What is the solution for x^2y^{''}+3xy^'-24y=0,y(1)=3,y^'(1)=-3 ?

The solution for x^2y^{''}+3xy^'-24y=0,y(1)=3,y^'(1)=-3 is y=(3-3/2)x^4+3/(2x^6)

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x^2y^{''}+3xy^'-24y=0,y(1)=3,y^'(1)=-3

x^{2} y^{^{''}} + 3 x y^{^{'}} - 24 y = 0, y (1) = 3, y^{^{'}} (1) = - 3

y = (3 - \frac{3}{2}) x^{4} + \frac{3}{2 x ^{6}}

Solution steps

x^{2} y^{''} (x) + 3 x y^{'} (x) - 24 y = 0

Solve linear ODE:

y = (3 - \frac{3}{2}) x^{4} + \frac{3}{2 x ^{6}}

y = (3 - \frac{3}{2}) x^{4} + \frac{3}{2 x ^{6}}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x} + \frac{1}{y} + x y)

\int \frac{1}{1 + sin ( θ )} d θ

\int 2 x^{2} - x^{3} d x

\frac{\partial}{\partial x} (\frac{x}{2 x ^{2} + 2 y ^{2} + 8})

x \to 0 lim (\frac{1}{2 ^{\frac{1}{x}} - 1})

What is the solution for x^2y^{''}+3xy^'-24y=0,y(1)=3,y^'(1)=-3 ?
The solution for x^2y^{''}+3xy^'-24y=0,y(1)=3,y^'(1)=-3 is y=(3-3/2)x^4+3/(2x^6)