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

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

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

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

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

Solution steps

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

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}

\frac{d}{d x} (5 arctan (3 x))

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

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

\frac{\partial}{\partial x} (30 x^{4} y^{5} + 24 x^{5} y^{7})

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