# 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:

• For each function they give you, just compute its first and second derivatives and then compute if the equation is satisfied Commented Mar 17, 2019 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. Commented Mar 17, 2019 at 18:29
• These equations are perfectly linear and in no way "exponential" ! They are even linear with constant coefficients, which is the kind for which resolution is the most systematic.
– user65203
Commented Dec 20, 2020 at 20:14

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.

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.

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 $$v(t)=C_1e^{-9t}+C_2e^{-7t}.$$
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.$$\qquad C_1e^{-9t}+C_2e^{-7t}$$
• However other answers are also solutions, if not general ones. $C_{1,2}$ can take any values including, for example, zero. Commented Dec 20, 2020 at 19:56