What is the solution for x^2y^{''}-4xy^'=x^5 ?

The solution for x^2y^{''}-4xy^'=x^5 is y=c_{1}x^5+c_{2}+1/5 x^5ln(x)-(x^5)/(25)

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x^2y^{''}-4xy^'=x^5

x^{2} y^{^{''}} - 4 x y^{^{'}} = x^{5}

y = c_{1} x^{5} + c_{2} + \frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}

Solution steps

x^{2} y^{''} (x) - 4 x y^{'} (x) = x^{5}

Solve linear ODE:

y = c_{1} x^{5} + c_{2} + \frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}

y = c_{1} x^{5} + c_{2} + \frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}

d er i v a t i v e f (x) = (3 x^{2} - x + 5) (x^{2} + 3 x - 9)

x \to 2 + lim (\frac{- 2}{x ^{2} - 4})

x \to 0 lim (- \frac{3}{x ^{2}})

d er i v a t i v e arctan (sin (4 x))

\frac{\partial}{\partial x} (2 + x ln (x y - 7))

What is the solution for x^2y^{''}-4xy^'=x^5 ?
The solution for x^2y^{''}-4xy^'=x^5 is y=c_{1}x^5+c_{2}+1/5 x^5ln(x)-(x^5)/(25)