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

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

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y^'+2y=te^{-2t},y(1)=0

y^{^{'}} + 2 y = t e^{- 2 t}, y (1) = 0

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

Solution steps

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

Solve linear ODE:

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

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

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