Applying Euler's Eq to $x^2y^{\prime\prime}-3xy^\prime+4y=0$

Question

I was looking over the solution to this problem and they came out with $y(x) = c_1x^2 + c_2x^2\log x$. I don't know where the $\log x$ came from. I got this as my final solution: $c_1e^{1 + i(12^{1/2})/2} + c_2 c_1e^{1 - i(12^{1/2})/2}$?

Answer 1

I'm not sure how you got your own answer (perhaps you could edit it into your question), but with $t:=\ln x$ we have $\dot{y}=xy^\prime$ so$$\ddot{y}=x\left(xy^\prime\right)^\prime=x^2y^{\prime\prime}+xy^\prime=x^2y^{\prime\prime}+\dot{y}.$$Hence $$\ddot{y}-4\dot{y}+4=0\implies y=(c_1+c_2t)e^{2t}=(c_1+c_2\ln x)x^2.$$The $\ln x$ is due to a repeated root in our characteristic polynomial. if you're unfamiliar with what these yield, let me explain. Defining $D:=\frac{d}{dt},\,z:=e^{-2t}y$, we can rewrite the problem as $$0=e^{-2t}(D-2)^2(e^{2t}z)=e^{-2t}(D-2)(e^{2t}\dot{z})=e^{-2t}(e^{2t}\ddot{z})=\ddot{z},$$which of course has a linear-in-$t$ solution.