$$y(x) = \int_0^x (x - \tau)y(\tau) d\tau + x^3$$
$$y'(x) = x y(x) + 3x^2$$
$$y''(x) = y(x) + xy'(x) + 6x = y(x) + x^2 y(x) + 3x^2 + 6x$$
Let
$$u = (u_1, u_2) = (y, y')$$
$$u' = (u_1', u_2') = (y', y'')$$
So we get:
$$\begin{pmatrix}u_1'\\ u_2'\end{pmatrix}=\begin{pmatrix}0&1\\ -1&x\:+\frac{2}{x}\end{pmatrix}\begin{pmatrix}u_1\\ u_2\end{pmatrix}$$
But apparently, this cannot be the right solution, because our matrix $\begin{pmatrix}0&1\\ -1&x\:+\frac{2}{x}\end{pmatrix}$ is not Lipschitz for $x=0$ (which is the other part of our exercise to prove that it's Lipschitz for $x=0$). The reason it's not Lipschitz is because the matrix norm would be equal to infinity
So where is my mistake?
$$y(x) = \int_0^x (x - \tau)y(\tau) d\tau + x^3$$ Apply Leibniz integral rule: $$y'(x) = \int_0^x y(\tau) d\tau + 3x^2$$ $$y''(x) = y(x) + 6x$$