Nonhomogeneous Linear Differential Equations with Constant Coefficients

[highlander2k5]

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]

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.

 

[highlander2k5]

If I use (x^3)(e^-x) I get y_p = (1/6)(x^3)e^(-x) - 2 + x does that sound right?

 

[HallsofIvy]

Yes, that is correct.

 

[highlander2k5]

Thanks... I understand why now.

 

[FunkyDwarf]

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]

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".

 

[FunkyDwarf]

ok fair point :P I am lazy though, i freak out in exams and make mistakes, variation of parameters, for me, leaves less room for error as its more methodical

 

[CPL.Luke]

it involves less work in a lot of cases as well