1
$\begingroup$

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:

question

$\endgroup$
3
  • 4
    $\begingroup$ For each function they give you, just compute its first and second derivatives and then compute if the equation is satisfied $\endgroup$ Commented Mar 17, 2019 at 18:27
  • $\begingroup$ 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. $\endgroup$
    – GReyes
    Commented Mar 17, 2019 at 18:29
  • $\begingroup$ 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. $\endgroup$
    – user65203
    Commented Dec 20, 2020 at 20:14

3 Answers 3

1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ However other answers are also solutions, if not general ones. $C_{1,2}$ can take any values including, for example, zero. $\endgroup$ Commented Dec 20, 2020 at 19:56
  • 1
    $\begingroup$ @MarkBennet I see how the question is worded... A. and I. are also solutions. This is why I called it general integral in my answer although it is not mentioned elsewhere here...thank you for remarking this :) $\endgroup$ Commented Dec 20, 2020 at 20:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .