On $I:=(0,\infty)$ let's consider the following Cauchy's problem: $$y'=\frac{y}{2x^2}+1\qquad y(1)=0$$
a) Taking $\varphi_0:=0$ construct the first two Picard's approximations $\varphi_1,\;\varphi_2$
b) Let $\varphi$ be the exact solution to this problem. Using $\varphi_1,\;\varphi_2$ give an approximate value for $\varphi(\sqrt{2})$ and estimate the error of approximation.
So using the recursive Picard's method: $$\varphi_{n+1}(x):= z + \int_{x_0}^x f\left(s,\varphi_{n}(s)\right)\;ds \qquad\text{where}\;\varphi_0(x)=z$$
In our case $f(x,y)=\frac{y}{2x^2}+1$ and if I didn't make any silly mistakes: $$\varphi_1(x) = x-1$$ $$\varphi_2(x) = \frac{1}{2}\left(\log(x)+\frac{1}{x}+x-3\right)$$
I can approximate $\varphi(\sqrt{2})$ by computing $\varphi_2(\sqrt{2})$ but how can I evaluate the error ?
I found somewhere in my lecture that if we define the following application on a closed ball around $z$: $$(T\phi)(x)=z+\int_{x_0}^x f(s,\phi(s))\;ds$$ We can show (under certain condition but it applies to my case) that $T$ is $L$-Lipschitz and so we have the following Lemma: $$\forall p\geq 1:\quad\|(T^p\phi)(x)-(T^p\psi)(x)\|\leq\frac{1}{p!}L^p\cdot|x-x_0|^p\cdot\|\phi-\psi\|\qquad\text{Using}\;\|\cdot\|_{\infty}$$
But then I have to main issues. Firstly my upper limit of $L$ is: $L\leq\frac{b}{8}$ where $0<b\in\mathbb{R}$.
Then even if I assume I didn't make errors (which is clearly not clear) and continue on this road I arrive at the expression: $$\|\varphi_2-\varphi\|_{\infty}\leq \frac{L^2}{2}\alpha^2\|\varphi\|\qquad(\alpha=\frac{1}{2})$$
So I cannot evaluate the error without knowing the maximum of the exact solution $\varphi$, so it is useless. I also tried to apply the fixed point theorem which state that: $$\|x_\infty-x_n\|\leq\dfrac{c^n}{1-c}\|x_1-x_0\|\qquad\text{where $c$ is the contraction factor}$$ But totaly failed to find a way to make it work. I Googled ways to approximate the error on Picard's iteration but mostly found Master's level papers that are way above my level. Which path should I take to find this error ?