MHB Integral equation by successive approximation

[Suvadip]

if

, then what will be

. In fact I was solving the integral equation

by the method of successive approximation.

 

[Siron]

I was not familiar with the name of the method until I noticed it's the same as Picard Iterations (after some googling). I tried some ways to see a pattern in the polynomials but I failed. Especially the leading terms are very unpredictable in my opinion. Perhaps it's not possible to find an explicit form for $y_n(x)$.

What's the background of this problem?

 

[chisigma]

if

, then what will be

. In fact I was solving the integral equation

by the method of successive approximation.

Under appropriate conditions the solution of the differential equation ...

$\displaystyle y^{\ '} = f(x,y),\ y(0)= y_{0}\ (1)$

... must satisfy the following integral equation...

$\displaystyle y = y_{0} + \int_{x_{0}}^{x} f\{t, y(t)\}\ d t\ (2)$

If You define...

$\displaystyle y_{1} = y_{0} + \int_{x_{0}}^{x} f\{t, y_{0}\}\ d t\ (3)$

... and...

$\displaystyle y_{n} = y_{0} + \int_{x_{0}}^{x} f\{t, y_{n-1}\}\ d t\ (4)$

... then the sequence of $y_{n}$ converges to the solution $y(x)$...

In Your case is $y_{0}=1$, $x_{0}=0$ and $y_{1} = (1 + x)^{2}$, so that is $f(x,1) = 2\ (1+x)$...

Kind regards

$\chi$ $\sigma$