What is the solution for y^'=32x^3y^3 ?

The solution for y^'=32x^3y^3 is y=sqrt(-1/(2(8x^4+c_{1)))},y=-sqrt(-1/(2(8x^4+c_{1)))}

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y^'=32x^3y^3

y^{^{'}} = 32 x^{3} y^{3}

y = - \frac{1}{2 ( 8 x ^{4} + c _{1} )}, y = - - \frac{1}{2 ( 8 x ^{4} + c _{1} )}

Solution steps

y^{'} (x) = 32 x^{3} y^{3}

Solve separable ODE:

y = - \frac{1}{2 ( 8 x ^{4} + c _{1} )}, y = - - \frac{1}{2 ( 8 x ^{4} + c _{1} )}

y = - \frac{1}{2 ( 8 x ^{4} + c _{1} )}, y = - - \frac{1}{2 ( 8 x ^{4} + c _{1} )}

\int_{- 2}^{2} x^{2} e^{x^{3} + 8} d x

5 ln (x)

x^{2} lo g_{3} (x^{3})

\int (8 x - \frac{7}{x}) d x

y = cos (x), at x = \frac{3 π}{2}

What is the solution for y^'=32x^3y^3 ?
The solution for y^'=32x^3y^3 is y=sqrt(-1/(2(8x^4+c_{1)))},y=-sqrt(-1/(2(8x^4+c_{1)))}