"Find a particular solution for $y'' + 2y' - 3y = 1 + xe^x$"

My Work So Far

I'm getting the wrong solution for this (keep ending up with 2 equations to solve 3 variables), and the process is quite long, so I'd just like to get some verification for whether my setup is correct. If it isn't, then that's the problem solved. If it is, then I'm probably adding a term wrong somewhere, not surprising with how tedious particular solutions are. I'm using the Method Of Undetermined Coefficients:

$y_{complementary} = C_1e^x+C_2xe^x$ (simply found with $r^2 + 2r - 3 = 0$)

For $f(x) = 1 + xe^x,$ I need to break it into two parts: $1$, and $xe^x$

$1:$ This is a polynomial with zeroes for the $x$ terms, so its contribution will be $A$.

$xe^x:$ This fits into the form $($polynomial of $x$$)e^{rx}$, so its contribution to the solution will be $x^s(Bx+C)e^x$, where $s$ is the smallest non-negative integer that avoids any terms being duplicates of $y_{complementary}.$ In this case, $s = 1$, and therefore:

$y_{particular} = A + (Bx^2 + Cx)e^x $.

Now, is this setup correct?

  • $\begingroup$ I think you should check the roots of your characteristic polynomial again. $\endgroup$ – Mike Mar 29 '17 at 7:03
  • $\begingroup$ Oh geez, so much looking over the calculus that I missed a basic factoring error. Thank you, what a stupid mistake. $\endgroup$ – Chase Mar 29 '17 at 7:15
  • $\begingroup$ Even if your reasoning was slightly faulty, the parametrization of the particular solution is correct. Please add additional steps if the original problem (2 equations for 3 variables) persists. $\endgroup$ – LutzL Mar 29 '17 at 10:41

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