What is the solution for x^2y^{''}-2xy^'-40y=x^4 ?

The solution for x^2y^{''}-2xy^'-40y=x^4 is y=c_{1}x^8+(c_{2})/(x^5)-(x^4)/(36)

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x^2y^{''}-2xy^'-40y=x^4

x^{2} y^{^{''}} - 2 x y^{^{'}} - 40 y = x^{4}

y = c_{1} x^{8} + \frac{c _{2}}{x ^{5}} - \frac{x ^{4}}{36}

Solution steps

x^{2} y^{''} (x) - 2 x y^{'} (x) - 40 y = x^{4}

Solve linear ODE:

y = c_{1} x^{8} + \frac{c _{2}}{x ^{5}} - \frac{x ^{4}}{36}

y = c_{1} x^{8} + \frac{c _{2}}{x ^{5}} - \frac{x ^{4}}{36}

\int \frac{1}{t ^{2} 225 - t ^{2}} d t

x \to 1 + lim (- 6)

(7 x + 8)^{2}

\frac{\partial}{\partial x} (\frac{- y + y ^{2} + 4 x ^{2}}{2})

\frac{d}{d x} (6 - x)

What is the solution for x^2y^{''}-2xy^'-40y=x^4 ?
The solution for x^2y^{''}-2xy^'-40y=x^4 is y=c_{1}x^8+(c_{2})/(x^5)-(x^4)/(36)