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.
Hello, sc8nt4u!
I agree with Dan; Variation of Parameters doesn't apply here.
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 $\displaystyle e^{t^2}\!:\;\;y \;= \;Ce^{t^2} - \frac{1}{2}$