Nonhomogeneous problems

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.

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)
 

HallsofIvy

Science Advisor
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e-x, xe-x and x2e-x are solutions to the homogeneous solution so of course y_p1= e-x gives you nothing. Try y_p1= x3e-x.
 
If I use (x^3)(e^-x) I get y_p = (1/6)(x^3)e^(-x) - 2 + x does that sound right?
 

HallsofIvy

Science Advisor
41,618
821
Yes, that is correct.
 
Thanks... I understand why now.
 
i find in this situation variation of parameters works better than undetermined coefficients, when you get to having to guess the form of somin like x^3e^-x i just freak out :P
 

HallsofIvy

Science Advisor
41,618
821
I don't see why you would have to "guess". Obviously, something of the form e-x is called for but e-x, xe-x, and x2e-x are already "taken". x3e-x is obviously "next in line".
 
ok fair point :P im lazy though, i freak out in exams and make mistakes, variation of parameters, for me, leaves less room for error as its more methodical
 
it involves less work in alot of cases as well
 

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