A deceptively simple problem?

Question

Suppose the IVP

y' = y, y(0) = 1

has a solution defined on the real line.

Prove that y(s + t) = y(s) y(t) for all real s and t.

Note: I'm asking people to prove this property of the exponential function using only the structure of the IVP to which it is the solution. Simpy saying 'e^x is the solution' will earn you nil points.

Answer 1

First, we prove a lemma: ∀t, y(t)≠0.

Proof: Consider the function y(t)y(-t). The derivative of this function is y'(t)y(-t) - y(t)y'(-t), which, since y=y', is equal to y(t)y(-t) - y(t)y(-t), which is 0. Therefore, since the derivative of y(t)y(-t) is everywhere 0, y(t)y(-t) is a constant function. Since it equals 1 at t=0, y(t)y(-t)=1 everywhere. Now, suppose for some t y(t)=0. Then y(t)y(-t)=0≠1, a contradiction. Therefore, y(t) is never equal to zero.

Now, since we know y is nonzero, we can meaningfully divide by it. Consider now the quotient:

y(s+t)/y(t)

Holding s constant, differentiation with respect to t yields that:

∂(y(s+t)/y(t))/∂t = (y'(s+t)y(t) - y(s+t)y'(t))/y(t)² = (y(s+t)y(t) - y(s+t)y(t))/y(t)² = 0/y(t)² = 0

So y(s+t)/y(t) = C, where C is constant in terms of t. Evaluating at t=0 yields that C=y(s)/y(0)=y(s). So y(s+t)/y(t) = y(s), meaning y(s+t)=y(t)y(s), as required. Q.E.D.