What is the solution for x^'=(x^2+x)/(2x+1) ?

The solution for x^'=(x^2+x)/(2x+1) is x=(-1+sqrt(1+4e^{t+c_{1)}})/2 ,x=(-1-sqrt(1+4e^{t+c_{1)}})/2

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x^'=(x^2+x)/(2x+1)

x^{^{'}} = \frac{x ^{2} + x}{2 x + 1}

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

Solution steps

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

Solve separable ODE:

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

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

What is the solution for x^'=(x^2+x)/(2x+1) ?
The solution for x^'=(x^2+x)/(2x+1) is x=(-1+sqrt(1+4e^{t+c_{1)}})/2 ,x=(-1-sqrt(1+4e^{t+c_{1)}})/2