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

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

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x^2y^'+x(x+2)y=e^x

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

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

Solution steps

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

Solve linear ODE:

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

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

x = 2 y, x = y^{2} - 8

x^{a} (1 - x)^{b}

t \to \infty lim (ln (t))

x \to 0 lim (\frac{5}{1 + x})

n = 0 \sum \infty (0.5 + 0.5 i)^{n}

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