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$$ f(x) + \int_0^1 (xy+x^2y^2) f(y) dy = x^3 +\frac16x^2+ \frac15x $$

I have this fredholm integral equation of the second kind and am not sure how to answer this equation. I know that is has to be written in the form $$ \sum a_jx^{j-1} $$

I do not know how to integrate with the function $f(y)$ in the integral.

Any help would be appreciated.

Thanks

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can you start with: $$f(x)+\int_0^1(xy+x^2y^2)f(y)dy=x^3+\frac16x^2+\frac15x$$ $$f'(x)+(x+x^2)f(1)=3x^2+\frac13x+\frac15$$ $$f'(x)=(3-f(1))x^2+\left(\frac13-f(1)\right)+\frac15$$ so: $$f(x)=\int\left[(3-f(1))x^2+\left(\frac13-f(1)\right)+\frac15\right]dx+C$$

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  • $\begingroup$ Once integrating that final integral would it be able to write it as a sum of a function. Also, how would I integrate f(1) $\endgroup$ – p s Dec 2 '18 at 21:00
  • $\begingroup$ f(1) is just a constant, but we would need a way to work out its value, which is an issue as the function is defined with the function itself $\endgroup$ – Henry Lee Dec 2 '18 at 21:02
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Note that we can write the equation as $$ f(x) = x^3 + \left[\frac{1}{6} - \int \limits_0^1 y^2 f(y) \, \mathrm{d} y\right] x^2 + \left[\frac{1}{5} - \int \limits_0^1 y f(y) \, \mathrm{d} y\right] x \, , \, x \in [0,1] \, . $$ Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $a, b \in \mathbb{R}$ such that $$ f(x) = x^3 + a x^2 + b x \, , \, x \in [0,1] \, . $$ We can then plug this form of $f$ into the previous equation and compute the integrals to obtain $$ x^3 + a x^2 + b x = x^3 - \left(\frac{a}{5}+\frac{b}{4}\right) x^2 - \left(\frac{a}{4}+\frac{b}{3}\right)x \, , \, x \in [0,1] \, . $$ This yields the linear system $$ \begin{pmatrix} \frac{6}{5} && \frac{1}{4}\\ \frac{1}{4} && \frac{4}{3} \end{pmatrix} \begin{pmatrix} a \\b \end{pmatrix} = \begin{pmatrix} 0 \\0 \end{pmatrix} \, ,$$ which only has the trivial solution $a=b=0$ . Therefore $$ f(x) = x^3 \, , \, x \in [0,1] \, , $$ is the unique solution to the integral equation.

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  • $\begingroup$ Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$ \sum a_jx^{j-1} $$ how would I change what you have solved above into this format? $\endgroup$ – p s Dec 2 '18 at 23:22
  • $\begingroup$ @ps You can write $f(x) = x^3 = \sum_{j=2}^4 a_j x^{j-1} $ with $a_4 = 1$ and $a_3 = a_2 = 0$ . $\endgroup$ – ComplexYetTrivial Dec 2 '18 at 23:32
  • $\begingroup$ how did you get to $$ x^3 - \left(\frac{a}{5}+\frac{b}{4}\right) x^2 - \left(\frac{a}{4}+\frac{b}{3}\right)x \, , \, x \in [0,1] \, . $$ $\endgroup$ – p s Dec 9 '18 at 19:00

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