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#### karush

##### Well-known member

- Jan 31, 2012

- 2,928

$\displaystyle y^\prime - 2y =3te^t, \\$

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -2 \, dx =e^{-2t} \\$

Multiply thru with $e^{-2t}$

$e^{-2t}y^\prime + 2e^{-2t}y= 3te^{-t} \\$

Simplify:

$(e^{-2t}y)'= 3te^{-t} \\$

Integrate:

$\displaystyle e^{-2t}y=\int 3te^{-t} dt =-3e^{-t}(t+1)+c_1 \\$

{Divide by $e^{-2t}$

$y=-3e^t t - 3e^{-t} +c_1 e^t \\$

Answer from text book

$y=\color{red}{c_1 e^{2t}-3e^t}$

ok sumtum went wrong somewhere?

$$\tiny\color{blue}{\textbf{Text book: Elementary Differential Equations and Boundary Value Problems}}$$