Homogenous, Second order linear differential equations

Question

all.

Just begun learning about Second order differential equations. I would just like to clarify a few things and ask a few questions. So, I don't quite understand working out the general solutions for the case when $b^2=4ac$. So, take the arbitrary second order, homogenous linear differential equation with constant coefficients: $$ ay''+by'+cy = 0$$ $y(x) = e^{rt}$ is always a solution to these types of differnetial equations. So, $$y(x) = e^{rt}$$ $$ y'(x) = re^{rt}$$$$y''(x) = r^2e^{rt}$$ $$ a(r^2e^{rt})+ b(re^{rt})+c(e^{rt}) = 0 $$ $$ e^{rt} (ar^2+br+c)=0$$ Clearly, $e^{rt} \not=0$ $\therefore ar^2+br+c=0$ but $$b^2=4ac$$ so, $r=-\frac{b}{2a}$ must be a solution to the quadratic. Thus, $y(x) = e^{-\frac{b}{2a}x}$ must be a solution to the differential equation. But in order to give a general solution to a differential equation of the form $$y(x) = c_1y_1(x)+C_2y_2(x)$$ Where $y_1(x),y_2(x)$ are two solutions to the differential equation.

My question is how do I find a second solution? I already have one, $e^{-\frac{b}{2a}x}$, but how do I find another solution?

Answer 1

It turns out that if you try the solution $y(x)=v(x)y_1(x)$ in your differential equation, where $y_1=e^{-\frac{b}{2x}x}$, then you get that $v''(x)=0$.

Obviously then, $v=ax+b$, so our second and linearly independent solution must be $y_2(x) = xe^{-\frac{b}{2x}x}$.