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Unformatted text preview: ACM 95b/100b Solutions for the Midterm Sean Mauch sean@caltech.edu February 20, 2002 1 Problem 1 Problem. Find the general solution to each of the following first order ODEs. a. (6 points) xy + xy = 1 y b. (6 points) (3 x 2 y 2 + x 2 y ) y + 2 xy 3 + y = 0 c. (6 points) y + y tan x = sec x d. (6 points) xy = y + x e y/x e. (6 points) y = e x + y Solution. a. This is a linear differential equation. We find the integrating factor and then integrate to determine the solution. x = 0 is a singular point of the equation. We assume that x is nonzero. xy + xy = 1 y y + 1 + 1 x y = 1 x I = e R (1+1 /x )d x = e x +ln  x  =  x  e x We choose I = x e x . d d x ( x e x y ) = e x x e x y = e x + c y = 1 x + c x e x b. We consider the differential equation ( 3 x 2 y 2 + x 2 y ) y + 2 xy 3 + y = 0 . We note that the equation is exact. d d x ( 3 x 2 y 2 + x 2 y ) = d d y ( 2 xy 3 + y ) = 6 xy 2 + 1 1 We integrate to implicitly determine the solution. d d x ( x 2 y 3 + xy y 2 ) = 0 x 2 y 3 y 2 + xy + c = 0 One could factor the cubic equation for y , but its not easy without a computer and its not a nice answer. c. This is a linear differential equation. We find the integrating factor and then integrate to determine the solution. y + y tan x = sec x I = e R tan x d x = e ln  cos x  =  sec x  We choose I = sec x . d d x y cos x = 1 cos 2 x y cos x = tan x + c y = sin x + c cos x d. We consider the equation xy = y + x e y/x . Since the coefficients are homogeneous functions of order 1, this is a homogeneous coefficient differential equation. We make the change of dependent variable u = y/x to make the equation separable. y = y x + e y/x xu + u = u + e u e u u = 1 x e u = ln  x  + c e y/x + ln  x  = c We can solve for y to obtain an explicit solution. e u = ln c x u = ln ln c x y = x ln ln c x e. This is a separable equation. y = e x + y e y y = e x e y = e x + c e x + e y = c We can solve for y to obtain an explicit solution. y = ln( c e x ) 2 2 Problem 2 Problem. a. (15 points) Transform the following ordinary differential equation into a first order linear system and find a fundamental matrix y 000 y 00 y + y = 0 . b. (15 points) Solve the following initial value problem using the variation of parameters d x d t = 4 1 1 2 x + t t , x (0) = 1 2 Solution. a. We consider the differential equation y 000 y 00 y + y = 0 . We transform the problem to a first order system of differential equations. y y 1 y 2 y 3 y y y 00 y = y 2 y 3 y 3 + y 2 y 1 y = 1 0 0 1 1 1 1 y The eigenvalues of the matrix are 1 = 1, 2 = 3 = 1. The corresponding eigenvectors are 1 = 1 1 1 , 2 = 1 1 1...
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This note was uploaded on 01/08/2011 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.
 Winter '09
 NilesA.Pierce

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