Recently I have been studying Fredholm theory to solve integral equations, and as I am trying to familiarize myself with the theory I have been trying to solve certain problems. This one is one of them:
Find the solution of the following equation: $$\ f(x) - \lambda\int\limits_{0}^{1} 5x^{2}t^{2}f(t) dt = 4x +bx^{2}\ \in \ C_1[0,1].\ $$ Regarding this problem, there are a lot of similar problems in this platform, as the following one: 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 tried to reproduce the solution of this problem adapting it to mine, but I got stuck with this part in specific:
Since $\int_0^1 9 f(t) \mathrm{d} t = \frac{9}{4} a+ 3b+ \frac{9}{2} c - 9$, we get the system: $$ c = 0 \qquad b (1-3 \lambda) - \frac{9}{4} \lambda \left( 2 c + a -4\right) = 0 $$
My first question is, where and how did he got this equality $\int_0^1 9 f(t) \mathrm{d} t = \frac{9}{4} a+ 3b+ \frac{9}{2} c - 9$ and the subsequent system of equations, maybe I am being dumb, but I don't really understand where is he getting this result from.
My second question, is wether I was doing a good job on solving the original problem I had.
My attempt:
First, we have to differentiate the equation till the integral is not dependent on $x$:
$$f'(x)= 4 + 2bx + \lambda\int^{1}_{0} 10xt^{2} f(t) dt$$ then, $$f''(x)= 2b + \lambda 10 \int^{1}_{0} t^{2} f(t) dt.$$ So, now that the integral is not dependent of x, we can see that $$f(x)=bx^{2}+\frac{1}{2}k_{1}x^{2}+k_{2}x.$$ Now we substitute this on the original equation and we have $$\ bx^{2}+\frac{1}{2}k_{1}x^{2}+k_{2}x - \lambda\int\limits_{0}^{1} 5x^{2}t^{2}f(t) dt = 4x +bx^{2}$$
And this is the part where I am stuck, any help would be really appreciated.
For the purpose of completion I post the answer to my own question:
First, we have to differentiate the equation till the integral is not dependent on $x$:
$$f'(x)= 4 + 2bx + \lambda\int^{1}_{0} 10xt^{2} f(t) dt$$ then, $$f''(x)= 2b + \lambda 10 \int^{1}_{0} t^{2} f(t) dt.$$ So, now that the integral is not dependent of x, we can see that $$f(x)=bx^{2}+\frac{1}{2}k_{1}x^{2}+k_{2}x.$$ Now we substitute this into the original equation and we have $$\ bx^{2}+\frac{1}{2}k_{1}x^{2}+k_{2}x - \lambda5x^{2}\int\limits_{0}^{1} t^{2}f(t) dt = 4x +bx^{2}.\tag{1}$$ On the other hand we note that the following also is true: $$\int\limits_{0}^{1} 5t^{2}f(t) dt = \int\limits_{0}^{1}(bt^{4}+\frac{1}{2}k_{1}t^{4}+k_{2}t^{3})= b+\frac{1}{2}k_{1}+\frac{5}{4}k_{2}.\tag{2}$$ Substituting (2) in (1) we get: $$ bx^{2}+\frac{1}{2}k_{1}x^{2}+k_{2}x = 4x +bx^{2} +\lambda b x^{2}- \frac{\lambda}{2}k_{1}x^{2}-\frac{5}{4}k_{2}x^{2}.$$ So, $$ \frac{1}{2}x^{2}(k_{1}-2\lambda b-\lambda k_{1}-\frac{5\lambda}{2}k_{2})+x(k_{2}-4)=0.$$ So we get the following system of equations: $$k_{2}=4,\qquad k_{1}-2\lambda b -\lambda k_{1} -\frac{5}{2}k_{2}=0.$$ So, from this we obtain that the solution is: $$f(x)= bx^{2} + x^{2}\lambda(\frac{b+5}{1-\lambda})+4x$$