Math 217 Fall 2000 Exam 1Notational Remark: In this exam, the symbol ∂∂x( )y x means dydx.1. Suppose that ( )y x is a solution to the differential equation , = = ∂∂∂∂x( )y x ( )F ,x ( )y x = = ( )y x0y0. Then y'(x0) must equal: a) y0 b) x0 c) ( )F ,x0( )y x d) ( )F ,x y0 e) ( )F ,x ( )y x f) ( )F ,x0y0 g) ( )Fx,x0y0 h) ( )Fy,x0y0 i) Can only be determined from the given information when the initial value problem is known to have a unique solution. j) Cannot be determined from the given information.Solution: f y'(x0) = F(x0 , y(x0)) = F(x0 , y0) 2. For which ( )F ,x y is the direction field plot of the equation = = ∂∂∂∂x( )y x ( )F ,x ( )y x in the window [-2,2]x[-2,2] the following plot:a) y2 b) x2 c) − − x2y2 d) + + x2y2 e) + + x y f) − − x y g) xy h) − − 1 y i) + + 1 x j) x3 Solution: bThis is easiest done by elimination. The slopes evidently depend x on but not on y. That eliminates all answers but (b) and (j). All the slopes are positive. That eliminates answer (j) which is negative for negative x. Only (b) remains. With a little examination it becomes clear that the slopes in view are consistent with this answer.3. Suppose that ( )y x is the solution of the initial value problem , = = x∂∂∂∂x( )y x 2 ( )y x = = ( )y 1 3. Then ( )y 2 is equal to: a) 24b) 12 c) 8 d) 6 e) 4 f) 2 g) 1 h) 0 i) -2 j) -6Solution: bThe equation is separable: = x∂∂x( )y x 2 ( )y x = ∂∂x( )y x( )y x21x = d⌠⌡1yy + d⌠⌡21xx C = ( )ln y + 2 ( )ln x C = ( )y x eCx2 Simplify the constant and solve using = ( )y 1 3:= ( )y x C x2 = 3 C = ( )y x 3 x2 = ( )y 2 124. Find the general solution ( )y x of the differential equation = = ∂∂∂∂x( )y x + + ( )y x1( )y x3( )cos x . (The answer is in implicit form.) a) = = ( )ln + + ( )y x414( )sin x b) = = + + − − ( )y x41 C e( )4 ( )sin x0 c) = = + + − − ( )y x31 C e( )3 ( )sin x0 d) = = ( )ln + + ( )y x313( )sin x e) = = ( )y x3 − − + + ( )y x4( )cos x C f) = = + + − − ( )y x41 C e( )4 ( )cos x0 g) = = + + − − ( )y x31 C e( )3 ( )cos x0 h) = = ( )y x3 − − + + ( )y x4( )sin x C i) = = ( )y x3 − − ( )y x4C ( )sin x j) = = ( )y x3 − − ( )y x4C ( )cos xSolution: b = ∂∂x( )y x + ( )y x1( )y x3( )cos xIn Bernoulli form with n = -3: = − ∂∂x( )y x ( )cos x ( )y x( )cos x( )y x3Substitute = ( )w x ( )y x( ) − 1 ( )−3: = ∂∂x( )w x 4 ( )y x3∂∂x( )y xSubstitute the original equation: = ∂∂x( )w x 4 ( )y x3 + ( )y x1( )y x3( )cos x = ∂∂x( )w x 4 ( ) + ( )y x41 ( )cos xSubstitute for ( )y x = ∂∂x( )w x 4 ( ) + ( )w x 1 ( )cos xThis is a separable equation: = d⌠⌡1 + w 1w + d⌠⌡4 ( )cos x x C = ( )ln + w 1 + 4 ( )sin x C = ( )w x − e( ) + 4 ( )sin x C1Substitute = ( )w x ( )y x4 = ( )y x4 − e( ) + 4 ( )sinx C1This answer is equivalent to b.5. Consider the differential equation = = + + ∂∂∂∂x( )y x2 ( )y xxe( )x3. What is a sensible first step towards its solution? a) Give up. The Existence Theorem guarantees that there is no solution. b) Make the change of variable = = ( )v x( )y xx. c) Make the change of variable = = ( )v x ( )y x2.d) Make the change of variable = = ( )v x ( )y x( )−−2. e) Multiply both sides of the equation by 0. f) Multiply both sides of the equation by x. g) Multiply both sides of the equation by 2 x. h) Multiply both sides of the equation by x2. i) Multiply both sides of the equation by e( )−−x3. j) Substitute = = ( )y x C e( )x3 and solve for C.Solution: hThe equation is linear. A sensible first step is to multiply by the integrating factor = ed⌠⌡2xxx2. 6. Find an integrating factor for the equation = = + + x∂∂∂∂x( )y x ( )y x 3 x ( )y x . a) e( )3 xx b) x3e( )3 x c) x d) e( )−−3 x e) e( )3 x f) 3 xex g) 13 x h) xe( )3 x i) 3 x2 j) ( )ln x Solution: hIn linear form the equation is: = + ∂∂x( )y x − 1x3 ( )y x 0 := Integrating_factor = ed⌠⌡ − 1x3 xx( )ex3 7. Find ( )y 1 if ( )y t is the solution of the initial value problem , = = + + ∂∂∂∂t( )y t 2 ( )y t 2 e( )−−2 t = = ( )y 0 2.a) 3 e b) 3 c) 3e d) 3e2 e) e2 f) 2 e2 g) 2 e h) 4 i) 4e j) 4e2 Solution: jThe given equation is linear with integrating factor = e( )d⌠⌡2 te( )2 t. = + ∂∂t( )y t 2 ( )y t 2 e( )−2 t = e( )2 t + ∂∂t( )y t 2 ( )y t 2 e( )2 te( )−2 t = + e( )2 t∂∂t( )y t 2 e( )2 t( )y t 2 = ∂∂t[ ]e( )2 t( )y t 2 = e( )2 t( )y t + d⌠⌡2 t C = e( )2 t( )y t + 2 t C = ( )y t + 2 t Ce( )2 tUse = ( )y 0 2 to solve for the constant: = C 2 = ( )y t + 2 t 2e( )2 t = ( )y 1 41e28. Let = = ( )u x1( )y x and let = = ( )w x ( )y x2 in conjunction with the differential equation = = + + ∂∂∂∂x( )y x x2( )y x ( )y x2. Choose the correct statement. (Only one statement is correct. If you …
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