Functional Equation : If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x).

Question

Problem :

If $(x-y)f(x+y) -(x+y)f(x-y) =4xy(x^2-y^2)$ for all x,y find f(x).

My approach :

The given equation can be written as $$(x-y)f(x+y) -(x+y)f(x-y) =4xy(x-y)(x+y)$$

$$\Rightarrow \frac{(x-y)f(x+y)}{(x-y)(x+y)} -\frac{(x+y)f(x-y)}{(x-y)(x+y)} =4xy$$

$$\Rightarrow \frac{f(x+y)}{x+y} -\frac{f(x-y)}{x-y} =4xy$$

Now we know that $$(x+y)^2 -(x-y)^2 = 4xy$$

$\Rightarrow \frac{f(x+y)}{x+y} =(x+y)^2$....(i)

& $\frac{f(x-y)}{x-y} = (x-y)^2$...(ii)

Now putting y =0 in (i) and (ii) we get :

$\frac{f(x)}{x} =x^2$ $\Rightarrow f(x) =x^3$

But the answer is $f(x) =x^3 +kx$ ( where k is any constant ) please clarify this part thanks...

Answer 1

$\Rightarrow \frac{f(x+y)}{x+y} -\frac{f(x-y)}{x-y} =4xy=(x+y)^2-(x-y)^2$

or, $\frac{f(x+y)}{x+y}-(x+y)^2=\frac{f(x-y)}{x-y}-(x-y)^2$

make the substitutions, $x\to \frac{x+y}{2}$ and $y \to \frac{x-y}{2}$

Then the above becomes $\frac{f(x)}{x}-(x)^2=\frac{f(y)}{y}-(y)^2=k(say)$

There you have $f(x)=x^3+kx$

Answer 2

The additional term comes from the relation $(x-y)k(x+y)-(x+y)k(x-y)=0$ for any $k$