Picards Method
Example:
\[\frac{dy}{dx}=f(x,t), \; y(x_0)=y_0\]
. The iterative solution is found by applying \[y_{n+1}=y_0+ \int^x_{x_0} f(x, y_n )dx\]
repeatedly.Example
\[\frac{dy}{dx}=y+x, \; y(0)=1\]
.\[y_1=1+\int^x_0 (1+x+x)dx=1+ x+ \frac{2x^2}{2}=1+x+x^2\]
\[y_2=1+\int^x_0 (1+ x+x^2+x)dx=1+ x+ x^2+ \frac{x^3}{3}\]
\[y_3=1+\int^x_0 (1+ x+ x^2+ \frac{x^3}{3}+x)dx=1+ x+ x^2+ \frac{x^3}{3}+ \frac{x^4}{12}\]
By adding
\[1+x\]
to both sides of the above equations we get\[y_1+1+x=2(1+x+\frac{x^2}{2})\]
\[y_2+1+x=2(1+ x+ \frac{x^2}{2}+ \frac{x^3}{6}\]
\[y_3+1+x=2(1+ x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24}\]
Continuing in this way we get
\[y_n+1+x=2 \sum^n_0 \frac{x^n}{n!}\]
and letting \[n \rightarrow \infty\]
gives \[y+1+x= 2 \sum^{\infty}_0 \frac{x^n}{n!}=e^x \rightarrow y=2e^x-1-x\]
. 
