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

**Belowzero78** · Oct 30th 2010, 02:11 PM

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

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!

**TheEmptySet** · Oct 30th 2010, 04:30 PM

The equation is of the Cauchy Euler type.

Make the substitution

$z=\ln(t)$ then

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

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

Can you finish from here?

This gives the ODE

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

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

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