What is the solution for y^'+2y=e^{-2t} ?

The solution for y^'+2y=e^{-2t} is y= t/(e^{2t)}+(c_{1})/(e^{2t)}

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y^'+2y=e^{-2t}

y^{^{'}} + 2 y = e^{- 2 t}

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

Solution steps

y^{'} (t) + 2 y = e^{- 2 t}

Solve linear ODE:

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

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

x \to 0 lim ((e^{x} + x)^{\frac{2}{x}})

\frac{d}{d t} (1 - t^{2})

\frac{\partial}{\partial x} (\frac{4}{x y})

\frac{d y}{d t} = e^{2 (t)} - 3 y, y (0) = 1

What is the solution for y^'+2y=e^{-2t} ?
The solution for y^'+2y=e^{-2t} is y= t/(e^{2t)}+(c_{1})/(e^{2t)}