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

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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 0 ( x ) = C ( x ) y 1 ( x ), which implies C y 1 = - 1 , C = - x 4 , C = - x 5 5 . The general solution y = y 0 + 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 + 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 + 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 ) . 1

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Therefore, y p = Re ( z p ) = 1 2 xe - 2 x sin x satisfies Ly p = e - 2 x cos x , and the general solution is y ( x ) = e - 2 x ( C 1 cos x + C 2 sin x ) + 1 2 xe - 2 x sin x.
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s2 - Math 5525 Spring 2010 Introduction to Ordinary...

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