What is the solution for xy^{''}+2y^'=20x^2 ?

The solution for xy^{''}+2y^'=20x^2 is y=c_{1}+(c_{2})/x+(5x^3)/3

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xy^{''}+2y^'=20x^2

x y^{^{''}} + 2 y^{^{'}} = 20 x^{2}

y = c_{1} + \frac{c _{2}}{x} + \frac{5 x ^{3}}{3}

Solution steps

x y^{''} (x) + 2 y^{'} (x) = 20 x^{2}

Solve linear ODE:

y = c_{1} + \frac{c _{2}}{x} + \frac{5 x ^{3}}{3}

y = c_{1} + \frac{c _{2}}{x} + \frac{5 x ^{3}}{3}

\int (2 - x)^{3} d x

lo g_{10} (x)

\frac{\partial}{\partial y} (y + z)

\frac{d}{d x} ((sin (x))^{3 x})

x \to 0 lim (\frac{sin ( 8 x )}{8})

What is the solution for xy^{''}+2y^'=20x^2 ?
The solution for xy^{''}+2y^'=20x^2 is y=c_{1}+(c_{2})/x+(5x^3)/3