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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

n = 0 \sum \infty n! (x - 2)^{n}

\frac{x}{x + 7}

\int sin^{2} (\frac{nπ x}{a}) d x

\int ln (x^{9}) d x

\frac{d}{d x} (\frac{8}{3})

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