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

The solution for (2y-x^2y)y^'=x(y+2)^2 is y=-2-2/(\W(-(2sqrt(x^2-2))/(c_{1)))}

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(2y-x^2y)y^'=x(y+2)^2

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

y = - 2 - \frac{2}{W ( - \frac{2 x ^{2} - 2}{c _{1}} )}

Solution steps

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

Solve separable ODE:

y = - 2 - \frac{2}{W ( - \frac{2 x ^{2} - 2}{c _{1}} )}

y = - 2 - \frac{2}{W ( - \frac{2 x ^{2} - 2}{c _{1}} )}

y = \frac{sin ( 2 x ) + 3 cos ( x )}{x + 1}

\int csch (x) d x

3 x + \frac{x ^{\frac{1}{3}}}{3}

F (X) = X - ln (1 + X)

\int \frac{66 x ^{2}}{x ^{4} - 65 x ^{2} + 784} d x

What is the solution for (2y-x^2y)y^'=x(y+2)^2 ?
The solution for (2y-x^2y)y^'=x(y+2)^2 is y=-2-2/(\W(-(2sqrt(x^2-2))/(c_{1)))}