What is the solution for xy^{''}+2y^'=4x^2 ?

The solution for xy^{''}+2y^'=4x^2 is y=c_{1}+(c_{2})/x+(x^3)/3

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xy^{''}+2y^'=4x^2

x y^{^{''}} + 2 y^{^{'}} = 4 x^{2}

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

Solution steps

x y^{''} (x) + 2 y^{'} (x) = 4 x^{2}

Solve linear ODE:

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

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

\frac{d}{d x} (lo g_{4} (x^{2} + 3))

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

x^{3} - 3 x^{2} + 2 x - 2, at x = 2

\frac{d}{d x} (2 - e^{x})

\frac{d y}{d x} - \frac{2}{x} y = x

What is the solution for xy^{''}+2y^'=4x^2 ?
The solution for xy^{''}+2y^'=4x^2 is y=c_{1}+(c_{2})/x+(x^3)/3