Question

If a differentiable function satisfies \[ (x-y)f(x+y) - (x+y)f(x-y) = 2(x^2y-y^3), \qquad \forall x,y\in\mathbb{R} \] and \(f(1)=2\), then:

• A. \(f(x)\) must be a polynomial function
• B. \(f(3)=13\)
• C. \(f(3)=12\)
• D. \(f(0)=0\)

Hint:

For functional equations containing both \(f(x+y)\) and \(f(x-y)\), the substitution \[ u=x+y, \qquad v=x-y \] usually converts the equation into a separable algebraic form.

Answer 1

The Correct Option is A

Concept: For functional equations involving expressions such as \(f(x+y)\) and \(f(x-y)\), the substitution \[ u=x+y, \qquad v=x-y \] greatly simplifies the equation. Also, \[ x=\frac{u+v}{2}, \qquad y=\frac{u-v}{2} \] Step 1: Transform the given equation.
Given: \[ (x-y)f(x+y)-(x+y)f(x-y)=2(x^2y-y^3) \] Using \[ u=x+y, \qquad v=x-y, \] we get: \[ v\,f(u)-u\,f(v) = 2(x^2y-y^3) \] Now, \[ x^2-y^2=(x+y)(x-y)=uv \] Hence, \[ x^2y-y^3 = y(x^2-y^2) = \frac{u-v}{2}\cdot uv \] Therefore, \[ 2(x^2y-y^3) = (u-v)uv \] So the functional equation becomes: \[ v\,f(u)-u\,f(v)=uv(u-v) \]

Step 2: Separate the variables.
Divide both sides by \(uv\) (\(u,v\neq0\)): \[ \frac{f(u)}{u}-\frac{f(v)}{v} = u-v \] Rearranging: \[ \frac{f(u)}{u}-u = \frac{f(v)}{v}-v \] The left side depends only on \(u\) and the right side only on \(v\). Therefore, both sides must be equal to a constant, say \(k\). Thus, \[ \frac{f(u)}{u}-u=k \] Hence, \[ \frac{f(u)}{u}=u+k \] Multiplying by \(u\): \[ f(u)=u^2+ku \] Therefore, \[ f(x)=x^2+kx \] So \(f(x)\) is a polynomial function. Hence, option (A) is correct.

Step 3: Use the condition \(f(1)=2\).
Substituting \(x=1\): \[ f(1)=1+k=2 \] Thus, \[ k=1 \] Therefore, \[ f(x)=x^2+x \]

Step 4: Check the remaining options.
For option (B) and (C): \[ f(3)=3^2+3=9+3=12 \] Thus, \[ f(3)=12 \] Hence, option (C) is correct and option (B) is incorrect. For option (D): \[ f(0)=0^2+0=0 \] Hence, option (D) is also correct. Therefore, the correct options are: \[ \boxed{(A),\ (C)\ \text{and}\ (D)} \]