What is the solution for t^3y^'+4t^2y=e^{-t} ?

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

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

t^{3} y^{^{'}} + 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} y^{'} (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}}

n = 1 \sum \infty \frac{1}{( n + 1 ) !}

\int 2 y + x e^{x y} d y

\frac{d}{d x} (\frac{1}{( x + 3 ) ^{5}})

y^{^{''}} + 2 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

x \to 0 lim (x sin (\frac{7}{x}))

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