What is the solution for y^'+4y=t^2e^{-4t} ?

The solution for y^'+4y=t^2e^{-4t} is y=(t^3)/(3e^{4t)}+(c_{1})/(e^{4t)}

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y^'+4y=t^2e^{-4t}

y^{^{'}} + 4 y = t^{2} e^{- 4 t}

y = \frac{t ^{3}}{3 e ^{4 t}} + \frac{c _{1}}{e ^{4 t}}

Solution steps

y^{'} (t) + 4 y = t^{2} e^{- 4 t}

Solve linear ODE:

y = \frac{t ^{3}}{3 e ^{4 t}} + \frac{c _{1}}{e ^{4 t}}

y = \frac{t ^{3}}{3 e ^{4 t}} + \frac{c _{1}}{e ^{4 t}}

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What is the solution for y^'+4y=t^2e^{-4t} ?
The solution for y^'+4y=t^2e^{-4t} is y=(t^3)/(3e^{4t)}+(c_{1})/(e^{4t)}