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

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

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

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

y = \frac{e ^{c_{1}} ( e ^{\frac{x ^{2}}{2}} + e ^{c_{1}} )}{e ^{x^{2}} - e ^{2 c_{1}}}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{c_{1}} ( e ^{\frac{x ^{2}}{2}} + e ^{c_{1}} )}{e ^{x^{2}} - e ^{2 c_{1}}}

y = \frac{e ^{c_{1}} ( e ^{\frac{x ^{2}}{2}} + e ^{c_{1}} )}{e ^{x^{2}} - e ^{2 c_{1}}}

\frac{d}{d x} (3 cos (\frac{x}{2}))

\frac{d y}{d x}, y = - \frac{x}{y}

\int e^{\frac{- x ^{2}}{2}} d x

\frac{d}{d t} (\frac{2 t}{1 + 4 t ^{2}})

\int_{0}^{1} π (2 x^{2})^{2} d x

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