example of solving a functional equation
Let’s determine all twice differentiable^{} real functions $f$ which satisfy the functional equation
$f(x+y)\cdot f(xy)={[f(x)]}^{2}{[f(y)]}^{2}$  (1) 
for all real values of $x$ and $y$.
Substituting first $y=0$ in (1) we see that $f{(x)}^{2}=f{(x)}^{2}f{(0)}^{2}$ or $f(0)=0$. The substitution $x=0$ gives $f(y)f(y)=f{(y)}^{2}$, whence $f(y)=f(y)$. So $f$ is an odd function^{}.
We differentiate both sides of (1) with respect to $y$ and the result with respect to $x$:
$${f}^{\prime}(x+y)f(xy){f}^{\prime}(xy)f(x+y)=2f(y){f}^{\prime}(y)$$ 
$${f}^{\prime \prime}(x+y)f(xy)+{f}^{\prime}(xy){f}^{\prime}(x+y){f}^{\prime \prime}(xy)f(x+y){f}^{\prime}(x+y){f}^{\prime}(xy)=0$$ 
The result is simplified to ${f}^{\prime \prime}(x+y)f(xy)={f}^{\prime \prime}(xy)f(x+y)$, i.e.
$${f}^{\prime \prime}(x+y)/f(x+y)={f}^{\prime \prime}(xy)/f(xy).$$ 
Denoting $x+y:=u$, $xy:=v$ we obtain the equation
$$\frac{{f}^{\prime \prime}(u)}{f(u)}=\frac{{f}^{\prime \prime}(v)}{f(v)}$$ 
for all real values of $u$ and $v$. This is not possible unless the proportion $\frac{{f}^{\prime \prime}(u)}{f(u)}$ has a on $u$. Thus the homogeneous linear differential equation ${f}^{\prime \prime}(t)/f(t)=\pm {k}^{2}$ or
$${f}^{\prime \prime}(t)=\pm {k}^{2}f(t),$$ 
with $k$ some , is valid.
There are three cases:

1.
$k=0$. Now ${f}^{\prime \prime}(t)\equiv 0$ and consequently $f(t)\equiv Ct$. If one especially $C$ equal to 1, the solution is the identity function (http://planetmath.org/IdentityMap) $f:t\mapsto t$. This yields from (1) the wellknown “memory formula”
$$(x+y)(xy)={x}^{2}{y}^{2}.$$ 
2.
${f}^{\prime \prime}(t)={k}^{2}f(t)$ with $k\ne 0$. According to the oddness one obtains for the general solution the sine function $f:t\mapsto C\mathrm{sin}kt$. The special case $C=k=1$ means in (1) the
$$\mathrm{sin}(x+y)\mathrm{sin}(xy)={\mathrm{sin}}^{2}x{\mathrm{sin}}^{2}y,$$ which is easy to verify by using the addition and subtraction formulae (http://planetmath.org/AdditionFormula) of sine.

3.
${f}^{\prime \prime}(t)={k}^{2}f(t)$ with $k\ne 0$. According to the oddness we obtain for the general solution the hyperbolic sine^{} (http://planetmath.org/HyperbolicFunctions) function^{} $f:t\mapsto C\mathrm{sinh}kt$. The special case $C=k=1$ gives from (1) the
$$\mathrm{sinh}(x+y)\mathrm{sinh}(xy)={\mathrm{sinh}}^{2}x{\mathrm{sinh}}^{2}y.$$
The solution method of (1) is due to andik and perucho.
Title  example of solving a functional equation 
Canonical name  ExampleOfSolvingAFunctionalEquation 
Date of creation  20130322 15:30:00 
Last modified on  20130322 15:30:00 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  14 
Author  pahio (2872) 
Entry type  Example 
Classification  msc 34A30 
Classification  msc 39B05 
Related topic  ChainRule 
Related topic  AdditionFormula 
Related topic  SubtractionFormula 
Related topic  DefinitionsInTrigonometry 
Related topic  GoniometricFormulae 
Related topic  DifferenceOfSquares 
Related topic  AdditionFormulas 