What is the solution for 15x^2y+3x^3y^'-24y=0,y(1)=1 ?

The solution for 15x^2y+3x^3y^'-24y=0,y(1)=1 is y=(e^{\frac{-4+4x^2)/(x^2)}}{x^5}

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15x^2y+3x^3y^'-24y=0,y(1)=1

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

y = \frac{e ^{\frac{- 4 + 4 x ^{2}}{x ^{2}}}}{x ^{5}}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{\frac{- 4 + 4 x ^{2}}{x ^{2}}}}{x ^{5}}

y = \frac{e ^{\frac{- 4 + 4 x ^{2}}{x ^{2}}}}{x ^{5}}

\frac{d y}{d x} y sin (x^{2}) = x sin (y^{2})

y^{^{'}} = 2 y + 4

y^{^{'}} = \frac{1}{2} y^{3}, y (1) = 4

\frac{d y}{d x} = \frac{3 x ^{2}}{1 - 2 y}

\frac{d y}{d x} = \frac{x ^{3} + 1}{y}

What is the solution for 15x^2y+3x^3y^'-24y=0,y(1)=1 ?
The solution for 15x^2y+3x^3y^'-24y=0,y(1)=1 is y=(e^{\frac{-4+4x^2)/(x^2)}}{x^5}