What is the solution for y^'+2y=e^{-t},y(0)= 3/2 ?

The solution for y^'+2y=e^{-t},y(0)= 3/2 is y= 1/(e^t)+1/(2e^{2t)}

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y^'+2y=e^{-t},y(0)= 3/2

y^{^{'}} + 2 y = e^{- t}, y (0) = \frac{3}{2}

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

Solution steps

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

Solve linear ODE:

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

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

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What is the solution for y^'+2y=e^{-t},y(0)= 3/2 ?
The solution for y^'+2y=e^{-t},y(0)= 3/2 is y= 1/(e^t)+1/(2e^{2t)}