What is the solution for y^'+4x(y^2+2y+1)=0 ?

The solution for y^'+4x(y^2+2y+1)=0 is y=-1/(-2x^2+c_{1)}+(2x^2)/(-2x^2+c_{1)}-(c_{1})/(-2x^2+c_{1)}

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y^'+4x(y^2+2y+1)=0

y^{^{'}} + 4 x (y^{2} + 2 y + 1) = 0

y = - \frac{1}{- 2 x ^{2} + c _{1}} + \frac{2 x ^{2}}{- 2 x ^{2} + c _{1}} - \frac{c _{1}}{- 2 x ^{2} + c _{1}}

Solution steps

y^{'} (x) + 4 x (y^{2} + 2 y + 1) = 0

Solve separable ODE:

y = - \frac{1}{- 2 x ^{2} + c _{1}} + \frac{2 x ^{2}}{- 2 x ^{2} + c _{1}} - \frac{c _{1}}{- 2 x ^{2} + c _{1}}

y = - \frac{1}{- 2 x ^{2} + c _{1}} + \frac{2 x ^{2}}{- 2 x ^{2} + c _{1}} - \frac{c _{1}}{- 2 x ^{2} + c _{1}}

y^{^{'}} = - 3 x^{2} y

\frac{d y}{d x} = y cos (x)

x y^{^{'}} = (1 - x) y

(x^{2} + 1) d x - x y d y = 0

x^{^{'}} = \frac{1 + t}{t ^{2} x ^{2}}

What is the solution for y^'+4x(y^2+2y+1)=0 ?
The solution for y^'+4x(y^2+2y+1)=0 is y=-1/(-2x^2+c_{1)}+(2x^2)/(-2x^2+c_{1)}-(c_{1})/(-2x^2+c_{1)}