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

The solution for x^'=(x^2+x)/(2x+1),x(0)=1 is x=(-1+sqrt(1+8e^t))/2

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x^'=(x^2+x)/(2x+1),x(0)=1

x^{^{'}} = \frac{x ^{2} + x}{2 x + 1}, x (0) = 1

x = \frac{- 1 + 1 + 8 e ^{t}}{2}

Solution steps

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

Solve separable ODE:

x = \frac{- 1 + 1 + 8 e ^{t}}{2}

x = \frac{- 1 + 1 + 8 e ^{t}}{2}

What is the solution for x^'=(x^2+x)/(2x+1),x(0)=1 ?
The solution for x^'=(x^2+x)/(2x+1),x(0)=1 is x=(-1+sqrt(1+8e^t))/2