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

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

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

t^{3} \frac{d y}{d t} + 4 t^{2} y = e^{- t}

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

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

\int \frac{t t + t}{t ^{2}} d t

R (a) = (3 a + 1)^{2}

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

\int \frac{2 x + 7}{9 - 4 x ^{2}} d x

\int_{6}^{8} \frac{60}{( x - 6 ) ^{3}} d x

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