Solving the second order differential equation $y''=y'(2y+1)$

Question

$$y'' = y'(2y+1)$$ with $y(0)=1$ and $y'(0)=2$.

Could you please help me with the steps for answering this question.

The answer is $$y(x)=-\frac{1}{2} \tanh \left( \frac{x}{2}-\arctan (3) \right) -\frac{1}{2}$$

Answer 1

Hint: Put $u = y’ \implies \dfrac{u’}{u} = 2y + 1\implies \ln(u) = y^2 + y + C$....

Answer 2

By inspection, $2y\cdot y' = (y^2)'$

So the entire equation can be integrated once to give $y' = y^2 + y + c$, which can be solved using methods you should already be familiar with.

Answer 3

Hint

At first, by integrating you will obtain$$y'=y^2+y+C$$so after applying the initial conditions you have$$y'=y^2+y$$