Question

How do you do the Picard Iterative process?

Answer 1

For example dy/dx=2y y(0)=5

Answer 2

\[y_0=5\]
\[y_1=5+\int\limits_{0}^{t}2\cdot5ds=10t+5\]
\[y_2=5+\int\limits_{0}^{t}2\cdot(10s+5)\,\,ds=10t^2+10t+5\]
\[y_3=5+\int\limits_{0}^{t}2\cdot(10s^2+10s+5)\,\,ds=\frac{20t^3}{3}+10t^2+10t+5\]
\[y_4=5+\int\limits_{0}^{t}2\cdot\left(\frac{20s^3}{3}+10s^2+10s+5\right)\,\,ds=\frac{10t^4}{3}+\frac{20t^3}{3}+10t^2+10t+5\]

\[\cdots\]

Answer 3

replace t with x if you want

the above is an iterative solution to

\[\frac{dy}{dt}=2y\hspace{.2cm};\hspace{.2cm}y(0)=5\]

Answer 4

Is there a form the answer is supposed to be in? Basically, the problem I have is to show that the problem converges to a unique solution...

Answer 5

If you do enough of the iterations you should be able to see the pattern. Then you can check to see if the sum converges.

This one converges to \[y=5e^{2x}\]

Answer 6

What's the pattern for the constant? It's 5,10,20,20,40. The 20 repeating throws it off, doesn't it?

Answer 7

\[5\frac{2^0}{0!}+5\cdot\frac{2^1}{1!}+5\cdot\frac{2^2}{2!}+5\cdot\frac{2^3}{3!}+\cdots=5+10+10+\frac{20}{3}+\cdots\]

Answer 8

Ah, I see. Yeah, I suck at those. Thanks for your help!