What is the solution for (dy)/(dx)=(2y(2x-1)e^x)/(2x+1) ?

The solution for (dy)/(dx)=(2y(2x-1)e^x)/(2x+1) is y=e^{integral of (2e^x(2x-1))/(2x+1)}

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(dy)/(dx)=(2y(2x-1)e^x)/(2x+1)

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

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

Solution steps

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

Solve separable ODE:

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

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

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

y^{^{'}} = \frac{( e ^{- x} - e ^{x} )}{( 3 + 4 y )}

y^{^{'}} = y (1 - y^{2})

9 t y + e^{t} y^{^{'}} = \frac{y}{t ^{2} + 81}

x^{^{'}} - e^{t} x = 5 e^{t}

What is the solution for (dy)/(dx)=(2y(2x-1)e^x)/(2x+1) ?
The solution for (dy)/(dx)=(2y(2x-1)e^x)/(2x+1) is y=e^{integral of (2e^x(2x-1))/(2x+1)}