Integral equation by successive approximation

[Suvadip]

I have to solve the integral equation $$y(x)=1+2\int_0^x(t+y(t))dt$$ by the method of successive approximation taking $$y_0(x)=1$$.

Sol: After simplification the given equation we have
$$y(x)=1+x^2+2\int_0^xy(t)dt$$. So comparing it with $$y(x)=f(x)+\lambda\int_0^x k(x,t)y(t))dt$$ we have

$$f(x)=1+x^2, \lambda=2, k(x,t)=1$$

Now let us use the relation
$$y_n(x)=f(x)+\lambda\int_0^x k(x,t)y_{n-1}(t))dt$$

Using this relation we have

$$y_1(x)=1+2x+x^2, y_2(x)=\frac{2}{3}x^3+3x^2+2x+1, y_3(x)=\frac{1}{3}x^4+2x^3+3x^2+x+1$$

Using these I am unable to find a general formula for $$y_n(x)$$ from which the required solution $$y(x)$$ can be found using $$y(x)=lim_{n \to \infty} y_n(x)$$

How should I proceed.

 

[chisigma]

I have to solve the integral equation $$y(x)=1+2\int_0^x(t+y(t))dt$$ by the method of successive approximation taking $$y_0(x)=1$$.

Sol: After simplification the given equation we have
$$y(x)=1+x^2+2\int_0^xy(t)dt$$. So comparing it with $$y(x)=f(x)+\lambda\int_0^x k(x,t)y(t))dt$$ we have

$$f(x)=1+x^2, \lambda=2, k(x,t)=1$$

Now let us use the relation
$$y_n(x)=f(x)+\lambda\int_0^x k(x,t)y_{n-1}(t))dt$$

Using this relation we have

$$y_1(x)=1+2x+x^2, y_2(x)=\frac{2}{3}x^3+3x^2+2x+1, y_3(x)=\frac{1}{3}x^4+2x^3+3x^2+x+1$$

Using these I am unable to find a general formula for $$y_n(x)$$ from which the required solution $$y(x)$$ can be found using $$y(x)=lim_{n \to \infty} y_n(x)$$

How should I proceed.

Your application of the successive approximation method is correct and the result is obvious if You take into account the the 'exact' solution of the integral equation is...

$\displaystyle y(x) = \frac{3\ e^{2\ x} - 2\ x - 1}{2} = 1 + x + 3\ x^{2} + 2\ x^{3} + ... + \frac{3}{2}\ \frac{(2\ x)^{n}}{n!} + ...\ (1)$

Kind regards

$\chi$ $\sigma$