What is the solution for x^2y^{''}+2xy^'-12y=0 ?

The solution for x^2y^{''}+2xy^'-12y=0 is y=c_{1}x^3+(c_{2})/(x^4)

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x^2y^{''}+2xy^'-12y=0

x^{2} y^{^{''}} + 2 x y^{^{'}} - 12 y = 0

y = c_{1} x^{3} + \frac{c _{2}}{x ^{4}}

Solution steps

x^{2} y^{''} (x) + 2 x y^{'} (x) - 12 y = 0

Solve linear ODE:

y = c_{1} x^{3} + \frac{c _{2}}{x ^{4}}

y = c_{1} x^{3} + \frac{c _{2}}{x ^{4}}

\frac{\partial}{\partial x} (42 x^{5} y^{6} - 30 x^{4} y^{5})

x^{2} y^{^{''}} - 5 x y^{^{'}} - 7 y = 0

d er i v a t i v e f (x) = e^{- 6 x}

\int e^{1.2 x} d x

d er i v a t i v e 4 - x + 16

What is the solution for x^2y^{''}+2xy^'-12y=0 ?
The solution for x^2y^{''}+2xy^'-12y=0 is y=c_{1}x^3+(c_{2})/(x^4)