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

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

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(dy)/(dx)=2xy^2+3x^2y^2,y(1)=-1

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

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

Solution steps

\frac{d y}{d 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 \frac{4 x + 3}{1 - x ^{2}} d x

\frac{d}{d x} (sin (4 x + 7 x^{4}) + e^{1 - x^{2}})

\int \frac{x ^{2} + 24 x - 9}{x ^{3} - 9 x} d x

x \to \frac{π}{2} lim (- cos (x))

y = \frac{6 x ^{2} + 6 x + 2}{x}

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