MHB Integral equation by successive approximation 2

[Suvadip]

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

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

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

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

$$y_1(x)=-2+cos(x)-x, y_2(x)=-\frac{1}{2}x^2-2x-2+cos(x)+sin(x)$$

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]

Re: integral equation by successive approximation

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

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

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

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

$$y_1(x)=-2+cos(x)-x, y_2(x)=-\frac{1}{2}x^2-2x-2+cos(x)+sin(x)$$

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.

Take into account that the integral equation has the 'exact' solution...

$\displaystyle y(x)= \frac{\sin x + \cos x - 3\ e^{x}}{2} = \frac{\sin x + \cos x}{2} - \frac{3}{2} - \frac{3}{2}\ x - \frac{3}{4}\ x^{2} - \frac{1}{4}\ x^{3} -... - \frac{3}{2\ n!}\ x^{n} - ...\ (1)$

... and at each iteration You obtain a better approximation to (1)...

Kind regards

$\chi$ $\sigma$