What is the solution for xy^{''}+2y^'=8x^2 ?

The solution for xy^{''}+2y^'=8x^2 is y=c_{1}+(c_{2})/x+(2x^3)/3

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xy^{''}+2y^'=8x^2

x y^{^{''}} + 2 y^{^{'}} = 8 x^{2}

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

Solution steps

x y^{''} (x) + 2 y^{'} (x) = 8 x^{2}

Solve linear ODE:

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

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

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

x \to - 1 - lim (- 4 x - 4 - 8)

y^{^{'}} + y = \frac{1 - x}{2 x ^{2}}

y^{6} + x^{3} = y^{2} + 12 x, (0, 1)

What is the solution for xy^{''}+2y^'=8x^2 ?
The solution for xy^{''}+2y^'=8x^2 is y=c_{1}+(c_{2})/x+(2x^3)/3