I assume, then, that you are talking about a Fredholm integral equation:
y(x)= \inf_a^b K(x,t)y(t)dt+ f(t)[/itex]<br />
<br />
Yes, if K(x,t) is "separable", K(x,t)= A(x)B(t), then the solution is easy. The equation becomes<br />
y(x)= A(x)\int_a^b B(t)y(t)dt+ f(x)[/itex]<br />
and <br />
\int_a^b B(t)y(t)dt<br />
is a number. Let <br />
X= \int_a^b B(t)y(t)dt<br />
and you have y(x)= X A(x)+ f(x) so it is just a matter of solving for X. Multiply both sides of that equation by B(x) and integrate from a to b.<br />
\int_a^b B(x)y(x)dx= X\int_a^b A(x)B(x)dx+ \int_a^b B(x)F(x)dx<br />
The integral on the left is just X again and the other two integrals are of known functions. Let <br />
K= \int_a^b A(x)B(x)dx<br />
and <br />
F= \int_a^b B(x)f(x)dx<br />
and X satisfies the equation X= KX+ F which is easily solvable for X.<br />
<br />
Now, suppose K(x,t) is not separable but is the sum of two separable functions: K(x,t)= A1(x)B1(t)+ A2(x)B2(t). we can still take the &quot;x&quot; dependence out of the equations:<br />
y(x)= A1(x)\int_a^b B1(t)y(t)dt+ A2(x)\int_a^b B2(t)y(t)dt+ f(x)<br />
Let<br />
X1= \int_a^b B1(t)y(t)dt<br />
and<br />
X2= \int_a^b B2(t)y(t)dt<br />
Now,<br />
y(t)= X1A1(x)+ X2A2(x)+ f(x)[/itex]&lt;br /&gt;
and we only need to find the numbers X1 and X2.&lt;br /&gt;
If we multiply that equation by B1(x) and integrate:&lt;br /&gt;
\int_a^b B1(x)y(x)dx= X1\int_a^b A1(x)B1(x)dx+ X2\int_a^b A2(x)B1(x)dx+ \int_a^b B1(x)f(x)dx[/itex]&amp;lt;br /&amp;gt;
Again, the left side is just X1 and the integrals on the right are numbers:&amp;lt;br /&amp;gt;
X1= K_{11}X1+ K_{12}X2+ F1&amp;lt;br /&amp;gt;
with the obvious notation. Doing the same with B2(x), &amp;lt;br /&amp;gt;
X2= K_{12}X1+ K_{22}X2+ F2&amp;lt;br /&amp;gt;
&amp;lt;br /&amp;gt;
That is, we have two linear equations to solve for X1 and X2. (And, in fact, the system is symmetric, guarenteeing real solutions.)&amp;lt;br /&amp;gt;
&amp;lt;br /&amp;gt;
The extension to any kernel which is a finite sum of separable functions should be obvious and, in fact, it can be extended to an infinite sum of such functions, giving an infinite series solution.
