Fredholm integral - degenerate kernel method I have started answering a fredholm integral equation of the second kind and do not know where to go from here.
The answer has to be written in the form 
$$ \sum a_jx^{j-1} $$
The Fredholm integral equation is
$$ x^3+\frac16x^2+\frac15x = g(x) + \int_0^1(x^2y+xy^2) f(y) dy$$.
My method so far: 
Let: $$C_1 = \int_0^1yf(y)dy$$ and $$C_2 = \int_0^1y^2f(y)dy$$
Then 
$$ x^3+\frac16x^2+\frac15x = (C_1x^2 +C_2x) + g(x)$$.
Eliminating f(y) to get 
$$C_1 = (\frac14C_1 + \frac13C_2) + \int_0^1yg(y)dy$$
and 
$$C_2 = (\frac15C_1 + \frac14C_2) + \int_0^1y^2g(y)dy$$
I don't know where to go from here to get it into the form 
$$ \sum a_jx^{j-1} $$
Do I put it into matrix form, and solve simultaneously, (not sure how to do this) 
If I have gotten anything wrong here please let me know. 
Or if you need any more information I may be able to provide. (Like how I got to a specific equation)
Any help will be appreciated 
Thank you very much 
 A: $$g(x) = x^3 - \frac{38}{1077} x^2 + \frac{58}{1795} x$$
By rearranging the original equation we can see that g(x) = x^3 + Ax^2 + Bx. This is because both of the integrals on the right evaluate to a constant multiple of x^2 or x respectively. By using this formula for g(x) in each of the integrals we get:
The integral from 0 to 1 of 
$$x^2y(y^3+Ay^2+By) dy = x^2[\frac{y^5}{5} + \frac{Ay^4}{4} + \frac{By^3}{3}]  = x^2(\frac{1}{5} + \frac{A}{4} + \frac{B}{3})$$
and the integral from 0 to 1 of 
$$xy^2(y^3+Ay^2+By) dy = x[y^6/6 + \frac{Ay^5}{5} + \frac{By^4}{4}] = x(\frac{1}{6} + \frac{A}{5} + \frac{B}{4})$$
So the equality becomes:
$$x^3 + \frac{1}{6} x^2 + \frac{1}{5} x = (x^3 + Ax^2 + Bx) + x^2(\frac{1}{5} + \frac{A}{4} + \frac{B}{3}) + x(\frac{1}{6} + \frac{A}{5} + \frac{B}{4})$$
Rearranging gives:
$$x^2(\frac{-1}{30} - \frac{5A}{4} - \frac{B}{3}) + x(\frac{1}{30} - \frac{A}{5}- \frac{5B}{4}) = 0$$
So, by comparing coefficients;
$$\frac{-1}{30} - \frac{5A}{4} - \frac{B}{3} = 0$$ and
$$\frac{1}{30} - \frac{A}{5} - \frac{5B}{4} = 0$$
Solving simultaneously gives
$$A = - 38/1077 $$ and
$$B = \frac{58}{1795} $$
$$ g(x)= x^3-\frac{38}{1077} x^2 + \frac{58}{1795}$$
