Solution to given Integral equation

Question

I want to determine the solution $f(x)$ of the integral equation \begin{equation} f(x)=\int_{0}^{1}(x-y) f(y) d y+x^{2} . \end{equation} I got $$ f(x)=x^2+ \int_{0}^{1}xf(y) d y-\int_{0}^{1}f(y)y d y $$ so $$ f(x)=x^2+ax+b $$ with $$ \begin{aligned} &a=\int_{0}^{1}f(y) d y\\ &b=-\int_{0}^{1}-yf(y) d y. \end{aligned} $$ Next, we plug in $x^2+ax+b$ in the equation and get $$ \begin{aligned} &a=\int_{0}^{1}(y^2+ay+b) d y\\ &b=\int_{0}^{1}y(y^2+ay+b) d y. \end{aligned} $$ which equals $$ \begin{aligned} &a=\dfrac{6b+3a+2}{6}\\ &b=\dfrac{6b+4a+3}{12} \end{aligned} $$ and this gives $$ a= \frac{3}{13} $$ and $$ b=- \frac{17}{78} $$ Many thanks to your help.

Answer 1

You cannot plug in $y$ for $x$ since $y$ is a variable of integration.

Hint for solution: Separating the two terms on the right side we see that $f(x)$ has the form $x^{2}+ax+b$. [ $a=\int_0^{1} f(y)dy, b=\int_0^{1} yf(y)dy$].

Now plug in $f(x)=x^{2}+ax+b$ into the given equation and compare the coefficients of $x$ as well as the constant terms. You get two equations for $a$ and $b$. Solve these two equations.