# Mathematical induction for Picard iteration

For the initial value problem, $$y'=2x(1+y),y(0)=2$$, the Picard iteration is defined by $$y_{n+1}=2+\int_0^x2x(1+y_n(x))\ dx \quad (1)$$

I claimed that $$y_n=-1+3\sum_{k=0}^n\frac{x^{2k}}{k!}$$ Now I need to use mathematical induction to prove this is true. Skip the base case and for induction hypothesis, we have $$y_n=-1+3\sum^{n}_{k=0}\frac{x^{2k}}{k!}=2+\int_0^x2x(1+y_{n-1})\ dx$$ For $$n+1$$, then \begin{align*} y_{n+1}&=-1+3\sum^{n}_{k=0}\frac{x^{2k}}{k!}+\frac{x^{2(n+1)}}{(n+1)!} \\ &=\int_0^x2x(1+y_{n-1})\ dx + \frac{x^{2(n+1)}}{(n+1)!}\quad \text{(induction hypothesis)} \end{align*} I stuck at this point, such as, how to 'combine' these two terms, to make it become $$\int_0^x2x(1+y_n)\ dx$$? Alternatively, I was simply using (1), but I also not feel it will achieve mathematical induction. Any helps will be great. Thanks in advance.

Use the induction hypothesis for $$y_n$$ inside the integral in the definition of $$y_{n+1}$$, a change of index in the sum and finally a "convenient $$0$$":
\begin{align*} y_{n+1} &= 2+\int_0^x 2s(1+y_n(s))\,\mathrm ds, \\ &= 2+\int_0^x 2s\left(1-1+3\sum_{k=0}^n \dfrac{s^{2k}}{k!}\right)\,\mathrm ds , \\ &= 2+\int_0^x 6\sum_{k=0}^n \dfrac{s^{2k+1}}{k!}\,\mathrm ds , \\ &= 2+ 6\sum_{k=0}^n\int_0^x \dfrac{s^{2k+1}}{k!}\,\mathrm ds , \\ &= 2+ 6\sum_{k=0}^n \dfrac{s^{2k+2}}{k!(2k+2)}, \\ &= 2+ 3\sum_{k=0}^n \dfrac{s^{2(k+1)}}{(k+1)!}, \\ &= 2+ 3\sum_{k=1}^{n+1} \dfrac{s^{2k}}{k!}, \\ &= 2\color{red}{-3+3} +3\sum_{k=1}^{n+1} \dfrac{s^{2k}}{k!}= -1 + 3\sum_{k=0}^{n+1} \dfrac{x^{2k}}{k!}. \end{align*} Good job on your claim!