Question

If f(x) $=x+{\int }_0^1[x{y}^2+s^{2}y]f\left(y\right)$ dy where x and y are independent varibles. Find f(x)

• A. $f\left(x\right)=x+\frac{61}{119}x+\frac{80}{119}{x}^2$
• B. $f\left(x\right)=\frac{80x^2+180x}{119}$
• C. $f\left(x\right)=\frac{80x+180x}{119}$
• D. $f\left(x\right)=\frac{80x^2+180x}{19}$

Answer 1

Answer: (B)

$f\left(x\right)=\frac{80x^2+180x}{119}$

$f\left(x\right)=x+x{\int }_1^0{y}^{2}f\left(y\right)dy+x^2{\int }_0^{1}yf\left(y\right)dy$
$=x(1+{\int }_0^1{y}^{2}f\left(y\right)dy)+x^2\left({\int }_0^{1}yf\left(y\right)dy\right)$$\Rightarrow f\left(x\right)$ is a quadratic expression of the form $(ax+b{x}^2)$
where $a=1+{\int }_0^1{y}^{2}f\left(y\right)dy$
$a=1+{\int }_0^1{y}^2(ay+b{y}^2)dy$
$a=1+\frac{a}{4}+\frac{b}{5}$
$\Rightarrow 15a-4b=20$....(i)
and $b={\int }_0^{1}yf\left(y\right)dy={\int }_0^{1}y(ay+b{y}^2)dy$
$b=\frac{a}{3}+\frac{b}{4}\Rightarrow 9b-4a=0$....(ii)
from (i) and (ii)
$a=\frac{180}{119},b=\frac{80}{119}$ so
$f\left(x\right)=\frac{80x^2+180x}{119}$