How many different functions $f(x)$ exist such that $f'(x) = f(x)$?
The ones I know of right now are $f(x) = 0$ and $f(x) = ne^x$, for any real number $n$. What other functions satisfy this property?
Asked
Active
Viewed 219 times
0
-
Note that $(\ln f(x))'=1$ (if $f(x)\ne0$) – Kenta S Jan 08 '21 at 18:43
-
1Your "first" example is just your second one with $n=0$. – rschwieb Jan 08 '21 at 21:33
2 Answers
5
This is a simple Ordinary differential equation. And all solutions are of the form $$f(x)=Ce^x$$ for some real constant $C$. Note that $f(x)=0$ as soon as you set $C=0$.
proof. \begin{align} f'(x)& =f(x), \forall x\\ f'(x)e^{-x}- f(x)e^{-x}&=0, \forall x\\ f'(x)e^{-x}+ f(x)(e^{-x})'&=0, \forall x\\ (f(x)e^{-x})'& =0, \forall x\\ f(x)e^{-x}&=C, \forall x\\ f(x)&=Ce^x, \forall x \end{align}
Leonard Neon
- 1,354
-
1
-
1@hamam_Abdallah Thanks. I love this method since we don't need to assume that $f(x)\neq 0$. – Leonard Neon Jan 08 '21 at 19:06
-
2It is better to write $ \forall x\in \Bbb R $ before and $ \implies $ at the end. – hamam_Abdallah Jan 08 '21 at 19:09
0
Write the equation as $\frac{dy}{dx} = y$.
Take the reciprocal of both sides, integrate and make $y$ the subject:
$\frac{dx}{dy} = \frac{1}{y}$
$x = \ln y + C$
$x - C = \ln y$
$e^{x-C} = y$
$y = C_1 e^x$
1123581321
- 954