# How to solve Fredholm integral equation of the second kind? ($f(x) - \lambda\int\limits_0^1 9xtf(t) dt = ax^2 - 4x^2$)

I have an equation : $\ f(x) - \lambda\int\limits_0^1 9xtf(t) dt = ax^2 - 4x^2\$ in $\ L_2[0,1]\$ space.

And I want to understand how to solve it, not just obtain an answer.

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For this particular equation, it is simple. Differentiate both sides of the equation with respect to $x$: $$f^\prime(x) = 2 (a-4) x + \lambda \int_0^1 9 t f(t) \mathrm{d} t$$ Since the integral is independent of $x$, the solution is a quadratic $f(x) = (a-4) x^2 + b x + c$. Substituting into the original equation we get a linear system of equations for $b$ and $c$: $$(a-4) x^2 + b x + c - \lambda x \int_0^1 9 t f(t) \mathrm{d} t = (a-4) x^2$$ Since $\int_0^1 9 f(t) \mathrm{d} t = \frac{9}{4} a+ 3b+ \frac{9}{2} c - 9$, we get the system: $$c = 0 \qquad b (1-3 \lambda) - \frac{9}{4} \lambda \left( 2 c + a -4\right) = 0$$ Therefore we obtain: $$f(x) = (a-4) x \left( x -\frac{9}{4} \frac{\lambda}{3 \lambda - 1} \right)$$
How to solve this if $\lambda = \frac{1}{3}$ –  Philipp G. Sinicyn Nov 28 '11 at 10:38
@PhilippG.Sinicyn Come back to equations for polynomial coefficients. They would say $c=0$, $a=4$ and would leave $b$ free, which is to say, equation will have no solutions if $a\not=4$, and if $a=4$, then $f(x) = b*x$ would be a solution –  Sasha Nov 28 '11 at 14:05