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

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

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xy^'+2y=-e^x

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

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

Solution steps

x y^{'} (x) + 2 y = - 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^{^{'}} = - 0.5 y

f (x) = x^{2} (- 3 x^{2} - 2)

\int \frac{1}{x ( x + 3 )} d x

8 e^{- x}

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

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