<s>[INST] If $f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$ where $x$ and$y$ are independent variable.Find $f(x).$ If $f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$ where $x$ and$y$ are independent variable.Find $f(x).$
I tried to solve it.
$f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$
$f(x)=x+x\int_{0}^{1}y^2f(y)dy+x^2\int_{0}^{1}yf(y)dy$
I applied integration by parts but the expression did not simplify.Please guide me to reach the answer $f(x)=x+\frac{61}{119}x+\frac{80}{119}x^2.$
 [/INST] $$
f(x)
= x + \int_{0}^{1}[xy^2+x^2y]\,f(y)\,dy
= x + x \int_{0}^{1}y^2 f(y)\,dy + x^2 \int_{0}^{1}y\,f(y)\,dy
= x + Ax + Bx^2
$$
where
$$
A = \int_{0}^{1}y^2 f(y)\,dy
\qquad\mbox{and}\qquad
B = \int_{0}^{1}y\,f(y)\,dy
$$
So,
$$
A = \int_{0}^{1}y^2 \underbrace{\left(y + Ay + By^2\right)}_{f(y)}\,\,dy
= \frac{A+1}{4} + \frac{B}{5}
\qquad\mbox{and}\qquad
B = \int_{0}^{1}y\, \underbrace{\left(y + Ay + By^2\right)}_{f(y)}\,\,dy
= \frac{A+1}{3} + \frac{B}{4}
$$
which can now be solved for $A$ and $B$, resulting in
$$
A = \frac{61}{119}
\qquad\mbox{and}\qquad
B = \frac{80}{119}
$$
Therefore,
$$
f(x) = x + Ax + Bx^2 = x + \frac{61}{119}\,x + \frac{80}{119}\,x^2
$$
</s>