What is the solution for 2xy^2+4=2(3-x^2y)y^',y(-1)=8 ?

The solution for 2xy^2+4=2(3-x^2y)y^',y(-1)=8 is y=(3+sqrt(-4x^2(x-3)+9))/(x^2)

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2xy^2+4=2(3-x^2y)y^',y(-1)=8

2 x y^{2} + 4 = 2 (3 - x^{2} y) y^{^{'}}, y (- 1) = 8

y = \frac{3 + - 4 x ^{2} ( x - 3 ) + 9}{x ^{2}}

Solution steps

2 x y^{2} + 4 = 2 (3 - x^{2} y) y^{'} (x)

Solve exact ODE:

y = \frac{3 + - 4 x ^{2} ( x - 3 ) + 9}{x ^{2}}

y = \frac{3 + - 4 x ^{2} ( x - 3 ) + 9}{x ^{2}}

\frac{d}{d x} (x^{3} + 3 x^{2} - 144 x)

(- 1.4) (- 4.2)

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

\frac{d}{d x} (x^{3} - \frac{1}{12 x})

\frac{5}{x ^{3}}, - 1

What is the solution for 2xy^2+4=2(3-x^2y)y^',y(-1)=8 ?
The solution for 2xy^2+4=2(3-x^2y)y^',y(-1)=8 is y=(3+sqrt(-4x^2(x-3)+9))/(x^2)