Differential Equations
Differential Equations Revision
Differential Equations
A differential equation is an equation with a derivative term in it, such as \dfrac{dy}{dx}.
We can solve them by treating \dfrac{dy}{dx} as a fraction then integrating once we have rearranged.
They are often used to model real life scenarios, in which case it might use x and t, rather than y and x, where t represents time.
Solving Differential Equations
All differential equations at A-level have the form \dfrac{dy}{dx}=f(x)g(y)
We can treat \dfrac{dy}{dx} as a fraction and rearrange:
\dfrac{dy}{dx}=f(x)g(y)
\dfrac{1}{g(y)}\dfrac{dy}{dx}=f(x)
\dfrac{1}{g(y)}dy=f(x)dx
Now that it is in this form, we can integrate the left side with respect to y and integrate the right side with respect to x.
\int\dfrac{1}{g(y)}dy=\int f(x)dxOnce we have integrated we can get our final answer by rearranging to get y in terms of x.
The answer will include a +c from the integration (you only need to include this on one side). Sometimes the question will contain extra information to help you determine the value of c.
Note: We cannot usually treat \dfrac{dy}{dx} as a normal fraction, but we can in this case.
Real-Life Problems
Sometimes, you will be required to form a differential equation based on a real life problem.
Example: The rate at which the size of a goldfish, S, is increasing is inversely proportional to the current size of the goldfish. Form a differential equation for this scenario.
\dfrac{dS}{dt} is the rate of change of S (the size of the goldfish) with respect to t (time). This is inversely proportional to S. Hence:
\dfrac{dS}{dt}=\dfrac{k}{S} for some constant k.
As we can see from the example, real life differential equations often do not use x and y, but other variables. However, they can be solved in the same way.
Real life problems will also sometimes contain extra information to help you determine the constant of integration. If a question involving time provides a “starting condition” as this extra information, this is the value of the parameter when t=0.
You may also be asked to list limitations of modelling a real life problem with a differential equation. These could include:
- Not enough information (if no information to determine the value of constants is provided)
- The model could break down at very large or very small values.
- The appropriateness of the model (for example a continuous variable to monitor a population which is discrete would be a drawback).
- Any other things that have not been included (will vary based on the question and the context).
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Example 1: Differential Equations
Find the solution to \dfrac{dy}{dx}=24y^{2}\sin(x).
[2 marks]
\dfrac{dy}{dx}=24y^{2}\sin(x)
\dfrac{1}{24y^{2}}\dfrac{dy}{dx}=\sin(x)
\dfrac{1}{24}y^{-2}dy=\sin(x)dx
\int\dfrac{1}{24}y^{-2}dy=\int \sin(x)dx
-\dfrac{1}{24}y^{-1}=-\cos(x)+c
\dfrac{1}{24y}=\cos(x)+c
\dfrac{1}{24(\cos(x)+c)}=y
y=\dfrac{1}{24\cos(x)+c}
Note: Since c is an arbitrary constant, multiplication by 24 does not change that it is an arbitrary constant, so we can still just write +c.
Example 2: Real-Life Problems
The population P of a herd of sheep increases according to \dfrac{dP}{dt}=0.2P where t is in years. There are 15 sheep at t=0. Find P in terms of t.
[3 marks]
\dfrac{dP}{dt}=0.2P
\dfrac{1}{P}\dfrac{dP}{dt}=0.2
\dfrac{1}{P}dP=0.2dt
\int\dfrac{1}{P}dP=\int0.2dt
\ln(P)=0.2t+c
P=e^{0.2t+c}
P=e^{c}e^{0.2t}
Since c is a constant, e^{c} is a constant, which we commonly call A.
P=Ae^{0.2t}At t=0, P=15
15=Ae^{0.2\times0}
15=Ae^{0}
A=15
P=15e^{0.2t}
Differential Equations Example Questions
Question 1: Solve \dfrac{dy}{dx}=4\cos(x)e^{y}
[2 marks]
e^{-y}\dfrac{dy}{dx}=4\cos(x)
e^{-y}dy=4\cos(x)dx
\begin{aligned}\int e^{-y}dy=\int4\cos(x)dx\end{aligned}
-e^{-y}=4\sin(x)+c
e^{-y}=-4\sin(x)+c
-y=\ln(c-4\sin(x))
y=-\ln(c-4\sin(x))
Question 2: Solve \dfrac{dy}{dx}=3y^{3}-4xy^{3}
[3 marks]
\dfrac{dy}{dx}=y^{3}(3-4x)
\dfrac{1}{y^{3}}\dfrac{dy}{dx}=3-4x
y^{-3}dy=\left( 3-4x\right) dx
\begin{aligned}\int y^{-3}dy=\int\left( 3-4x\right) dx\end{aligned}
-\dfrac{1}{2}y^{-2}=3x-2x^{2}+c
\dfrac{1}{2y^{2}}=2x^{2}-3x+c
\dfrac{1}{2(2x^{2}-3x+c)}=y^{2}
y^{2}=\dfrac{1}{4x^{2}-6x+c}
y=\dfrac{1}{\sqrt{4x^{2}-6x+c}}
Question 3: A company advertises their soft drinks on the sides of bus stops in London. They believe that the number of sales N increases with the number of advertisments a, according to \dfrac{dN}{da}=0.05N. Before the company started advertising on bus stops, they had had 1000 sales. Find N in terms of a.
[3 marks]
\dfrac{1}{N}\dfrac{dN}{da}=0.05
\dfrac{1}{N}dN=0.05da
\begin{aligned}\int\dfrac{1}{N}dN=\int0.05da\end{aligned}
\ln(N)=0.05a+c
N=e^{0.05a+c}
N=e^{c}e^{0.05a}
N=Ae^{0.05a}
At a=0, N=1000
1000=Ae^{0.05\times0}
1000=Ae^{0}
A=1000
N=1000e^{0.05a}
Question 4: A colony of bacteria in a petri dish increase in population in direct proportion to the current population. Given that at time 0 there is 1 bacterium, and at time 9 there are 68 bacteria, find the population as a function of time.
[4 marks]
Call population P and time t.
\dfrac{dP}{dt}\propto P
\dfrac{dP}{dt}=kP
\dfrac{1}{P}\dfrac{dP}{dt}=k
\dfrac{1}{P}dP=kdt
\int\dfrac{1}{P}dP=\int kdt
\ln(P)=kt+c
P=e^{kt+c}
P=e^{c}e^{kt}
P=Ae^{kt}
At t=0, P=1
1=Ae^{k\times0}
1=Ae^{0}
A=1
P=e^{kt}
At t=9, P=68
68=e^{9k}
9k=\ln(68)
k=\dfrac{\ln(68)}{9}
P=e^{\frac{\ln(68)}{9}t}
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