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$f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y) dy$. Find $f(x)$

I've tried taking $\int_{0}^{1} (xy^2 + x^2y) f(y) dy$ to be $k(x)$ since it comes out to be a function of $x$. That transforms our equation to $f(x) = x + k(x)$.

$f(y) = y + k(y) \implies (xy^2 + x^2y)f(y) = (xy^2 + x^2y)y + (xy^2 + x^2y)k(y)$ Now I tried integrating on both sides against $dy$ from $0$ to $1$ in an attempt to find $k(x)$

$\int_{0}^{1} (xy^2 + x^2y)f(y).dy = k(x) = \int_{0}^{1} ((xy^2 + x^2y)y + (xy^2 + x^2y)k(y)).dy$

But I got stuck trying to integrate the right hand side. Any solutions or ideas are appreciated.

Cyanide2002
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3 Answers3

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Write you equation under the form :

$$f(x) = x + x \underbrace{\int_{0}^{1} y^2 f(y) dy}_b+ x^2 \underbrace{\int_{0}^{1}y f(y) dy}_a\tag{1}$$

which mean that

$$f(x)=ax^2+(b+1)x \tag{2}$$

with

$$a=\int_{0}^{1}y f(y) dy \ \ (i) \ \ \ \ \text{and} \ \ \ \ b=\int_{0}^{1}y^2 f(y) dy \ \ (ii)\tag{3}$$

is (at most) a second degree polynomial in variable $x$.

It suffices now to plug (2) into (3)(i) and (3)(ii) to get 2 equations in the 2 unknowns $a$ and $b$.

Up to you for the final step.

Jean Marie
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$$f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y) dy$$ $$f'(x) = 1 + \int_{0}^{1} (y^2 + 2xy) f(y) dy$$ $$f''(x) = \int_{0}^{1} 2y f(y) dy=\text{constant}$$ $$f(x)=ax^2+bx+c$$ $$ax^2+bx+c=x+ \int_{0}^{1} (xy^2 + x^2y) (ay^2+by+c) dy$$ After calculus of the integral and simplification : $$(-\frac13 a+\frac13 b+\frac12 c)x^2+(1+\frac15 a-\frac34 b+\frac13 c)x-c=0$$ $$\begin{cases} -\frac13 a+\frac13 b+\frac12 c=0 \\ 1+\frac15 a-\frac34 b+\frac13 c=0 \\ c=0 \end{cases} \quad\implies\quad \begin{cases} a=\frac{80}{119} \\ b=\frac{180}{119} \end{cases}$$ $$\boxed{f(x)=\frac{80}{119}x^2+\frac{180}{119}x}$$

JJacquelin
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Let $A=\int_0^{1} y^{2}f(y)dy$ and $B=\int_0^{1} yf(y)dy$. Then $f(x)=x+xA+x^{2}B$. Multiply by $x$ and integrate to get $B=\frac 1 3 +\frac A 3 +\frac 1 4 B$. Similarly multiply by $x^{2}$ and integrate to get $A=\frac 1 4 +\frac A 4 +\frac 1 5 B$. Solve these two equations for $A$ and $B$.

Kavi Rama Murthy
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