Find m. $y=e^{mx},m\in\mathbb R,\frac{d^2y}{dx^2}-3\frac{dy}{dx}-4y=0$ If: $$y=e^{mx},m\in\mathbb R$$
Find m if: $$\frac{d^2y}{dx^2}-3\frac{dy}{dx}-4y=0$$
Differentiating and substituting gives:
$$m^2e^{mx}-3me^{mx}-4e^{mx}=0$$
Dividing across by $e^{mx}$ and solving the resulting quadratic gives:
$$m = -1,4$$
But I'm not happy about dividing across by $e^{mx}$ as there is an $m$ in there. 
Is there another solution somewhere?
How can this be solved without dividing across by $e^{mx}$?
If $m\in\mathbb R$ were not a requirement is $m=-\infty$ a solution?
 A: There is no harm in dividing by $e^{mx}$ since $e^{mx} > 0$ for all $x \in \mathbb{R}$. If you don't want to divide by $e^{mx}$, you can plug in $x = 0$ instead.
A: 
But I'm not happy about dividing across by $e^{mx}$ as there is an $m$ in there.

Dividing by $e^{mx}$ is perfectly logical since $e^{mx}>0$ for all real $x$ and $e^{mx}\to 0$ when $x\to \mp \infty$ according as $m$ is positive or negative respectively but $e^{mx}\not = 0$ for all real $m,x$. 
So it is absolutely correct to write that:

$$m^2e^{mx}-3me^{mx}-4e^{mx}=0$$ $$\implies e^{mx}(m^2-3m-4)=0$$
  $$\implies m^2-3m-4=0$$

A: 
Notice, when $y(x)=\exp\left[mx\right]$:
  
  
*
  
*$$y'(x)=m\exp\left[mx\right]$$
  
*$$y''(x)=m^2\exp\left[mx\right]$$
  

So, we get:
$$y''(x)-3y'(x)-4y(x)=0\Longleftrightarrow$$
$$m^2\exp\left[mx\right]-3m\exp\left[mx\right]-4\exp\left[mx\right]=0\Longleftrightarrow$$
$$\exp\left[mx\right]\left(m^2-3m-4\right)=0\Longleftrightarrow$$

Now:


*

*If $x>0$:
$$\exp\left[mx\right]>0$$

*If $x<0$:
$$\exp\left[mx\right]>0$$


$$m^2-3m-4=0\Longleftrightarrow m=\frac{3\pm\sqrt{25}}{2}$$
So, $m=-1$ or $m=4$.
