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

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

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

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

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

Solution steps

y^{'} (x) = 2 \frac{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} y = x^{4 x}

\frac{d y}{d x} y = cosh^{2} (3 x) - sinh (x^{2})

\frac{d y}{d x} = \frac{3 + 12 x}{x y ^{2}}, x > 0

y^{^{'}} - \frac{( t + 1 ) y}{t ^{2}} = \frac{1}{t} + \frac{1}{t ^{2}}

y^{^{'}} = - (- \frac{x ^{2}}{3 y})

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