Solution set of a simple differential equation.

Question

Let $f : \mathbb{R}\to\mathbb{R}$ be differentiable $\forall x \in \mathbb{R}$

Let $f'(x) = f(x) \;\forall x \in \mathbb{R}$

We are asked to find all the functions that satisfy the given constraints.

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Here's my attempt at the solution:

Firstly I assume that $f(x) \neq 0 \; \forall x \in \mathbb{R}$. Then:

$$f'(x) = f(x) \implies {f'(x) \over f(x)} = 1 \space\space\forall x \in \mathbb{R}$$

As the latter is true $ \forall x \in \mathbb{R}$, integrating both sides of the equation over $\mathbb{R}$

$$\ln{|f(x)|} = x + c \;\; \forall x \in \mathbb{R}; \text{ where $c \in \mathbb{R}$ is the constant of integration}$$

which implies that

$$|f(x)| = e^{x+c} \; \forall x \in \mathbb{R}$$

Thus,

$$f(x) = \alpha e^{x} \space\space \forall x \in \mathbb{R}; \text{ where } \alpha \in \mathbb{R} \setminus \{0\}$$

So the set of functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x) = \alpha e^{x} \space\space \forall x \in \mathbb{R}$ where $\alpha \in \mathbb{R} \setminus \{0\}$ is a solution set of the given question.

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However I came at this solution set under the assumption that $f(x) \neq 0 \; \forall x \in \mathbb{R}$, but $f(x)$ can be $0$ at some, or all, $x \in \mathbb{R}$

So I think there may be more solutions to the given question, but I'm at a loss as how to obtain those. I would appreciate any help in regards to this.

Thank you!

PS: Im still in High School so I would request keeping the calculus discussion not too beyond Calculus $1$ and Calculus $2$.

Answer 1

$f(x)$ is never $0$. Your solution is correct:$$f(x)=\alpha e^x$$ for all real numbers $x$ and all real numbers except $0$ for $\alpha$. Since $\alpha$ is never $0$, this means that for the function to be $0$, $e^x=0$. But, $e^x$ is never $0$, viz. $ln(0)$ doesn’t exist. For more on this, you might want to visit https://www.quora.com/Why-cant-e-x-0.

Answer 2

You have correctly solved the case for $f(x) \ne 0$ (for all $x \in \mathbb{R}$); this can be seen since $e^x \ne 0$ for any $x \in \mathbb{R}$, so the constraint that $\alpha \ne 0$ is appropriate.

Of course, there are only two possible forms for $f$: those where $f \not \equiv 0$ and where $f \equiv 0$.

Hence we have two further cases to look at (aside from that you solved), as you astutely observed:

• $f \not \equiv 0$ but $f(x) = 0$ for some (not all) $x \in \mathbb{R}$
• $f \equiv 0$

The latter case is trivial: the derivative of the zero function is again the zero function, so we could widen things up to admit $\alpha = 0$ in your solution.

Now consider the former. By way of contradiction, suppose $f'=f$, $f \not \equiv 0$, and $f$ is zero at some point.

Let $x_0$ be a point where $f(x_0) = 0$. Then $f'(x_0) = 0$. Hence, $x_0$ is an extreme value of $f$ in some neighborhood about $x_0$. This means that $f'(x) < 0$ to the left and $f'(x) > 0$ to the right of $x_0$, making $x_0$ a minimum (or the reverse, wherein $x_0$ is a maximum; we focus on the minimum case WLOG). However, then $f(x) < 0$ to the left and $f(x) > 0$ to the right (since $f'=f$), contradicting that $x_0$ is an extremum.

Hence, $f$ cannot have zeroes without being a constant function, and of course that requires $f \equiv 0$.

Hence the full solution set would be encapsulated by

$$f(x) = \alpha e^x$$

with the further condition that $\alpha \in \mathbb{R}$ (as opposed to omitting $0$).

(This could arguably be made easier in one context since some define $e^x$ as being the solution to $f'=f$ with $f(0) = 1$.)