I'm trying to solve:
$$y''-5y'+6y=12e^{2x}$$
The roots for the homogeneous equation are $r=2$ and $r=3$. Homogeneous solution is $Ae^{2x}+Be^{3x}$
Via method of undetermined coefficients I got the particular solution as: $$y_p=-12xe^{2x}$$
Now with variation of parameters I get the same answer along with a second term:
$$y_p=-12xe^{2x}-12e^{2x}$$
Why is there a second term with Variation of Parameters?
Here's my working out:
Let:
$y_1$ = $e^{2x}$,
$y_2$ = $e^{3x}$
Find the Wronskian:
W= $$ \left[
\begin{array}{ c c }
e^{2x} & e^{3x} \\
2e^{2x} & 3e^{3x}
\end{array} \right]
$$
$$W=e^{5x}$$
Variation of Parameters formula:
$$y_p=-y_1\int \frac{y_2 g(x)}{W} dx +y_2\int \frac{y_1 g(x)}{W} dx$$
Let: $g(x)=12e^{2x}$
Plug in our values:
$$y_p=-e^{2x}\int \frac{(e^{3x}) (12e^{2x})}{e^{5x}} dx +e^{3x}\int \frac{(e^{2x}) (12e^{2x})}{e^{5x}} dx$$
$$y_p=-e^{2x}\int \frac{12e^{5x}}{e^{5x}} dx +e^{3x}\int \frac{12e^{4x}}{e^{5x}} dx$$
$$y_p=-e^{2x}\int 12 \ dx +e^{3x}\int \frac{12}{e^{x}} dx$$
$$y_p=-12xe^{2x} +12e^{3x}\int e^{-x} \ dx$$
$$y_p=-12xe^{2x} +12e^{3x}(-e^{-x})$$
$$y_p=-12xe^{2x} -12e^{2x}$$
What is that second term doing here? Why isn't this WORKING!?
Please help me figure this out or I will cry manly tears from my tear ducts and it may wet my Stewart Calculus Textbook (5th Edition) thus reducing its selling value from 5 dollars to 4 when I sell it to the pirates my university's book store.
Best Answer
It doesn't matter that that second term is there. Why? I'll show you.
$y=\color{green}{y_c}+\color{red}{y_p}=\color{green}{Ae^{2x}+Be^{3x}}\color{red}{-12xe^{2x}-12e^{2x}}=\underbrace{(A-12)}_{\text{another arbitrary constant}}\cdot e^{2x}+Be^{3x}-12xe^{2x}.$
If we re-label $A-12$ as $C$, we get:
$$\boxed{y=Ce^{2x}+Be^{3x}-12xe^{2x}}.$$