Initial Value Problems? Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is a solution of the initial value problem $f'=f; f(0)=1$I need to answer the following questions:(i)Fix $y\in\mathbb{R}$ and set $h:\mathbb{R}\rightarrow\mathbb{R},$ $h(x):=f(-x)f(x+y).$ Prove that $h$ is constant.Do I just show that $h'(x)=0$?(ii)Hence show that $f(-x)f(x)=1$ and $f(x+y)=f(x)f(y)$  $\forall{}{}{}$ $x,y \in \mathbb{R}$.(iii)Prove uniqueness of the solution of the initial value problem: if $g$ also solves the IVP then $g=f$.I'm not sure how to go about the rest of the questions. Any help would be greatly appreciated.
 A: For (i), you just have to show that $h'(x)=0$. For (ii), denote $h_y$ the $h$ as in (i) corresponding to a $y\in\mathbb{R}$. For $y=0$, $f(-x)f(x)=h_0(x)=h_0(0)=1$. 
Now let any $y\in\mathbb{R}$. Then $f(x+y)=f(x)h_y(x)=f(x)h_y(0)=f(x)f(y)$.
For the uniqueness of solutions, let $g$ be another solution of the initial problem value. Then the items (i) and (ii) apply, hence $g(-x)=g(x)^{-1}$. Let $k(x)=g(-x)f(x)$. Since $k'(x)=0$, $k$ is constant equal to $k(0)=1$, therefore
$$f(x)=k(x)g(x)=g(x)$$
A: For the first part of ii, let $y=0$ from part i and use $f(0)=1$ to reach the conclusion. For the second part, note that $h(x)$ is constant and at $x=0$ we get $h(0)=f(y)$ and so $h(x)=f(y)$ for the given constant $y$. multiplying both sides by $f(x)$, we get $h(x)f(x)=f(y)f(x)$. Now, by the definition of $h$, we get $f(-x)f(x+y)f(x)=f(y)f(x)$ and, from the first part of ii, we can get rid of $f(x)f(-x)$ as this is equal to 1. Hence, $f(x+y)=f(x)f(y)$.
For part iii, suppose $g$ and $f$ are solutions to the IVP. Now, let $l(x)=g(-x)f(x)$ and so $g(x)l(x)=f(x)$. Differentiating, we get $l'(x)=f(x)g(-x)-g(-x)f(x)=0$ and so $l$ is a constant function with constant equal to $l(0)=g(0)f(0)=1$. If $l(x)=1$ for all $x$, then $g(x)l(x)=f(x)$ for all $x$ and so $g(x)=f(x)$.
