Problems concerning known physical laws often involve differential equations. Examples are population growth, radioactive decay, interest and Newton's law of cooling.
One of the most common types of differential equations involved is of the form ^{dy}⁄_{dx} = ky.
Differential Equations of the type: ^{dy}⁄_{dx} = ky
Because this type of differential equation is so common the general solution will be found.

Questions involving differential equations often provide additional information to allow the values of the constants A and k to be found.
Example
A biologist finds that for ants, the growth rate of a colony is given by the differential equation ^{dA}⁄_{dt} = 0.2 A where A = the number of ants in the colony and t = time in days Solve the equation to find the general solution 
Solution
^{dA}⁄_{dt} = 0.2 A = 0.2 dt (separating variables) ∫ = ∫ 0.2 dt (integrating both sides) ln A = 0.2t + c A = e^{0.2t + c} (definition of a logarithm) A = e^{c} . e^{0.2t} A = k . e^{0.2t} (the general solution) When t = 0, A = 5 000 (from the question) 5000 = k. e^{0}
k = 5000Therefore A = 5000 e^{0.2t} (the particular solution) Now when t = 25
A = 5000 e^{0.2 x 25} = 5000 e^{5}
A = 742 066 ants
There will be 742 066 ants after 25 days.
Finding the Differential Equation
More difficult problems involve having to first construct the differential equation. This is known as mathematical modelling.
This will often involve proportionality:
If y is proportional to x  y α x  y = kx 
If y is inversely proportional to x  y α ^{1}⁄_{x}  y = k⁄_{x} 
If the rate of a quantity is changing proportional to the amount left  ^{dy}⁄_{dt} α y  ^{dQ}⁄_{dt} = ky 
where k is a constant.
Example
The charge in a car battery is draining at a rate proportional to the amount of charge.
If the initial charge is 50μC, and after 2 hours this has dropped to 40 μC, estimate the charge after 5 more hours.
Solution
The differential equation is of the form:
^{dQ}⁄_{dt}= k. QBecause the rate of change of the charge is proportional to the amount of charge. (see table above)
Q is the charge (μC) and t is the time (hours)
^{dy}⁄_{dx} dQ = k.dt (separating variables) ∫ ^{1}⁄_{Q} dQ = ∫ k.dt (integrating both sides) ln Q = k.t + A Q = e^{kt + A} (definition of a logarithm) Q = e^{A} . e^{kt} Q = B.e^{kt} (the general solution) When t = 0, Q = 50
50 = B. e^{0}
B = 50(from the question) when t = 2, Q = 40 (from the question) 40 = 50.e^{2k}
e^{2k} = 0.8
2k = ln 0.8
k = 0.11Q =50.e^{}^{0.11t}
(the particular solution) After a further 5 hours when t = 7 Q = 50.e^{0.77} Q = 23 μC The charge after 5 more hours is 23 μC