2

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 $f(x)=x+\frac{61}{119}x+\frac{80}{119}x^2.$

Brahmagupta
  • 4,204
  • 2
    You're on the right track. Note that the 2 integrals you're left with, on $y$, are just constants. Call them $A$ and $B$. Then, $f(x)$ is a quadratic polynomial in $x$. Inserting that polynomial back into the integrals you're left with (but now as a function of $y$) will let you determine the values of $A$ and $B$. BTW, I wasn't the one who down-voted your question. – wltrup Aug 21 '15 at 14:43

1 Answers1

6

$$ f(x) = x + \int_{0}^{1}[xy^2+x^2y]\,f(y)\,dy = x + x \int_{0}^{1}y^2 f(y)\,dy + x^2 \int_{0}^{1}y\,f(y)\,dy = x + Ax + Bx^2 $$ 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, $$ f(x) = x + Ax + Bx^2 = x + \frac{61}{119}\,x + \frac{80}{119}\,x^2 $$

wltrup
  • 3,983