y"-4y'+3y = 8xe^{x}
I guessed that the solution to y was Axe^{x} but when I plugged it into y it did not get me the correct answer. Please don't work out the problem for me, right now I just need help understanding how I would guess y in this situation. Much appreciated.
On the right side of the ODE you have the product of a linear function and an exponential function, so you would expect the particular solution would take the form:
Note: while I have switched A and B from the form The EmptySet used, they are equivalent.
As stated, you don't want any of the terms of your particular solution to be part of the complementary solution, so you multiply the particular solution by where is the smallest integer so that no terms of the particular solution are part of the complementary solution. In this case, . Hence:
Okay I think I understand what I was confused on before. But when I plug in Ax^{2}e^{x}+Bxe^{x} I got stuck at the part where it turns out to be
-2A2xe^{x}+A2e^{x}+Bx4e^{x}-4Bx2e^{x}+3Bxe^{x}=8xe^{x}
Ax^{2}e^{x}-4Ax^{2}e^{x}+3Ax^{2}e^{x} cancels out so I did not include this part above
Using:
we find:
and substituting into the ODE, we get:
Divide though by , distribute constants and collect like terms:
Equating coefficients yields the system:
Now, solve this system to determine A and B, and then you have your particular solution.