Solve the differential equation $t^2y''+3ty'-3y=t$
Hint: consider the homogeneous part of the form $t^s$
Following the hint I can sub in and factorize the DE.
$$t^n(s(s-1)+3s-3)=0\Rightarrow s=-3,1$$
This means that my homogeneous solution is:
$y_h=At+B/t^3$
At this point I thought using method of undetermined coefficients or variation of parameters.
Try variation of parameters, which means considering a solution to be $$y=v_1(t)t+v_2(t)/t^3$$
Then using variation of parameters:
$v_1't+v_2'(t)/t^3=0$ and (differentiating t and 1/t^3) $v_1'-3v_2'(t)/t^4=t$. Solving these and integrating for $v_1$ and $v_2$ gives me:
$$v_1=t^2/8+c$$ and $$v_2=-t^5/4+k$$
When I plug everything in get:
$$y=(t^2/8+c)t+(-t^5/4+k)/t^3$$
but the correct solution is:
$$at+bt^{-3}+1/4t\log t$$
Best Answer
I would consider this to be Euler's differential equation and as such I would make the substitution $u=\ln t$ so $t=e^u$. Then $$y^{\prime}=\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}=\frac1t\frac{dy}{du}$$ $$y^{\prime\prime}=-\frac1{t^2}\frac{dy}{du}+\frac1{t^2}\frac{d^2y}{du^2}$$ So out differential equation now reads $$\frac{d^2y}{du^2}+2\frac{dy}{du}-3y=e^u$$ The homogeneous part is $$\frac{d^2y_h}{du^2}+2\frac{dy_h}{du}-3y_h=0$$ With solution $y_h=e^{ru}$ provided $r^2+2r-3=(r+3)(r-1)=0$ so the homogeneous part is $y_h=c_1e^{-3u}+c_2e^u$. Since $e^u$ is already a in the homogeneous solution, the particular solution will look like $y_p=Aue^u$. Then $\frac{dy_p}{du}=A(u+1)e^u$, $\frac{d^2y_p}{du^2}=A(u+2)e^u$ and $$\frac{d^2y_p}{du^2}+2\frac{dy_p}{du}-3y_p=4Ae^u=e^u$$ So $A=\frac14$ and $$y=\frac14ue^u+c_1e^{-3u}+c_2e^u=\frac14t\ln t+\frac{c_1}{t^3}+c_2t$$