What is the solution for y^'=3x^2y^2,y(-1)=a,y(0)=1 ?

The solution for y^'=3x^2y^2,y(-1)=a,y(0)=1 is y=-a/(ax^3-1+a)

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y^'=3x^2y^2,y(-1)=a,y(0)=1

y^{^{'}} = 3 x^{2} y^{2}, y (- 1) = a, y (0) = 1

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

Solution steps

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

Solve separable ODE:

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

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

3 d x + y d y = 0

\frac{d y}{d x} = \frac{x y - 4 x + y - 4}{x y - 6 x - 3 y + 18}

\frac{d y}{d x} = x^{2} (y - 2), y (0) = 4

y^{^{'}} = e^{- y}, y (0) = 0

\frac{d y}{d x} = \frac{2}{3} x 1 - 9 y^{2}, y (0) = 0

What is the solution for y^'=3x^2y^2,y(-1)=a,y(0)=1 ?
The solution for y^'=3x^2y^2,y(-1)=a,y(0)=1 is y=-a/(ax^3-1+a)