Differential equation are widely used to model engineering systems, real life situations like exponential decay/growth, half life etc.
1.
Exponential growth. Let P(t) be the population that grows with time and its rate of increase is proportional to the same quantity P, then
$\frac{dP}{dt}$ $\propto$ P
$\frac{dP}{dt}$ = kP
Here
$\frac{dP}{dt}$ is the first derivative of P and k > 0.
Solution for this first order differential equation is
P(t) = A $e^{kt}$
where A is a non zero constant and it equals $P_0$ at t=0.
t = 0, P(0) = A $e^0$ => A
The final form can be written as P(t) = P0 $e^{kt}$For exponential growth P0 and k are positive.
2.
Exponential Decay: half life of a radioactive material.
Let M(t) be the mass of a material that decreases with time and its rate of decrease is
directly proportional to the quantity M. Since the rate is for decreasing it will be negative.

$\frac{dM}{dt}$ $\propto$ M
$\frac{dM}{dt}$ = kM
Here
$\frac{dM}{dt}$ is the first derivative of M and k > 0.
Solution of this differential equation is
M(t) = A $e^{kt}$
where A $\neq$ 0. At t = 0 if M(t) = M$_0$ then
M$_0$ = A $e^0$
M$_0$ = A
Finally, the equation becomes
M(t) = M$_0$ $e^{kt}$Since k is positive, M(t) is decreasing exponentially.
3.
Falling object When an object is dropped from a height. If h(t) is the height of the object at ant time t, a(t) is the acceleration and v(t) its velocity .We can show the relationship between h, v and a as
a(t) =
$\frac{dv}{dt}$, v(t) =
$\frac{dh}{dt}$For a falling object a(t) is constant and equals gravitational acceleration = g = 9.8 m/s
^{2 }g = 
$\frac{dv}{dt}$ =
$\frac{d^2 h}{dt^2}$ Integrating both sides of
g = 
$\frac{d^2 h}{dt^2}$ $\frac{dh}{dt}$ = gt + $v_0$
Integrate again to get h(t)
h(t) = $\frac{1}{2}$g $t_2$ + $v_0t$ + $h_0$This equations gives height of a falling object falling from initial height h0 with initial velocity $v_0$
4.
Newton’s law of cooling The law states that the
rate of change of the temperature is proportional to the difference between the temperature T of the object and surrounding temperature $T_e$
$\frac{dT}{dt}$ = k (T  $T_e$)
if we plug in x = T  $T_e$,
$\frac{dx}{dt}$ =
$\frac{dT}{dt}$… since $T_e$ is constant.
$\frac{dx}{dt}$ = k x
Solution to this differential equation will be
x = A $e^{kt}$
T  $T_e$ = A $e^{kt}$
at t = 0, T = $T_0$
$T_0$ – $T_e$ = A $e^0$
T(t) = $T_e$ + ($T_0 – T_e$)$e^{kt}$ The equation gives the temperature of a cooling object.