Rewrite $y(x) = \int_0^x xy(\tau) d\tau + 2x + 1$ as an ODE in the form of $u'(t) = F(t, u(t))$ with the condition $u(t_0) = u_0$

Question

I was told to differentiate it twice by $x$ (why not once?), so that's what I did here:

$$y(x) = \int_0^x xy(\tau) d\tau + 2x + 1$$

$$y'(x) = xy(x) - xy(0) + 2$$

$$y'' (x) = y(x) + xy'(x) - xy(0)$$

If I do $u = (u_1, u_2) = (y, y')$, and write $t$ instead of $x$, then I'll get

$$u' = (y', y'') = (u_2, u_1 (t) + tu_2(t) - tu_1 (0)) = F(t, u(x))$$

But I was told that I should get a system of ODEs and thus some matrix equation. But I don't see how we should get it here?

It is important because we also would like to prove that there exists exactly one solution in the neighborhood of $t_0 = 0$. For this, we need to prove that the matrix we get in our matrix equation fulfills the Lipschitz condition (by introducing an arbitrary matrix norm to prove it).

So what am I doing wrong?

Answer 1

If $Y(x)=\int_0^xy(τ)\,dτ$ is a primitive, then the equation can be written as $$ y(x)=xY(x)+2x+1. $$ The derivatives of that are $$ y'(x)=xy(x)+Y(x)+2,\\ y''(x)=xy'(x)+2y(x) $$ You could also eliminate $Y(x)$ in the first equation by the original equation, but then $x=0$ would be a singular point of the resulting DE.