What is the solution for x^2y^{''}-4xy^'-36y=0 ?

The solution for x^2y^{''}-4xy^'-36y=0 is y=c_{1}x^9+(c_{2})/(x^4)

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x^2y^{''}-4xy^'-36y=0

x^{2} y^{^{''}} - 4 x y^{^{'}} - 36 y = 0

y = c_{1} x^{9} + \frac{c _{2}}{x ^{4}}

Solution steps

x^{2} y^{''} (x) - 4 x y^{'} (x) - 36 y = 0

Solve linear ODE:

y = c_{1} x^{9} + \frac{c _{2}}{x ^{4}}

y = c_{1} x^{9} + \frac{c _{2}}{x ^{4}}

\frac{\partial}{\partial x} (x y cos (3 x) + e^{x y})

(x + 1) y^{^{'}} - y = x

\int (2 x^{5}) d x

\frac{x d y}{d x} + y = - 8 x y^{2}, y (1) = 5

f (x) = \frac{x}{4} arcsin (x)

What is the solution for x^2y^{''}-4xy^'-36y=0 ?
The solution for x^2y^{''}-4xy^'-36y=0 is y=c_{1}x^9+(c_{2})/(x^4)