When solving first order higher degree ODE for example : $$y=-xy'+y'^2+\frac{x^2}{2}$$ I solved it for $~y~$ as explained in this link :http://www.solitaryroad.com/c652.html
My problem is NOT how to solve the DE , I know how to solve it and I show below how I solved it. ( My question is shown after the solution , my question is can we solve it in an alternative way ? the alternative way is shown after the solution )
The right solution ( as in the link) :
I called $~y'~$ as $~p~$ and differentiated the DE w.r.t $~x~$ which gives me :
$$0=(-2p+x)(1-\frac{dp}{dx})$$ so the general solution is obtained from solving the DE obtained from the second bracket : $$p=x+c$$ substitute in the DE by $~p~$ to get the general solution which is : $$y=-x(x+c)+(x+c)^2+\frac{x^2}{2}$$ The first bracket gives the singular solution : $$p=\frac{x}{2}$$ substituting by $~p~$ in the DE gives the singular solution which is : $$y=\frac{x^2}{4}$$
The alternative solution : why in the above solution , we did not use the fact that $$ p = \frac{dy}{dx}$$ for example in the singular solution after obtaining $$p=\frac{x}{2}$$ why we did not complete as following : $$\frac{dy}{dx}= \frac{x}{2}$$ $$y=\frac{x^2}{4}+k$$ The same for the second bracket containing the differentiation of $~p~$ , why we did not complete as following : $$\frac{dy}{dx}=x+c$$ $$y=\frac{x^2}{2}+cx+c_1$$
$$y=-xy'+y'^2+\frac{x^2}{2}$$ It is false to say that $\quad y=\frac{x^2}{2}+cx\quad$ is solution of the ODE.
Putting $y=\frac{x^2}{2}+cx$ into the ODE leads to :
$\frac{x^2}{2}+cx=-x(x+c)+(x+c)^2+\frac{x^2}{2} \quad\to\quad c^2=0$
Thus $y=\frac{x^2}{2}+cx$ is not solution of the ODE, except the particular case $c=0$.
The correct solving is :
$$p'=\frac{dy}{dx}=x+c \quad\to\quad y=\frac{x^2}{2}+cx+k$$
But, at first begining, the differentiation of the first order ODE transforms it into a second order ODE. In the set of solutions of the second order ODE, of course, the set of solutions of the first order ODE is included, but there are more solutions which are not solutions of the first order ODE. These inadequate solutions must be eliminated.
That is why we always have to bring back the raw result into the initial ODE and check what are the convenient solutions :
$$\frac{x^2}{2}+cx+k=-x(x+c)+(x+c)^2+\frac{x^2}{2} \quad\to\quad k=c^2$$ The correct solution is : $$y=\frac{x^2}{2}+cx+c^2$$
Do the same with $\quad p=\frac{dy}{dx}=\frac{x}{2} \quad\to\quad y=\frac{x^2}{4}+k\quad$ which is correct for the second order ODE, but not for the first order ODE. Putting it back into the first order ODE : $$\frac{x^2}{4}+k=-x(\frac{x}{2} )+(\frac{x}{2})^2+\frac{x^2}{2} \quad\to\quad k=0$$ So, the correct particular solution is $\quad y=\frac{x^2}{4}$