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

The solution for y^'=2xy^2+3x^2y^2,y(1)=-1 is y=-1/(x^3+x^2-1)

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y^'=2xy^2+3x^2y^2,y(1)=-1

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

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

Solution steps

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

Solve separable ODE:

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

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

\int 2 a x d x

x \to - 2 lim (\frac{x + 2}{4 - x ^{2}})

\frac{\partial}{\partial x} (\frac{( 3 x - y )}{( x + 2 y )})

\int \frac{7 x ^{2} + 2 x - 7}{x ^{3} - x} d x

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

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