Wikipedia references:
<quote>
Streamlines are a family of curves that are instantaneously tangent to the
velocity vector of the flow. These show the direction a massless fluid element
will travel in at any point in time. </quote>
Consider the velocity field $(u,v)$ of a two-dimensional incompressible flow.
Let the family of curves be given by $\;\psi(x,y) = c$ . The velocity vectors are
tangent to these as shown for one of them in the following picture.
Thus, along the curve $\psi(x,y) = c$ , the following equations hold:
$$
\left. \begin{array}{l} \frac{dy}{dx} = \frac{v}{u} \\
d\psi = 0 = \frac{\partial \psi}{\partial x} dx + \frac{\partial \psi}{\partial y} dy
\end{array} \right\} \qquad \Longrightarrow \qquad
\frac{dy}{dx} = - \frac{\partial \psi / \partial x}{\partial \psi / \partial y}
= \frac{v}{u}
$$
Hence, apart from a constant:
$$
u = \frac{\partial \psi}{\partial y} \qquad ;
\qquad v = - \frac{\partial \psi}{\partial x}
$$
But the flow is incompressible, so:
$$
\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0
\qquad \Longrightarrow \qquad
\frac{\partial^2 \psi}{\partial x\, \partial y} =
\frac{\partial^2 \psi}{\partial y\, \partial x}
$$
Herewith the
conditions for an exact differential equation are fulfilled.
Now solve $\psi$ from:
$$
v\, dx - u\, dy = 0
$$
Example.
As taken from :
Find the velocity of a flow .
$$
u = -\frac{y}{x^2+y^2} \qquad ; \qquad v = \frac{x}{x^2+y^2}
$$
Then:
$$
v\, dx - u\, dy = \frac{x\,dx + y\,dy}{x^2+y^2}
= \frac{d\left( x^2+y^2 \right)}{x^2+y^2} = 0
\qquad \Longrightarrow \qquad
x^2 + y^2 = c
$$
It is concluded that the streamlines of this flow are circles.
Example. Somewhat related to the above one.
$$
u = \lambda\,x \qquad ; \qquad v = \lambda\,y
$$
Then, assuming that $\; x\ne 0$ (i.e. $\,x=0\,$ as a special case) :
$$
v\, dx - u\, dy = 0 \quad \Longleftrightarrow \quad
\frac{y\,dx - x\,dy}{x^2} = - d(y/x) = 0
\quad \Longrightarrow \quad y = c\, x
$$
An integrating factor has been used.
It is concluded that the streamlines of this flow are straight lines
through the origin.
Wikipedia reference:
The electric potential at a point $\vec{r}$ in a two-dimensional
static electric field $\vec{E}$ is given by the line integral:
$$
V = - \int_C \vec{E}\cdot d\vec{r} = - \int_C \left(E_x\, dx + E_y\, dy\right)
$$
where $C$ is an arbitrary path connecting the point with zero potential
to $\vec{r}$. It follows that:
$$
E_x = - \frac{\partial V}{\partial x} \qquad ; \qquad
E_y = - \frac{\partial V}{\partial y}
$$
The integral is zero if the path is closed. Then Green's theorem tells us:
$$
\oint \left( E_x\, dx + E_y\, dy \right) =
\iint \left( \frac{\partial E_y}{\partial x} -
\frac{\partial E_x}{\partial y} \right) dx\,dy =
- \iint \left( \frac{\partial^2 V}{\partial x \, \partial y}
- \frac{\partial^2 V}{\partial y \, \partial x}\right) dx\,dy = 0
$$
Thus establishing once more the
conditions for solvability of the exact differential equation:
$$
E_x \, dx + E_y \, dy = 0
$$
Solving this ODE results in the iso-lines $\,V(x,y) = c\,$ of the electric potential $\,V$ .
Example. An infinitely long and infinitely thin
charged wire perpendicular to the plane and intersecting it
in the origin. Apart from constants:
$$
(E_x,E_y) = \frac{(x,y)}{r^2} \quad \Longrightarrow \quad
E_x\, dx + E_y\, dy = \frac{x\,dx + y\,dy}{r^2} = 0
\quad \Longrightarrow \quad x^2+y^2 = c
$$
The equipotential lines are circles.
Can of worms: special cases, and a singularity at the origin in all of the examples.
C-14 is just the name of the isotope it is not an amount of the substance.
The amount at any time is given by,
$$r(t) = Ce^{\lambda t}$$
The amount at a later time $t+\tau$ should be half as much (\tau is the half life).
$$\frac{1}{2} r(t)= r(t + \tau) = C e^{\lambda (t+\tau)} = C e^{\lambda t} e^{\lambda \tau} = r(t) e^{\lambda \tau} $$
From this we conclude that,
$$ \frac{1}{2} = e^{\lambda \tau}$$
$$ \frac{\ln(1/2)}{ \tau} = \lambda $$
Best Answer
I was not sure if you were looking for a linear or a nonlinear. If nonlinear, I would have to think about it more.
You can investigate a series resistor, an inductor, and a capacitor (RLC circuit) in an electrical circuit.
This leads to the differential equation:
$$\displaystyle Ri(t) + L \frac{di}{dt} + \frac{1}{C} \int_{-\infty}^{\tau = t} i(\tau)d\tau = v(t)$$
$$\displaystyle \frac{d^2i(t)}{dt^2} +\frac{R}{L}\frac{di(t)}{dt} + \frac{1}{LC}i(t) = 0$$
You can see this circuit and the analysis on the Wiki. It is a second order system and depending on the type of voltage source, leads to intergro-differential or just differential equation. It is easy to analyze and to understand how it works. This can also be used for projects.
You can see the real life circuit, calculations and the Laplace Transform method to solve on the wiki RLC circuit.
Then you can simulate the results using free tools:
Lastly, you could actually breadboard the circuit, get a signal generator, a digital multimeter and oscilloscope and measure it in an actual circuit to see how close the ideal world and real world line up (errors abound in nature).
This gives you a very nice approach because with no cost you can review theory, solve a DEQ, run a simulation in two different environments (or use Modelica) and then actually build and test the circuit for no to little cost.
After doing a series circuit, you can switch to a parallel circuit and then a series-parallel circuit.
These components and circuits are used in every electronic device (including appliances, houses, cars, phones, computers...) that we own.