<s>[INST] Solution to Differential equation 2nd order, x*e^x I'm trying to solve the following differential equation:
$y'' + y = x \cdot e^x$
I already have the homogeneous solution: $y_h = c_1 \cdot cos(x) + c_2 \cdot sin(x)$ but I'm struggeling to find the particular solution.
I tried using $y_p = a \cdot x \cdot e^x$ but that didn't quite work, or maybe I did something wrong... 
Can somebody explain to me how to solve this?
Thanks.
Edit: Here's the solution:
$y_p = (a \cdot x + b) \cdot e^x$
$y_p' = (a + a \cdot x + b) \cdot e^x$ and $ y_p'' = (2 \cdot a + a \cdot x + b) \cdot e^x$
$y_p'' + y_p = (2 \cdot a \cdot x + 2 \cdot a + 2 \cdot b) \cdot e^x$
$\Rightarrow 2 \cdot a = 1 \Rightarrow a = \frac{1}{2}$ and $2 \cdot a + 2 \cdot b = 0 \Leftrightarrow 2 \cdot a = 1 = - 2 \cdot b \Rightarrow b = - \frac{1}{2}$
$\Rightarrow y_p = (a \cdot x + b) \cdot e^x = \frac{x-1}{2} \cdot e^x$
All in all: $y = y_h + y_p = c_1 \cdot cos(x) + c_2 \cdot sin(x) + \frac{x-1}{2} \cdot e^x$
 [/INST] Try $y_p = (ax+b)e^x$.
[Additional characters to bypass the 30 characters rule]
</s>