$\mathcal C^1$ solution of $y'=y+x^2$ are in fact $\mathcal C^\infty $?

Question

Let $u$ a solution of $$y'=y+x^2.$$ Such solution are in fact $\mathcal C^\infty $, no ? Because since $u'=u+x^2$ and $u$ is $\mathcal C^1$, we get that $u'$ is $\mathcal C^1$ as well, and thus $u$ is $\mathcal C^2$. But then, we get $$u''=(u')'=(u+x^2)'=u'+2x$$ and since $u'$ is $\mathcal C^1$ then $u''$ is $\mathcal C^1$ as well and thus $u\in \mathcal C^3$, and we can continue like that... so such solution must be $\mathcal C^\infty $, right ? If not, why my argument is wrong ?

Answer 1

Your reasoning is correct. In many cases the solution will even be analytic.

You should have taken a general case, $y'(x) = f(x, y),$ to have people look at your reasoning instead of just presenting the solution of the given equation.