Thread: Finding the General Solution of a 2nd Order Linear Ordinary Differential Equation

Question: Find the general solution to the equation: $\displaystyle t^2y'' - 2y = t $.

Attempt at solution: The main catch to this question that causes me problems is the fact that the coefficients are not numbers but rather variables such as t in the question. I know to use the method of undetermined coefficients to solve this question, but i am just stuck on that part!

Thanks to all who contribute!

The equation is of the Cauchy Euler type.

Make the substitution

$\displaystyle z=\ln(t)$ then

$\displaystyle \frac{dy}{dt}=\frac{dy}{dz}\frac{dz}{dt}=\frac{1}{ t}\frac{dy}{dz}$

$\displaystyle y''=\frac{d}{dt}\frac{dy}{dt}=-\frac{1}{t^2}\frac{dy}{dz}+\frac{1}{t^2}\frac{dy^2 }{dz^2}$

Can you finish from here?

This gives the ODE

$\displaystyle t^2(-\frac{1}{t^2}\frac{dy}{dz}+\frac{1}{t^2}\frac{dy^2 }{dz^2})-2y=e^{z} $

$\displaystyle y''-y'-2y=e^{z}$