Math 215 January 9, 2015 SOME SOLUTIONS TO THE HOMEWORK HW1 Note: All derivatives, y 0 , y 00 , u0 , u00 , are with respect to x. (1) Suppose that y is a function of x. Express the following in terms of x, y 0 and y 00 . d 2 (a) (y ) Answer: 2yy 0 dx d2 2 (y ) Answer: 2(y 0 )2 + 2yy 00 (b) dx2 d 2 (c) (x y) Answer: x2 y 0 + 2xy dx (2) Suppose that u is a function of x and y = ux2 . Express the following in terms of x, u0 and u00 . (a) y 0 Answer: x2 u0 + 2xu. This is Question 1(c) again. (b) y 00 Answer: x2 u00 + 4xu0 + 2u (3) Find all solutions of the following differential equations: (a) y 0 = 0 Answer: y = C with C a constant. (b) y 00 = 0 Answer: y = c0 + c1 x with c0 and c1 constants. (c) y 0 = x Answer: y = x2 /2 + C with C a constant. (d) y 00 = x Answer: y = x3 /6 + c0 + c1 x with c0 and c1 constants. (4) Consider the differential equation xy 0 − 2y = 0 with x > 0. (a) Suppose y = ux2 is a solution of the equation for some function u of x. Show that u0 = 0. Answer: From y = ux2 we get y 0 = x2 u0 +2xu. Hence xy 0 −2y = x(x2 u0 +2xu)−2x2 u = x3 u0 . Since x > 0, y = ux2 is a solution if and only if u0 = 0. (b) Using this fact, find all solutions of the differential equation. Answer: If u0 = 0, then u = C for some constant C, and hence y = x2 u = Cx2 is the general solution. (5) For what integers n is y = xn a solution of x2 y 00 + xy 0 − y = 0? Answer: If y = xn then y 0 = nxn−1 and y 00 = n(n − 1)xn−2 . Plugging these expressions into the differential equation we get x2 y 00 + xy 0 − y = x2 (n(n − 1)xn−2 ) + x(nxn−1 ) − y = (n2 − 1)xn . This is the zero function if and only if n2 − 1 = 0, that is, if n = ±1. HW2 (1) Verify that√the following functions are solutions of the corresponding differential equations. √ (a) y = 2 x + c, y 0 = 1/ x. √ Answer: Taking the derivative of both sides of y = 2 x + c gives y0 = √ d (2x1/2 + c) = x−1/2 = 1/ x. dx (b) y 2 = e2x + c, yy 0 = e2x . Answer: Taking the derivative of both sides of y 2 = e2x + c using the chain rule we get 2yy 0 = 2e2x , which simplifies to the given differential equation. 1 2 p d (c) y = arcsin xy, xy 0 + y = y 0 1 − x2 y 2 . Hint: arcsin x = sin−1 x and (sin−1 x) = dx 1 √ . 1 − x2 Answer: Using the chain rule, we first calculate the derivative of arcsin(xy) with respect to x: 1 d 1 dy d (arcsin(xy)) = p (xy) = p y+x . dx dx 1 − (xy)2 dx 1 − (xy)2 Now suppose that y satisfies y = arcsin(xy). Taking the derivative of both sides of this equation we get the differential equation dy dy 1 y+x =p , dx dx 1 − (xy)2 p which simplifies to xy 0 +y = y 0 1 − x2 y 2 . Hence y is a solution of the given differential equation. d (d) x + y = arctan y, 1 + y 2 + y 2 y 0 = 0. Hint: arctan x = tan−1 x and (tan−1 x) = dx 1 . 1 + x2 Answer: Suppose that y satisfies x + y = arctan y. Taking the derivative of both sides of this equation with respect to x using the chain rule we get the differential equation dy 1 dy 1+ = dx 1 + y 2 dx which simplifies to 1 + y 2 + y 2 y 0 = 0. Hence y is a solution of the given differential equation. (2) Use separation of variables to solve the following equations: (a) x5 y 0 + y 5 = 0 Answer: 1/y 4 + 1/x4 = C (b) y 0 − y tan x = 0 Answer: Rewrite as dy/y = tan x dx. Solution is y = C sec x. (c) (y + yx2 + 2 + 2x2 ) dy = dx Answer: y 2 /2 + 2y = tan−1 x + C (d) y 0 /(1 + x2 ) = x/y and y = 3 when x = 1. Answer: The general solution of the differential equation is 2y 2 = 2x2 + x4 + C. The solution such that y = 3 when x = 1 is 2y 2 = 2x2 + x4 + 15. (e) y 0 = x2 y 2 and the graph of the solution passes through (−1, 2). Answer: −1/y = x3 /3 − 1/6 dy (3) The differential equation 2y = x2 + y 2 − 2x is not separable. Supposing that y is a soludx tion of this equation, show that v = x2 + y 2 is a solution of a separable differential equation. Use this fact to find all solutions of the original differential equation. dy dv Answer: First we find the relationship between and by differentiating v = x2 + y 2 : dx dx dv d 2 dy = (x + y 2 ) = 2x + 2y . dx dx dx dv Plugging this into the original differential equation gives = v. This separable equation dx x 2 2 for v has solutions √ v = Ae where A is a constant. From this equation and v = x + y we x 2 find that y = ± Ae − x , with A > 0, are all solutions of the original differential equation.