What is the solution for y^'=(2x)/(y+x^2y),y(0)=-2 ?

The solution for y^'=(2x)/(y+x^2y),y(0)=-2 is y=-sqrt(2ln(1+x^2)+4)

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y^'=(2x)/(y+x^2y),y(0)=-2

y^{^{'}} = \frac{2 x}{y + x ^{2} y}, y (0) = - 2

y = - 2 ln (1 + x^{2}) + 4

Solution steps

y^{'} (x) = \frac{2 x}{y + x ^{2} y}

Solve separable ODE:

y = - 2 ln (1 + x^{2}) + 4

y = - 2 ln (1 + x^{2}) + 4

\frac{d y}{d x} = 55 y x^{10}

\frac{d v}{d t} = 9 v, v (0) = 11.3

y^{^{'}} = cot (y) sin (x)

(x y^{3} - x) d x + x y^{2} d y = 0

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

What is the solution for y^'=(2x)/(y+x^2y),y(0)=-2 ?
The solution for y^'=(2x)/(y+x^2y),y(0)=-2 is y=-sqrt(2ln(1+x^2)+4)