Integral equation by successive approximation

[Suvadip]

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

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

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

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

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

Using these I am unable to find a general formula for \(\displaystyle y_n(x)\) 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]

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

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

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

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

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

Using these I am unable to find a general formula for \(\displaystyle y_n(x)\) 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.

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

$\chi$ $\sigma$