I was wondering if anyone could check my work on this to make sure I'm doing this right for finding a particular solution to y''' + 3y'' + 3y' + y = e^(-x) + 1 + x. First I split the problem into 2 halfs y_p1 and y_p2.(adsbygoogle = window.adsbygoogle || []).push({});

y_p1 = Ce^(-x)

-Ce^(-x) + 3Ce^(-x) - 3Ce^(-x) + Ce^(-x) = e^(-x)

0*Ce^(-x) = e^(-x)

y_p1 = 0

y_p2 = C1 + (C2)x

0 + 3(0) + 3(C2) + C1 + (C2)x = 1 + x

get like terms together so...

(C2)x = x and 3(C2) + C1 = 1 therefore C2 = 1 and C1 = -2

y_p2 = -2 + x

particular solution = y_p1 + y_p2 = -2 + x

and then the general solution to the problem would be:

C1e^(-x) + c2(x)e^(-x) + C3(x^2)e^(-x)

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# Nonhomogeneous problems

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