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: https://imgur.com/EoNd2XR

• For each function they give you, just compute its first and second derivatives and then compute if the equation is satisfied – J. W. Tanner Mar 17 at 18:27
• Apparently, your problem is about checking if the given functions are solutions of your DE. All you have to do is, given a candidate $y=f(t)$, find $f'(t)$, $f''(t)$, etc. plug into the equation and see if you get an identity, true for all $t$. Your equation is linear and homogeneous, and all of its solutions are exponentials, exponentials multiplied by polynomials or sines or cosines, possibly multiplied by exponentials and polynomials. – GReyes Mar 17 at 18:29

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.
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.
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 $$v(t)=C_1e^{-9t}+C_2e^{-7t}.$$