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Unformatted text preview: GE 207K Method of Undetermined Coefficients – PS October 03, 2011 Problem 1 Find the general solution to the following differential equation y 00 y = 1 . (1) Solution : The corresponding homogeneous differential equation is y 00 y = 0 , and the characteristic equation along with roots are r 2 r = 0 , r ( r 1) = 0 , ⇒ r 1 = 0 ,r 2 = 1 . Therefore, the homogeneous solution is y h = c 1 + c 2 e t . (2) For the given f ( t ) = 1 , we would choose y p = A . But a constant already appears in y h , therefore we apply the modification rule and multiply y p by t , hence obtaining ! y p = At. (3) Next we recheck to make sure our new y p doesn’t have any terms which appear in y h . 3 The derivatives of y p are, y p = A, y 00 p = 0 . Plugging these into the ODE we find (0) {z} y 00 p A {z} y p = 1 . ⇒ A = 1 . Therefore, the particular solution is y p = t, (4) and the general solution is ∴ y = c 1 + c 2 e t t . (5) [email protected] 1 c hf, 2011 GE 207K Method of Undetermined Coefficients – PS October 03, 2011 Problem 2 Find the general solution to the following differential equation y 00 3 y 4 y = 2 sin t (6) Solution : The corresponding homogeneous differential equation is y 00 3 y 4 y = 0 , and the characteristic equation along with roots are r 2 3 r 4 = 0 , ( r 4)( r + 1) = 0 , ⇒ r 1 = 4 ,r 2 = 1 . Therefore, the homogeneous solution is y h = c 1 e 4 t + c 2 e t . (7) For the given f ( t ) = 2 sin t , we choose y p = A cos t + B sin t. (8) By inspecting, we find that none of the terms in our choice for y p appear in y h . 3 The derivatives of y p are, y p = A sin t + B cos t y 00 p = A cos t B sin t. Plugging these into the ODE we find ( A cos t B sin t )  {z } y 00 p 3 ( A sin t + B cos t )  {z } y p 4 ( A cos t + B sin t )  {z } y p = 2 sin t. Rearranging above equation by factoring our the functions, we have ( A 3 B 4 A ) cos( t ) + ( B + 3 A 4 B ) sin t = 2 sin t + 0 cos t. (9) Comparing the coefficients of sin t and cos t on left and right sides of the equation, we note that B + 3 A 4 B = 2 , A 3 B 4 A = 0 . [email protected] 2 c hf, 2011 GE 207K Method of Undetermined Coefficients – PS October 03, 2011 Solving for A , and B we find that A = 13 17 , and B = 5 17 . The particular solution is then y p = 13 17 cos t 5 17 sin t, (10) and the general solution is y = y h + y p ∴ y = c 1 e 4 t + c 2 e t + 13 17 cos t 5 17 sin t . (11) [email protected] 3 c hf, 2011 GE 207K Method of Undetermined Coefficients – PS October 03, 2011 Problem 3 Find the general solution to the following differential equation y 00 + 4 y = 8 x 2 . (12) Solution : The corresponding homogeneous differential equation is y 00 + 4 y...
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 Fall '08
 Fonken
 Differential Equations, Equations, yp, undetermined coefficients

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