What is the solution for xy^'+2y=-1e^x ?

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

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xy^'+2y=-1e^x

x y^{^{'}} + 2 y = - 1 e^{x}

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

Solution steps

x y^{'} (x) + 2 y = - 1 e^{x}

Solve linear ODE:

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

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

y = \frac{1}{3} x^{\frac{3}{2}}

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

\frac{d}{d x} (\frac{x}{9 - 4 x})

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

y^{^{''}} + 9 y = - 24 sin (3 t)

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