Prove that the solution of y'=x^3-y^3 with an arbitrary…

Prove that the solution of $y'=x^3-y^3$ with an arbitrary initial condition $y(x_0)=y_0$ exists for all $x_0\le x<\infty$.

Step 1: Local existence

$f(x,y)=x^3-y^3$ is a polynomial, hence $C^1$ (in particular locally Lipschitz in $y$) on all of $\mathbb{R}^2$. By the Picard–Lindel\\\"of theorem, the IVP has a unique solution on some maximal interval $[x_0,X)$, and if $X<\infty$ then necessarily $|y(x)|\to\infty$ as $x\to X^-$ (the standard blow-up alternative for ODEs with globally defined, locally Lipschitz right-hand side).

Step 2: Fix any finite target interval $[x_0,X_0]$ and choose a bound

Let $X_0>x_0$ be arbitrary but finite, and choose $M=\max(|x_0|,|y_0|,|X_0|)+1$, so in particular $M>X_0\ge x$ for every $x\in[x_0,X_0]$, and $M>|y_0|$.

Step 3: Show $y(x)$ can never reach $M$ from below

Suppose, for contradiction, that $y(x)$ first reaches the level $M$ at some $x^\ast\in(x_0,X_0]$ (possible since $y(x_0)

Step 4: Symmetric bound from below

An identical argument with $-M$ shows $y(x)>-M$ throughout $[x_0,X_0]$: if $y$ first touched $-M$ at some $\bar x\le X_0$, then $y'(\bar x)=\bar x^3-(-M)^3=\bar x^3+M^3>0$ (since $\bar x^3>-M^3$, as $|\bar x|

Step 5: Conclude global existence

So $|y(x)|X_0$. Since $X_0$ was an arbitrary finite target, $X=\infty$: the solution exists for all $x\ge x_0$.