$(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2)$ and $f(1)=2$

Question

Let $f$ be a differentiable function such that $(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2)$ and $f(1)=2$. Then area enclosed by $$\frac{|f(x)-x|^{1/3}}{17}+\frac{|f(y)-y|^{1/3}}{2}\le\frac14$$

I rearranged the equation to $$\frac{f(x+y)}{x+y}-\frac{f(x-y)}{x-y}=4xy$$ Involving some intution I guessed $$\frac{f(x)}x=x^2\text{ since} (x+y)^2-(x-y)^2=4xy$$ But the resulatant $f(x)=x^3$ does not satisfy $f(1)=2$ so seeing what to find, I again guessed $f(x)=x^3+x$ which "eureka!" is what we want. Now the rest part is easy, now what would the proper method to arrive at this function probably using "Let $f$ be a differentiable function..."

Answer 1

Rewrite the equation as:

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

Define $g(z)=\frac{f(z)}{z}-z^2$. Then $g$ is defined and differenetiable on $\mathbb R \backslash\{0\}$.

The Equation yields $$g(x+y)=g(x-y) \, (*)$$ Now pick $s,t \in \mathbb R \backslash\{0\}$ and set $$ x=\frac{s+t}{2}\\ y=\frac{-t}{2}$$

Then the equation gives $$g(s)=g(t)$$

This shows that $g$ is constant. Let $C$ denote the constant, then we get $$f(x)=x^3+cx \,,$$ for all $x \in \mathbb R \backslash\{0\}$. By differentiability this also holds for $x=0$.

Now $f(1)=2$ gives the constant $c$.