I have the following ODE:

$x=t(1+x')+(x')^2;t,x\in \mathbb{R}$

I made the substitution $p=\frac{dx}{dt}$

The equation after differentiating with respect to $t$ is $1+(t+2p)\frac{dp}{dt}=0$

This new equation I solved using the integrating factor method. In this case, I computed the integrating factor to be $e^p$ After multiplying by the integrating factor we get $$e^pdt+(te^p+2pe^p)dp=0$$

This equation has an exact solution. Let that solution be some function $F(t,p)=c$ $$F(t,p)=\int_{t_0}^{t}e^pds + \int_{p_0}^{p}(t_0e^y+2ye^y)dy$$

After integrating and picking some convenient values for $p_0,t_0$, namely $0,0$ we get $$F(t,p)=te^p+2e^pp-2e^p+c=0$$

If I make the substitution to find x I must solve $$e^{x'}(t+2x'-2)+c=0$$

This according to Wolfram alpha has separable variables and is solvable, but I don't have a pro account and I can't see a step by step solution. Can anyone help me? Also I would like to know if my method of solving this was ok. Maybe there was an easier way of getting the answer?


$$1+(t+2p)\frac{dp}{dt}=0$$ $$\frac{dt}{dp}+2p+t=0$$ This is a linear ODE which solution is : $$t=c_1e^{-p}-2p+2$$ $$p=W\left(\frac{c_1}{2}e^{\frac{t}{2}-1} \right)-\frac{t}{2}+1$$ $W$ is the LambertW function.

$\frac{dx}{dt}=p \quad\to\quad x=\int \left(W\left(\frac{c_1}{2}e^{\frac{t}{2}-1} \right)-\frac{t}{2}+1 \right)dt$

$$x(t)=\left(W\left(\frac{c_1}{2}e^{\frac{t}{2}-1}\right)+1\right)^2 -t^2+t+c_2$$

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