What is the solution for y^'=x^2y^3 ?

The solution for y^'=x^2y^3 is y=sqrt(-3/(2(x^3+c_{1)))},y=-sqrt(-3/(2(x^3+c_{1)))}

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y^'=x^2y^3

y^{^{'}} = x^{2} y^{3}

y = - \frac{3}{2 ( x ^{3} + c _{1} )}, y = - - \frac{3}{2 ( x ^{3} + c _{1} )}

Solution steps

y^{'} (x) = x^{2} y^{3}

Solve separable ODE:

y = - \frac{3}{2 ( x ^{3} + c _{1} )}, y = - - \frac{3}{2 ( x ^{3} + c _{1} )}

y = - \frac{3}{2 ( x ^{3} + c _{1} )}, y = - - \frac{3}{2 ( x ^{3} + c _{1} )}

y = x + 4, y = \frac{( x )}{( 2 ) + 2}

f (x) = (- 16 x^{2} + 1)^{2}

\int 4 x^{2} + 9 d x

\int 2 (1 - \frac{1}{x ^{2}}) d x

What is the solution for y^'=x^2y^3 ?
The solution for y^'=x^2y^3 is y=sqrt(-3/(2(x^3+c_{1)))},y=-sqrt(-3/(2(x^3+c_{1)))}