Worksheet11_1B_Solutions

# Undetermined coefficients y p ax b ce x i ln x

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Undetermined coefficients; y p = Ax + B + Ce x . i) ln x . Variation of parameters. 2. Use undetermined coefficients to find the general solution: a) y 00 + 2 y 0 - 3 y = e x . First, we solve the homogeneous equation y 00 + 2 y 0 - 3 y = 0. The characteristic equation is r 2 +2 r - 3 = 0, which factors as ( r - 1)( r +3) = 0, so r = 1 , - 3. The general solution of the homogeneous equation is y = c 1 e x + c 2 e - 3 x . Now, we try to find a particular solution of the non-homogeneous equation. Since G ( x ) = e x , then a good candidate is y p = Ae x . However, this is a solution of the homogenous equation, so let’s try y p = Axe x . Thus, y 0 p = A ( e x + xe x ) and y 00 p = A (2 e x + xe x ). So, we have A (2 e x + xe x ) + 2 A ( e x + xe x ) - 3 Axe x = e x (2 A + 2 A ) e x + ( A + 2 A - 3 A ) xe x = e x 4 Ae x = e x Therefore, A = 1 / 4, and the general solution of the non-homogeneous equation is y + y p = c 1 e x + c 2 e - 3 x + 1 / 4 xe x . b) y 00 + 2 y 0 + y = e x cos x . First, we solve the homogeneous equation y 00 + 2 y 0 + y = 0. The characteristic equation is r 2 + 2 r + 1 = 0, which factors as ( r + 1) 2 = 0, so r = - 1. The general solution of the homogeneous equation is y = c 1 e - x + c 2 xe - x . Now, we try to find a particular solution of the non-homogeneous equation. Since G ( x ) = e x cos x , then a good candidate is y p = Ae x cos x + Be x sin x . Thus, y 0 p = ( A + B ) e x cos x + ( B - A ) e x sin x and y 00 p = 2 Be x cos x - 2 Ae x sin x . So, we have 2 Be x cos x - 2 Ae x sin x + 2(( A + B ) e x cos x + ( B - A ) e x sin x ) + Ae x cos x + Be x sin x = e x cos x (3 A + 4 B ) e x cos x + (3 B - 4 A ) e x sin x = e x cos x Therefore, 3 A + 4 B = 1 and 3 B - 4 A = 0. Solving this system of equations yields

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• Fall '08
• Reshetiken
• u2, Heterogeneity, Homogeneity, general solution, 1 2 g

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