General solution of a second order differential equation

Question

I am given the differential equation: $t^2\cdot x''-t\cdot x'+4x=\log(t)$ and I have to find it's general solution. I have made the substitution $t=e^s$ and this resulted in this equation: $y''-2y'+4y=s$. Since it's not your normal form of a second order differential equation (i.e. $y''+ay'+by=0$) I don't know how to continue from here.

Answer 1

$$t^2 x''-t x'+4x=\ln(t)$$ You can deduce directly the solution to the inhomogeneous equation

Substitute $$x_p=A \ln (t)+B, x'_p=\frac A t, x''_p=- \frac A {t^2}$$ We have that $$-A-A+4A \ln(t)+4B=\ln(t)$$ $$\implies A=\frac 14 , B=\frac 1 8$$ So the particular solution is $$x_p=\frac 14 \ln (t)+\frac 18$$

For the general solution we substitute $t=e^s, s= \ln t$

Note that $$x'=\frac {dx}{dt}=\frac {dx}{ds}\frac {ds}{dt}=x'_s \frac 1t$$ $$x''=\frac {dx'}{dt}=-\frac 1 {t^2}x'_s+\frac 1 {t^2}x''_s$$ the equation becomes $$x''_x-2x_s+4x=0$$ The characteristic polynomial is $$r^2-2r+4=0 \implies (r-1)^2-3i^2=0 \implies r=1 \pm \sqrt 3 i$$ Therefore the general solution of the homogeneous equation is

$$ x_h(s)=e^s(c_1 \cos (\sqrt 3 s)+c_2 \sin (\sqrt 3 s))$$ Substitute back t variable $$ x_h(t)=t(c_1 \cos (\sqrt 3 \ln |t|)+c_2 \sin (\sqrt 3 \ln|t|))$$ The general solution : $$x(t)=x_h(t)+x_p(t)$$ $$ x(t)=t(c_1 \cos (\sqrt 3 \ln |t|)+c_2 \sin (\sqrt 3 \ln|t|))+\frac {\ln t}4 +\frac 1 8$$