midterm3

midterm3 - Math 108 Third Midterm Examination November 28,...

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Unformatted text preview: Math 108 Third Midterm Examination November 28, 2007 NAME (Please print) Page Possible Score 2 20 3 20 4 10 5 10 6 20 7 20 Total 100 Instructions: 1. Do all computations on the examination paper. You may use the backs of the pages if necessary. 2. Put answers inside the boxes (when applicable). 3. Provide arguments for functions, and bounds on integrals. 4. Write equations and inequalities, not disconnected mathematical expressions. 5. Please signify your adherence to the honor code: I, , have neither given nor received unauthorized help on this exam and I have conducted myself within the guidelines of the Duke Community Standard. 1 (20 Points) Score Find the general solution of y 00 ( x ) + 4 y ( x ) + 4 y ( x ) = x- 2 e- 2 x , x > Undetermined coefficients does not apply because the right-hand side is not of the form of a polynomial in x times an exponential. Instead, use variation of parameters. The fundamental solutions of the homogeneous equation y 00 ( x )+4 y ( x )+4 y ( x ) = 0 are y 1 ( x ) = e- 2 x and y 2 ( x ) = xe- 2 x . We can form the matrix of derivatives of the fundamental solutions Y ( x ) = y 1 ( x ) y 2 ( x ) y 1 ( x ) y 2 ( x ) = 1 x- 2 1- 2 x e- 2 x and its inverse Y ( x )- 1 = 1- 2 x- x 2 1 e 2 x Then we compute u ( t ) = Z x Y ( s )- 1 e 2 s- 2 e- 2 s 1 ds = Z x 1- 2 x- x 2 1 1 s- 2 ds = Z x- s- 1 s- 2 =- log x- x- 1 Then a particular solution to the differential equation is given by y p ( x ) = e > 1 Y ( x ) u ( x ) = 1 0 1 x- 2 1- 2 x e- 2 x- log x- x- 1 = 1 x- log x- x- 1 e- 2 x =- (1 + log x ) e- 2 x The general solution is obtained by adding an arbitrary linear combination of the fundamental solutions to the particular solution....
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midterm3 - Math 108 Third Midterm Examination November 28,...

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