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Differential Equations                                        Quiz #7                                                      April 13, 2005

Name____________________                   R.  Hammack                                                 Score ______
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(1)    Find the general solution of the differential equation  y' '- y = x e^x+ 4

Looking at the associated homogeneous equation   y''-y = 0,
you can see that its auxiliary equation is m^2-1 = 0, with roots 1 and -1.  
 It follows that the complementary function is y_c = c_1e^(-x)+c_2e^x.


Now, the D.E. that we want to solve is (D^2-1)y = x e^x+ 4,
so we apply the annihilator operator D(D - 1)^2 to get rid of the function  x e^x+ 4:

(D^2 - 1) y = x e^x + 4  D(D - 1)^2 (D^2 - 1) y = D(D - 1)^2 (x e^x + 4)  D(D - 1)^2 (D^2 - 1) y = 0

the roots are -1, 1, 1, 1, 0 so the general solution is
y = c_1e^(-x)+c_2e^x+ A x e^x+ B x^2e^x+ C,
with y_c = c_1e^(-x)+c_2e^x and y_p=A x e^x+B x^2e^x+C.


Note y_p' = A x e^x+ A e^x+ B x^2e^x+ 2B x e^x
And   y_p'' = A x e^x+ A e^x+ A e^x+ B x^2e^x+ 2B x e^x+ 2B x e^x+ 2B e^x
Or  y_p'' = (2A + 2B)e^x+ (A x+4B) x e^x+ B x^2e^x


Plugging y_p into  y''-y = x e^x+ 4 gives
(2A + 2B)e^x+(A x + 4B) x e^x+ B x^2e^x-(A x e^x+ B x^2e^x+C) = x e^x+ 4
(2A + 2B)e^x+ 4B x e^x-C = x e^x+ 4

From this we see C = -4, B = 1/4, and A = -1/4.

SOLUTION: y = c_1e^(-x)+c_2e^x-1/4 x e^x+1/4 x^2e^x-4