What is the solution for x^3y^'+4x^2y=e^{-x} ?

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

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x^3y^'+4x^2y=e^{-x}

x^{3} y^{^{'}} + 4 x^{2} y = e^{- x}

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

Solution steps

x^{3} y^{'} (x) + 4 x^{2} y = e^{- x}

Solve linear ODE:

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

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

\int \frac{1}{cot ( x )} d x

g (x) = x + x^{3}

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

f (x) = \frac{sec ( x ) + tan ( x )}{sec ( x ) - tan ( x )}

\frac{d}{d x} (\frac{4 x}{( x + 1 )})

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