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

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

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y^'=2xy(1+4y)

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

y = \frac{e ^{x^{2} + c_{1}}}{1 - 4 e ^{x^{2} + c_{1}}}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{x^{2} + c_{1}}}{1 - 4 e ^{x^{2} + c_{1}}}

y = \frac{e ^{x^{2} + c_{1}}}{1 - 4 e ^{x^{2} + c_{1}}}

\frac{d}{d θ} (r (θ)) = \frac{r ( θ ) ^{2}}{θ ^{2}}

\frac{d y}{d x} = \frac{2 x + 5}{x ( 1 + 2 y )}

a y^{^{'}} + b y = k

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

y^{^{'}} - y = 2, t e^{2 t} y (0) = 1

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