Find all differentiable function for which $f'(x)=177f(x)$ My try:
 I think functions that meet the requirements of the task are: $$f(x)=c\cdot e^{177x}$$
Then I have:
$$f'(x)=c\cdot (\ln e) \cdot e^{177}\cdot 177=177c\cdot e^{177}=177f(x)$$
However I don't know how to prove that these are the only functions that I am looking for.
Have you some hints for me?
 A: $$\frac{f'(x)}{f(x)}=177$$
Integrating both sides with respect to $x$ gives
$$\ln{(f(x))}=177x+C_1$$
$$f(x)=e^{177x+C_1}=C_2e^{177x}$$
Where $C_2=e^{C_1}$
A: Hint: If $f$ is differentiable and positive, then the derivative of $ln \circ f$ is $\frac{f'}{f}$.
A: This can be split into two types of solutions;
Case 1:
Non Trivial solution. In this case, assume that $f(x)\neq 0$ then proceed by separating variables and integrating as done by @Peter Foreman  
Case 2:
The obvious solution is $f(x)=0$ this is trivial. 
From the two observation, the functions are given by $f(x)=0$ and $f(x)=f_{0} e^{177x}$ where $f_{0}$ is a constant.  
A: Except for $f(x)\equiv 0$, write $${f'(x)\over f(x)}=177$$ therefore $$\int {f'(x)\over f(x)}dx=\int {1\over f(x)}df(x)=\int 177d x$$which leads to $$\ln f(x)+C_1=177x+C_2$$therefore $$f(x)=C_3 \exp(177x)$$the only solutions ever!
A: Recall the uniqueness theorem for first-order differential equations, which says that for a differential equation $y’=F(x,y)$, if both $F$ and $F_y$ are continuous functions in a rectangle around a point $(x_0,y_0)$, then there is a unique solution to $y’=F(x,y)$ passing through the point $(x_0,y_0)$. Now note that $F$ and $F_y$ are continuous in your case. What can you conclude about the solution that you obtained?
