What is the solution for (dy)/(dx)=3x^2y^2,y!=0 ?

The solution for (dy)/(dx)=3x^2y^2,y!=0 is y=-1/(x^3+c_{1)}

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(dy)/(dx)=3x^2y^2,y!=0

\frac{d y}{d x} = 3 x^{2} y^{2}, y! = 0

y = - \frac{1}{x ^{3} + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{x ^{3} + c _{1}}

y = - \frac{1}{x ^{3} + c _{1}}

y^{^{'}} + x \cdot y = x \cdot y^{2}

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

f^{^{'}} (x) - 2 f (x) = 0

y (81 - x^{2})^{\frac{1}{2}} d y = (81 + y^{2})^{\frac{1}{2}} d x

\frac{d y}{d x} = \frac{x y}{ln ( y )}, y = e, x = 7

What is the solution for (dy)/(dx)=3x^2y^2,y!=0 ?
The solution for (dy)/(dx)=3x^2y^2,y!=0 is y=-1/(x^3+c_{1)}