3
$\begingroup$

I'm trying to analyze the transient state of a RC circuit. My book gives me the following differential equation:

$$\frac{d(v(t))}{dt} + av(t) = c$$

for some constants $a$ and $c$.

The book thens proceeds to solve it, and says that:

$$v(t) = K_1 + K_2e^{-t/\tau}$$

for some constants $\tau, K_1, K_2$. We haven't learned differential equations yet, so I wasn't able to follow along the solution of the differential equation.

However, doesn't:

$v(t) = c/a$

also satisfy the differential equation? Why isn't this a valid solution to the equation? If it is a valid question, what could possibly motivate the book to not include it?

$\endgroup$
  • 1
    $\begingroup$ This solution is actually embedded in the solution they give you. They just swallowed it up into the $K_1$ (constant) term. Note that $c/a$ is constant as well. $\endgroup$ – Cameron Williams Apr 14 '14 at 19:29
  • $\begingroup$ But for $c/a$ to be embedded in the solution, $K_2e^{-t/\tau}$ has to be zero for some $t$. And that isn't possibly for $t < \infty$ is it? $\endgroup$ – dfg Apr 14 '14 at 19:31
  • 1
    $\begingroup$ @dfg According to that reasoning then it would fail for $t\neq 0$, not just negative $t$. The thing here is that the solution is given like this: there exist constants $K_1, K_2$ such that $\forall t\in \mathbb R(v(t) = K_1 + K_2e^{-t/\tau})$. Now consider the function $f$ such that $f(t)=\dfrac c a$ for all $t\in \mathbb R$. Do there exist constants $K_1, K_2$ such that $\forall t\in \mathbb R(v(t) = K_1 + K_2e^{-t/\tau})$? Yes, they do with $K_1=\dfrac c a$ and $K_2=0$. $\endgroup$ – Git Gud Apr 14 '14 at 19:35
6
$\begingroup$

Given a non-trivial interval $I$ and $a,b$ continuous functions (therefore integrable) defined on $I$, consider the differential equation $y′+ay=b$. Let $A$ be an antiderivative of $a$. There exists $K\in \mathbb R$ such that for all $t\in I$ the following holds: \begin{align} y'(t)+a(t)y(t)=b(t)&\iff e^{A(t)}(y'(t)+a(t)y(t))=e^{A(t)}b(t)\\ &\iff \dfrac{d}{dt}\left(t\mapsto e^{A(t)}y(t)\right)(t)=e^{A(t)}b(t)\\ &\iff e^{A(t)}y(t)=\int e^{A(t)}b(t)\,\mathrm dt + K\\ &\iff y(t)=e^{-A(t)}\int e^{A(t)}b(t)\,\mathrm dt+Ke^{-A(t)},\end{align} where $\displaystyle \int e^{A(t)}b(t)\mathrm dt$ denotes an antiderivative of $t\mapsto e^{A(t)}b(t)$ on $I$.

As mentioned the solution $v(t)=\dfrac c a$ for all $t$ is given in the general formula.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Makes sense. Thanks, appreciate the help. $\endgroup$ – dfg Apr 14 '14 at 20:03
  • $\begingroup$ @dfg You're welcome. $\endgroup$ – Git Gud Apr 14 '14 at 20:03
0
$\begingroup$

"Why isn't this a valid solution to the equation?": it absolutely is, setting $K_1=\frac{c}{a}$ and $K_2=0$ !

You just found a particular case of the general solution.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

To solve a first order differential equation, you can use integral factor. This video walks you through the process(for RC circuits). https://www.youtube.com/watch?v=th39yALievI&t=3s

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.