Solutions for $f'=\lambda f$

Question

I am trying to figure out the following problem: Show that $f'=\lambda f$ for a real constant $\lambda$ has only $ce^{\lambda x}$ solutions.

My work: We take a look at $g(x)=f(x)\exp(\lambda x)$. We know $exp(x)$ never vanishes and f is differentiable, and g is differentiable, so we compute $g'(x)=\frac{f'(x)exp(\lambda x)-\lambda f(x)exp(\lambda x)}{(exp(x))^2}$. Since $f'=\lambda f$, $g'(x)=0$, so g is a constant, which means that $f(x)=g(x) \exp(\lambda x)=c\exp(\lambda x)$.

This follows rather nicely, but what I am confused about is if and how this exhausts all possible solutions for $f'=\lambda f$ for a real constant $\lambda$. Is this a complete proof?

Thanks in adnvace!

Answer 1

Your computation of $g'$ is not consistent with your definition of $g.$

Better define $g$ as $g(x)=e^{-\lambda x}f(x)$ (not $e^{+\lambda x}f(x)$), so that $g'(x)=e^{-\lambda x}(f'(x)-\lambda f(x)).$

After that, reason by equivalences to be sure that you "exhaust all possible solutions".

$$\begin{align}f'=\lambda f&\iff g'=0\\&\iff\exists c\quad\forall x\in\Bbb R\quad g(x)=c\\&\iff\exists c\quad\forall x\in\Bbb R\quad f(x)=ce^{\lambda x}.\end{align}$$

The second $\iff$ is justified by the fact that the domain of $f$ and $g$ is implicitely $\Bbb R,$ hence connected.

Answer 2

It just need to prove this:

Given $g(x)$ derivable, if $g'(x) \equiv 0$, then $g(x)$ is a constant function.

It's obvious. If $\exists x_1,x_2, g(x_1)\neq g(x_2)$, by mean value theorem, $\exists a\in(x_1,x_2),g'(a)\neq 0$. Contradiction.

In addition, any constant function satisfies the condition.

Answer 3

If f was a monovariable real function, then the differential equation solves by integration: $$\frac{df}{dx}=\lambda f\Leftrightarrow \frac{df}{f}=\lambda dx|\int\Leftrightarrow\int{\frac{1}{f}df}=\int{\lambda dx}\Leftrightarrow \log |f| +C_1=\lambda x +C_2\Rightarrow f(x)=e^{\lambda x +C}$$ $$ f(x)=C\cdot e^{\lambda x}(\therefore)$$

Note than $C_1, C_2, C$ are constants so moving them around won’t change the fact the a constant appears in the final result.

Solving by integration ensures the uniqueness of the solution in a straightforward and intuitive manner. However, once we know the expression of f is easy to build an alternate solution that apparently didn’t require integration: $$\begin{align} g\prime(x)=&\lambda g(x)|\cdot e^{-\lambda x}\\ g\prime(x) \cdot e^{-\lambda x}=&\lambda g(x) \cdot e^{-\lambda x}\\ (g(x)\cdot e^{-\lambda x})\prime=&0\\g(x)\cdot e^{-\lambda x}=&c\\g(x)=&c\cdot e^{\lambda x}\text{,c constant}\end{align}$$

I personally find this approach deceiving because is a matter of stumbling onto what to multiply by in order to find the solution. However it seems to be the mostly sought off method so is worth mentioning.