Math Book Cambodia: Find all function `f(x)` if `(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2)`

Thursday, December 22, 2022

Solution

Let's $u = x + y$ $u=x+y$ and $v = x - y$ $v=x-y$

Then, $x = \frac{u + v}{2}$ $x=(u+v)/2$ and $y = \frac{u - v}{2}$ $y=(u-v)/2$

From the equation : $f (x)$ $f(x)$ if $(x - y) f (x + y) - (x + y) f (x - y) = 4 x y (x^{2} - y^{2})$ $(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2)$

We will get: $v f (u) - u f (v) = (u^{2} - v^{2}) u v$ $vf(u)-uf(v)=(u^2-v^2)uv$

\frac{f (u)}{u} - u^{2} = \frac{f (v)}{v} - v^{2}

$f(u)/u-u^2=f(v)/v-v^2$ for all

u, v

$u,v$ different from

0

$0$

Let: $v = 1$ $v=1$ $\to \frac{f (u)}{u} - u^{2} = f (1) - 1$ $rightarrowf(u)/u-u^2=f(1)-1$

Therefore: $f (u) = u^{3} + a u$ $f(u)=u^3+au$ for all $u \neq 0$ $u!=0$ and $(a = f (1) - a)$ $(a=f(1)-a)$

If $x = y = 0$ $x=y=0$ $\to 2 f (0) = 0$ $rightarrow2f(0)=0$ Then $f (0) = 0$ $f(0)=0$

Hence, $f (x) = x^{3} + a x$ $f(x)=x^3+ax$ .

Math Book Cambodia