Solving a differential equation? 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?
 A: 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.
A: "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.
A: 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
