s2 - Math 5525: Spring 2010. Introduction to Ordinary...

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Unformatted text preview: Math 5525: Spring 2010. Introduction to Ordinary Differential Equations: Homework #2. Problems and Solutions #1. Find the solution of the problem xy + 4 y + x = 0 . satisfying the initial condition y (1) = 0. Solution. The corresponding homogeneous equation y 1 + 4 x- 1 y 1 = 0 has a solution y 1 ( x ) = x- 4 . One can find a particular solution of the given equation in the form y ( x ) = C ( x ) y 1 ( x ), which implies C y 1 =- 1 , C =- x 4 , C =- x 5 5 . The general solution y = y + C 1 y 1 =- x 5 + C 1 x- 4 . Since y (1) = 0, we have C 1 = 1 5 , and y = 1 5 ( x- 4- x ). #2. Find the general solution of the equation dy dx = x + y 2 2 xy . Solution. Set z = y 2 . Then dy dx = x + y 2 2 xy = dz dx = 2 y dy dx = x + z x = 1 + z x , v = z x = v + x dv dx = 1 + v, dv = dx x , v = ln | x | + C, y 2 = xv = (ln | x | + C ) x. #3. Find the general solution of the equation y 00 + 4 y + 5 y = e- 2 x cos x. Solution. The characteristic equation r 2 +4 r +5 = 0 has zeros r 1 =- 2+ i and r 2 =- 2- i . Let us first find a particular solution of the equation Lz = ( D 2 + 4 D + 5) z = z 00 + 4 z + 5 z = e r 1 x = e- 2 x (cos x + i sin x ) . We take z p ( x ) = xAe r 1 x . Then Lz p = ( D- r 1 )( D- r 2 )[ e r 1 x Ax ] = e r 1 x D ( D + r 1- r 2 )( Ax ) = e r 1 x ( D 2 +2 iD )( Ax ) = 2 iAe r 1 x = e r 1 x = A =- i 2 , z p =- i 2 xe r 1 x = 1 2 xe- 2 x (sin x- i cos x ) ....
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This note was uploaded on 02/15/2012 for the course MATH 5525 taught by Professor Staff during the Spring '08 term at Minnesota.

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s2 - Math 5525: Spring 2010. Introduction to Ordinary...

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