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

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

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

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

y = - \frac{e ^{- x}}{x ^{2}} - \frac{e ^{- x}}{x ^{3}} + \frac{c _{1}}{x ^{3}}

Solution steps

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

Solve linear ODE:

y = - \frac{e ^{- x}}{x ^{2}} - \frac{e ^{- x}}{x ^{3}} + \frac{c _{1}}{x ^{3}}

y = - \frac{e ^{- x}}{x ^{2}} - \frac{e ^{- x}}{x ^{3}} + \frac{c _{1}}{x ^{3}}

\int 2 x^{2} d x

x + \frac{1}{x ^{3}} (1.2)

\frac{\partial}{\partial y} (3 x^{3} y^{2})

\frac{d}{d x} (x (1 - x))

\frac{d}{d t} (sin (t^{2}))

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