Suvadip
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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.
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.