What is the solution for x^3y^'-(8x^2-5)y=0 ?

The solution for x^3y^'-(8x^2-5)y=0 is y=e^{5/(2x^2)+c_{1}}x^8

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x^3y^'-(8x^2-5)y=0

x^{3} y^{^{'}} - (8 x^{2} - 5) y = 0

y = e^{\frac{5}{2 x ^{2}} + c_{1}} x^{8}

Solution steps

x^{3} y^{'} (x) - (8 x^{2} - 5) y = 0

Solve separable ODE:

y = e^{\frac{5}{2 x ^{2}} + c_{1}} x^{8}

y = e^{\frac{5}{2 x ^{2}} + c_{1}} x^{8}

(x^{2} + 3) \frac{d y}{d x} = x e^{y}

\frac{d y}{d x} = 2 x y + 5 y, y (0) = 3

x ln (x) d y - y d x = 0

\frac{d y}{d x} = e^{x - y}, y (0) = ln (4)

\frac{d y}{d x} = \frac{tan ( y )}{x}

What is the solution for x^3y^'-(8x^2-5)y=0 ?
The solution for x^3y^'-(8x^2-5)y=0 is y=e^{5/(2x^2)+c_{1}}x^8