Consider the initial value problem $$x'=4t+tx^2,\quad x(0)=0$$ Find an interval $I_0$ starting from $D=[-1,1]\times [-2,2]$ so that the Picard-Lindelöf Theorem holds.
I know I need to show $f(t,x)=4t+tx^2$ is Lipschitz in $x$ with $L$ the lipschitz constant can be chosen independently from $t$, and continuous in $t$.
First I solved the differential equation $$\frac{dx}{dt}\frac{1}{t}=4+tx^2,~ \text{for $t\neq0$}$$ $$\int t dt=\int \frac{1}{x^2+4}dx \iff \frac{t^2}{2}+C=\frac{1}{2}\arctan{\left(\frac{x}{2}\right)}$$ Which then gives $x(t)=2\tan(t^2+C)$ after some algebra and taking $\tan$ of both sides.
$f$ is continious in $t$ since it is a polynomial in $t$?, not sure how to show Lipschitz continuity
For $(t,x),(t,y) \in D$ we have
$|f(t,x)-f(t,y)|= |t||x^2-y^2|=|t||x+y||x-y| \le $
$|t|(|x|+|y|)|x-y| \le 1(2+2)|x-y|=4|x-y|$.