(t) = , g(t) = . Then the points t(t − 4) t(t − 4) t(t − 4) of discontinuity of the coefficients are t = 0 and t = 4. Therefore, the and p(t) = longest open interval, containing the initial point t = 3, in which all the coefficients are continuous, is 0 < t < 4. (12). The original solution can written as y 00 + y0 + (tan x)y = 0 x−2 Then the only points of discontinuity of the coefficients are x = 2, and x = 1 kπ + π, k ∈ Z. Therefore, the longest open interval, containing the 2 initial point x = 4, in which all the coefficients are continuous, is 3 2 < x < π. 2 Problem14 on page 156 1 Verify that y1 (t) = 1 and y2 (t) = t 2 are solutions of the differential 00 0 1 equation yy + (y )2 = 0 for t > 0. Then show that c1 + c2 t 2 is not, in 10 general, a solution of this equation. Explain why this result does not contradict Theorem 3.2.2. Answer: It is easy to verify y1 and y2 are solutions of the differential 00 1 0 equation yy + (y )2 = 0 for t > 0, and y = c1 + c2 t 2 is not a solution(in general) of this equation. This result does not contradict Theorem 3.2.2 because this equation is nonlinear. Problem 15 on page 156 00 Show that if y = φ(t) is a solution of the differential equation y + 0 p(t)y + q(t)y = g(t), where g(t) is not always zero, the y = cφ(t), where c is any constant other than 1, is not a solution. Explain why this result does not contradict the remark following Theorem 3.2.2. Answer: 00 0 [cφ(t)] + p(t)[cφ(t)] + q(t)[cφ(t)] 00 0 = c[φ(t) + p(t)φ(t) + q(t)φ(t)] = cg(t) 6= g(t) if c is a constant other than 1, and g(t) is not always zero. This result does not contradict Theorem 3.2.2 because this equation is not homogeneous. Problem 16 on page 156 Can y = sin t2 be a solution on an interval containing t = 0 of an equation y 00 + p(t)y 0 + q(t)y = 0 with continuous coefficients? Explain your answer. Answer: By direct computing, we have y = sin t2 , y 0 = 2t cos t2 , y 00 = −4t2 sin t2 + 2 cos t2 we have (−4t2 + q(t)) sin t2 + (2tp(t) + 2) cos t2 = 0 (∗) 11 Assume that y = sin t2 is a solution on an interval containing t = 0 and coefficients are continuous, substitute t = 0 into (*), we get 2 cos 0 = 0 which is not true. So y = sin t2 can not satisfy all the requirements. Problem 17 on page 156 If the Wronskian W of f and g is 2e6t , and if f (t) = e3t , find g(t). Answer: 0 0 0 W = f (t)g (t) − f (t)g(t) = e3t g (t) − 3e3t g(t) Let W = 2e6t , we get the following equation 0 g − 3g(t) = 2e3t . From the above equation, g(t) = 2te3t + ce3t . Proble 18 on page 156 If the Wronskian W of f and g is 2t2 et , and if f (t) = t, find g(t). Answer: 0 W = f (t)g 0 (t) − f 0 (t)g(t) = tg (t) − g(t) Let W = 2t2 et , we get the following equation 0 tg − g(t) = 2t2 et . From the above equation, g(t) = 2tet + Ct, where C be a constant. Problem 20 on page 156 If the Wronskian of f and g is t cos t−sin t and if u = f +3g, v = f −2g, find the Wronskian of u and v. 12 Answer: 0 0 = uv − u v W (u, v) 0 0 0 0 = (f + 3g)(f − 2g ) − (f + 3g )(f − 2g) 0 0 = −5f g + 5f g = −5W (f, g) = −5(t cos t − sin t). Problem 21 on page 156 Answer: W (y3 , y4 ) = y3 y40 − y30 y4 = (a1 y1 + a2 y2 )(b1 y10 + b2 y20 ) − (a1 y10 + a2 y20 )(b1 y1 + b2 y2 ) = (a1 b2 − a2 b1 )(y1 y20 − y10 y2 ) = (a1 b2 − a2 b1 )W (y1 , y2 ) Problem 23 on page 156 Find the fundamental set of solutions specified by Thm 3.2.5 for the given differential equation and initial point. y 00 + 5y 0 + 4y = 0, t0 = 1. Answer: The characteristic equation is r2 + 5r + 4 = 0, we have r = −1 and r = −4. The solution is y1 (t) = e−t , y2 (t) = e−4t . By computation, we have W (t0 ) = −3e−5 6= 0. So the set of fundamental solutions is {e−t , e−4t } 13 Problem 25 on page 156 Answer: y100 − 4y10 + 4y1 = 4e2t − 8e2t + 4e2t = 0 y200 − 4y20 + 4y2 = 4e2t + 4te2t − 8te2t − 4e2t + 4te2t = 0 Heace y1 y2 are the solutions of the ODE. W (y1 , y2 ) = y1 y20 − y10 y2 = e4t 6= 0 Hence they constitute the fundamental set of solutions. Problem 28 on page 156 Answer: (a) Since the Wronskian W = 3et , they form a fundamental set of solutions. (b) Since they are all linear combinations of y1 and y2 , they are also solutions of the given equation. (c) By calculating the respective Wronskians, we see that [y1 , y3 ] and [y1 , y4 ] [y4 , y5 ] are fundamental set of solutions while the others are not. Problem 31 on page 156 Answer: 14 1 v2 Divide by x2 on both sides: y 00 + y 0 + (1 − 2 )y = 0 x x By Abel’s theorem: Z 1 dx] W (y1 , y2 ) = c exp[− x = c exp[− ln |x|] c = |x| Problem 33 on page 157 Answer: Convert the ODE into: y 00 + p0 q + y=0 p p By Abel’s Theorem: p0 dt] p = c exp[− ln p(t)] Z W (y1 , y2 )(t) = c exp[− = c/p(t) Problem 34 on page 157 Answer: 2 y 00 + y 0 + et y = 0 t Z 2 dt] t = c exp[−2 ln t] c = 2 t W (y1 , y2 )(t) = c exp[− Since W (y1 , y2 )(1) = c = 3, W (y1 , y2 )(5) = 3 . 25 15 Problem 36 on page 157 Answer: Since W (y1 , y2 )(t) = c exp − R p(t)dt = c0 ,thus p(t) = 0 for all t. Problem 38 on page 157 Answer: Since W (y1 , y2 )(t) = y1 y20 −y10 y2 ,suppose that y1 (t0 ) = y2 (t0 ) = 0,then W (y1 , y2 )(t0 ) = 0. Thus then cannot constitute the fundamental set of solutions. Problem 39 on page 157 Answer: If so then exist a point t0 such that y1 (t0 )0 = y2 (t0 )0 = 0, hence W (y1 , y2 )(t0 ) = 0 Problem 40 on page 157 Answer: W (y1 , y2 )(t0 )0 = y1 y200 (t0 ) − y100 (t0 )y2 = 0 On the other hand: 0 Z W (y1 , y2 )(t0 ) = −c exp[− p(t)dt]p(t0 ) = 0 Then if Wronskian determinant is not zero then p(t0 ) = 0, For arbitrary solution φ(t) of the ODE which is the linear combination of y1 , y2 ,at the point t0 we have: q(t0 )φ(t0 ) = 0 If y1 (t0 ) = y2 (t0 ) = 0,q(t0 ) is arbitrary, otherwise q(t0 ) has to be zero.
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