Proof that the initial value problem $$y'(x)=arccot(y(x))+x^2y(x)+y^2(x)+1\quad y(0)=0$$ has got an unique solution $y:[0,\frac{1}{10}]\to\Bbb R$
Now this looks like picard-lindelöf, but I struggle finding the Lipschitz-constant. My attempt:
Let $$f(x,y) =arccot(y(x))+x^2y(x)+y^2(x)+1$$ than for $y\neq z:$ $$\begin{align}|f(x,y)-f(x,z)|&=|arccot(y)+x^2y+y^2+1-arccot(z)-x^2z-z^2-1 |\\ &=| arccot(y)-arccot(z)+x^2(y-z)+y^2-z^2|\\&\le(|\frac{arccot(y)-arccot(z)}{y-z}|+x^2+|y+z|)|y-z||\\ &\le \bigl(\underbrace{\frac{1}{y^2+1}}_{\le sup\{\partial f/\partial y\}:(x,y)\in\Bbb R}+x^2+|y+z|\bigr)|y-z|\end{align}$$
I got confused , did I anything wrong ? How should I define my constant ?
There is nothing wrong. From your calculations, we can deduce that
$$ |f(x,y)-f(x,z)|\le\bigl(1+\frac{1}{100}+2\,M\bigr)|y-z|,\quad0\le x\le\frac{1}{10},\quad |y|,|z|\le M. $$ This gives local existence and uniqueness in an interval $[0,a]$, with $a$ depending on $M$. You will have to show that $M$ can be chosen so that the interval of existence guaranteed by the Picard-Lindelöf theorem is at least $[0,1/10]$.