Given the equation y'' + 2y' + y = 0 and the solution y1 = ${xe}^{-x}$. Solve for a second solution $y_2$.
I've solved it twice and get $y_2 = {xe}^{-x}(-{x}^{-1}+c)$ , but that's not being counted as correct. I distributed the ${xe}^{-x}$ in my actual answer. Can someone please walk me through how to get the solution?
I solved once using $y_2 = y_1(x)\int\frac{e^{-\int P(x) dx}}{y_1^2}$ and again using $y_2 = y_1(x)u(x)$.
Consider $y_2 = c(x) x e^{-x}$. After substitution we arrive at
$$ (2c'(x)-x c''(x))e^{-x}=0 $$
and solving the ode $2c'(x)-x c''(x)=0$ we have
$$ c(x) = c_1+\frac{c_2}{x} $$
note that $y_2 = c_1x e^{-x}+c_2 e^{-x}$ covers the solutions (two independent constants) so $y_2 = e^{-x}$