# Solving functional equation with partial differentiation

Here's a question I found while studying functional equations. $f(x+y)=f (x)f (y)$ (for all $x,y$ belonging to the set of real numbers) and $f'(0)=2.$ We have to find $f (x).$

So the method we've been taught in class goes like this

1. Partially differentiate both sides of functional equation w.r.t. $x$ or $y$ $f'(x+y)=f'(x)f (y)$ (Partial differentiating w.r.t. $x$)

2. Put $x=0$ to get $f'(y)=f'(0)f (y).$

Now since $f '(0)$ is known, we have a differential equation which we can easily solve to give the answer which comes out to be $f (x)=e^{2x}.$

But I am not getting it why this method works?

Secondly, in the first step while partial differentiating we assumed that $y$ is a constant. But in the second step we are integrating w.r.t. $y.$ This doesn't make sense. How can $y$ be treated as a constant and a variable in a single problem?

• This is the classic homomorphism $x\to e^x$ which can be composed with the continuous bijection $x\to ax$, $a$ constant, giving $x\to e^{ax}$ whose derivative gives $a=2$. May 30, 2018 at 15:23
• The partial differentation should be $f'(x+y) = f'(x)f(y) + f(x)f'(y) \frac{dy}{dx}$. Assume that y is independent of x, i.e $\frac{dy}{dx}=0$ May 30, 2018 at 15:26

We have $$f(x+y) = f(x)f(y)$$ Partial differentiate respect to x $$f'(x+y) = f'(x)f(y) + f(x)f'(y) \frac{dy}{dx}$$ Since there is no restriction on x or y. y is independent of x. $\frac{dy}{dx}=0$ $$f'(x+y) = f'(x)f(y)$$ The definition of function is for all x,y it applies the above $\forall x,y$. We can choose any x we want. Since we know $f'(0)$, we choose $x=0$ $$f'(y) = f'(0)f(y)=2f(y)$$ So, $f'(y)=2f(y)$ solve using seperation of variables $$\frac{df(y)}{dx}=2f(y)$$ $$\frac{df(y)}{f(y)}=2dx$$
$$ln(f(y))=2x+C$$
$$f(y)=e^{2x+C}=Ke^{2x}$$
This can be solved in the following way. Because $f$ is continues at least at $x=0$, and satisfies Cauchy equation $f(x+y)=f(x)f(y)$, it must be $f(x)=a^x$ or $f(x)=0$, for some $a>0$, but it can't be $f(x)=0$, since $f'(0)=2$. Then $$f'(x) = a^x\ln(a)$$ and putting $x=0$ $$f'(0)=2=\ln(a) \implies a=e^2$$ So we have $$f(x)=e^{2x}$$