Integral equation by successive approximation 2

[Suvadip]

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

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

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

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

Using this relation we have

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

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

How should I proceed.

 

[chisigma]

Re: integral equation by successive approximation

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

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

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

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

Using this relation we have

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

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

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

$\chi$ $\sigma$