What is the solution for (dy)/(dt)+2y=te^{-2t} ?

The solution for (dy)/(dt)+2y=te^{-2t} is y=(t^2)/(2e^{2t)}+(c_{1})/(e^{2t)}

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(dy)/(dt)+2y=te^{-2t}

\frac{d y}{d t} + 2 y = t e^{- 2 t}

y = \frac{t ^{2}}{2 e ^{2 t}} + \frac{c _{1}}{e ^{2 t}}

Solution steps

\frac{d y}{d t} + 2 y = t e^{- 2 t}

Solve linear ODE:

y = \frac{t ^{2}}{2 e ^{2 t}} + \frac{c _{1}}{e ^{2 t}}

y = \frac{t ^{2}}{2 e ^{2 t}} + \frac{c _{1}}{e ^{2 t}}

\frac{\partial}{\partial x} (- 3 e^{x} ln (y))

\int \frac{x ^{3}}{x ^{2} - 4} d x

x \to 1 + lim (\frac{( 2 )}{x - 1})

\int_{- 2}^{3} (x^{2} - 4) d x

\int \frac{2 x}{x ^{2} + 6 x + 13} d x

What is the solution for (dy)/(dt)+2y=te^{-2t} ?
The solution for (dy)/(dt)+2y=te^{-2t} is y=(t^2)/(2e^{2t)}+(c_{1})/(e^{2t)}