What is the solution for t^3(dy)/(dt)+4t^2y=e^{-t},y(-1)=0 ?

The solution for t^3(dy)/(dt)+4t^2y=e^{-t},y(-1)=0 is y=-(e^{-t})/(t^3)-(e^{-t})/(t^4)

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t^3(dy)/(dt)+4t^2y=e^{-t},y(-1)=0

t^{3} \frac{d y}{d t} + 4 t^{2} y = e^{- t}, y (- 1) = 0

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

Solution steps

t^{3} \frac{d y}{d 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}}

f (x) = \frac{2}{x - 1}

- \frac{5}{s ( 50 + 20 s )}

\frac{d}{d x} (ln (\frac{1 - x ^{2}}{1 + x ^{2}}))

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{9}{2}}} d x

- 4 (t - 1)

What is the solution for t^3(dy)/(dt)+4t^2y=e^{-t},y(-1)=0 ?
The solution for t^3(dy)/(dt)+4t^2y=e^{-t},y(-1)=0 is y=-(e^{-t})/(t^3)-(e^{-t})/(t^4)