How to solve an exponential differential equation? My electrical engineering class just moved into differential equations from linear algebra, which is a topic I've never touched on before. The professor doesn't work hardly any examples on the board, so I've found myself a bit confused on how to solve differential equations that aren't linear. Below I've attached a homework problem that I'm having quite a bit of trouble solving. Is there any way I could get some help with the steps on how to solve an exponential differential equation (in simplistic terms too, please!) from an example in this homework problem? I don't want the entire problem answered, just one or two with an explanation so I can wrap my head around how to solve these things. Thanks!
Problem:

 A: Here is the first problem on the list.  Let me know if you need further explanation.
If $$v(t)=C_1e^{-9t},$$ then $$\frac {dv}{dt}(t)=-9C_1e^{-9t}$$ and $$\frac {d^2v}{dt^2}(t)=81C_1e^{-9t}, $$ so $$\frac {d^2v}{dt^2}(t)+16\frac {dv}{dt}(t)+63v(t)=81C_1e^{-9t}+16\times(-9C_1e^{-9t})+63C_1e^{-9t}\equiv0,$$
since $81-144+63=0$.  Therefore, $v(t)=C_1e^{-9t}$ is a solution.  
Some others on the list will not be solutions.
Summary for other answers:

 You should be learning that solutions of $(D+9)(D+7)v(t)=0,$ where $D=\frac d {dt}$ is the differential operator, are of the form $v(t)=C_1e^{-9t}+C_2e^{-7t}.$

A: For a constant-coefficient linear ODE, in this case one of the form $y’’+ay’+by=0$, solutions can be easily obtained without working backwards from the solutions. The first step is to compute the roots of the characteristic polynomial $r^2+ar+b=0$. Let the roots of this equation be $r_1,r_2$. Then solutions to the differential equation take the form $y_1=e^{r_1x},y_2=e^{r_2x}$. If the root is repeated (i.e. $r_1=r_2$, the solutions are $y_1=e^{r_1x},y_2=xe^{r_1x}$. If the roots are complex, use Euler’s formula $e^{it}=\cos t+i\sin t$ to simplify solutions.
A: The algebraic equation associated to your ODE is
$$z^2+16z+63=0.$$
It has positive discriminant $\Delta=4$ and (real) roots:
$${z_1=\frac{-16-2}{2}=-9,}\quad{z_2=\frac{-16+2}{2}=-7.}$$
Hence, the general integral is the one in answer

B.$\qquad C_1e^{-9t}+C_2e^{-7t}$

