# 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.

Once 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).

**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.

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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