Let's take for example the equation: $$y=1+y'^2$$ Now I see that this equation is seperable but I don't understand why using the general method for implicit ODEs can lead to the following strange outcome. First we substitute: $$y'=p(x)\, \,\,\,\,\,\,\, \text{ so} \,\,\,\,\,\,\, y=y(x,p(x))$$ Then we take the full derivative of $y$ to find $p(x)$: $$\frac{dy}{dx}=\frac{\partial y}{\partial x}+\frac{\partial y}{\partial p}\frac{dp}{dx}=p(x)$$ And so we get: $$p(x)=2y'p'\Rightarrow p'=\frac{1}{2}$$ But $p'=y''$ So we have two options:

One is to integrate twice and get an expression for $y$: $$\int p' dx=p=\frac{x}{2}+C_1\\ \int p \,dx=\int y'dx=y=\frac{x^2}{4}+xC_1+C_2 \\ \Rightarrow y(x)=\frac{x^2}{4}+xC_1+C_2$$ Second is to solve for $p(x)$ and plug it in the original ODE: $$\int p' dx=p=\frac{x}{2}+C \\ y=1+p^2=1+(\frac{x}{2}+C)^2$$ Why do the two methods give us different answers and what is the intuitive meaning behind the second constant? I just can't wrap my head around it. Or maybe there is a mistake in the first method that I don't catch. Please explain.

  • 3
    $\begingroup$ I would be very suspicious if any method gave me two free constants for a first order ODE... $\endgroup$
    – rafexiap
    Jul 31 at 10:24
  • $\begingroup$ @A.P. Yes I mean the two different options that give us similar solutions but with different constants. Somehow using the first option loses more information on the function but i can't understand why is that. I'm not interested in the fact that it is seperable but why is there a solution with two constants to a first order ODE. $\endgroup$ Jul 31 at 10:43
  • $\begingroup$ I deleted my comment because I had misunderstood the problem. We have solutions $y_{\pm}=x^2/4\pm Ax+A^2+1$. $\endgroup$
    – A. P.
    Jul 31 at 10:57

1 Answer 1


So basically, for this problem the method boils down to differentiating the ODE once which yields:

$$ y'= 2y'y'' $$ This implies that $$ y'=0 \text{ or } y''=\tfrac{1}{2}\\ y=C\text{ or } y=\tfrac{x^2}{4}+C_1x+C_2 $$ (It's important to note that a solution can also "jump" between different solution branches if you can glue them together in a sufficiently smooth way).

These two equations are easy to solve but by differentiating we introduced an additional degree of freedom. Indeed if we would have started with the ODE: $$ y=1+y'^2+C $$ we would get the exact same solution because the $C$ would just vanish.

Therfore, to get really solutions of tue original ODE and not a more generic class of solutions we need to (at some point) evaluate our original ODE which gives $$ y=1\text{ or } y=1+(\tfrac{x}{2}+C)^2 $$ Again, please note that a specific solution can also consist of multiple sections with different constants and branches.

Does this answer your question sufficiently?


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