# Variation of Parameters

• Dec 6th 2006, 03:18 AM
sc8nt4u
Variation of Parameters
I must use Variation of Parameters to solve the following eqn:

dy/dt − 2ty = t

I don't know where to move the -2t so that I can solve for the homogeneous solution to find y1 and y2.
• Dec 6th 2006, 04:28 AM
topsquark
Quote:

Originally Posted by sc8nt4u
I must use Variation of Parameters to solve the following eqn:

dy/dt − 2ty = t

I don't know where to move the -2t so that I can solve for the homogeneous solution to find y1 and y2.

For starters, since this is a first order differential equation, there is only one homogeneous solution, not two. I'm not sure you can do a variation of parameters on a first order equation (at least, my text doesn't discuss it.)

-Dan
• Dec 6th 2006, 06:36 AM
Soroban
Hello, sc8nt4u!

I agree with Dan; Variation of Parameters doesn't apply here.

Quote:

I must use Variation of Parameters to solve: .$\displaystyle \frac{dy}{dt} - 2ty \:=\:t$

Integrating factor: .$\displaystyle I \:=\:e^{\int(-2t)dt} \:=\:e^{-t^2}$

We have: .$\displaystyle e^{-t^2}\frac{dy}{d1} -2te^{-t^2}y \:=\:te^{-t^2}$

. . which is: .$\displaystyle \frac{d}{dt}\left(e^{-t^2}y\right) \:=\:te^{-t^2}$

Integrate: .$\displaystyle e^{-t^2}y \;= \;-\frac{1}{2}e^{-t^2} + C$

Multiply by $\displaystyle e^{t^2}\!:\;\;y \;= \;Ce^{t^2} - \frac{1}{2}$