I try to find a general solution for the equation below $$y'' = (x + y')^2$$
This way I wanted to transform it to a first order ode $$y' = p$$ $$y'' = p'$$
This gives me the equation below which looks like a ricardi equation.
$$p' = p^2 + 2xp + x^2$$
The problem I am having here is that there is no solution given for riccardi. I am stuck at this step. How can I advance further? Or am I solving this wrong? Thanks
$$y''=(x+y')^2$$ Change of function : $y'=u(x)-x$ $$u'-1=u^2$$ This is a separable ODE. $$\frac{du}{u^2+1}=dx$$ $$u=\tan(x+c_1)$$ $y'=\tan(x+c_1)-x$
$y=\int \left(\tan(x+c_1)-x\right)dx$ $$y=-\ln|\cos(x+c_1)|-\frac12 x^2+c_2$$