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.

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- Dec 6th 2006, 03:18 AMsc8nt4uVariation 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 AMtopsquark
- Dec 6th 2006, 06:36 AMSoroban
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 $\displaystyle e^{t^2}\!:\;\;y \;= \;Ce^{t^2} - \frac{1}{2}$