What is the solution for t^3y^'+4t^2y=e^{-t},y(-1)=0 ?

The solution for t^3y^'+4t^2y=e^{-t},y(-1)=0 is y=-(e^{-t})/(t^3)-(e^{-t})/(t^4)

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t^3y^'+4t^2y=e^{-t},y(-1)=0

t^{3} y^{^{'}} + 4 t^{2} y = e^{- t}, y (- 1) = 0

y = - \frac{e ^{- t}}{t ^{3}} - \frac{e ^{- t}}{t ^{4}}

Solution steps

t^{3} y^{'} (t) + 4 t^{2} y = e^{- t}

Solve linear ODE:

y = - \frac{e ^{- t}}{t ^{3}} - \frac{e ^{- t}}{t ^{4}}

y = - \frac{e ^{- t}}{t ^{3}} - \frac{e ^{- t}}{t ^{4}}

What is the solution for t^3y^'+4t^2y=e^{-t},y(-1)=0 ?
The solution for t^3y^'+4t^2y=e^{-t},y(-1)=0 is y=-(e^{-t})/(t^3)-(e^{-t})/(t^4)