# Differential Equations

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## 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.

A Level

## 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 $x$.

$\int\dfrac{1}{g(y)}dy=\int f(x)dx$

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.

A Level

## 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).
A Level
A Level

## 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 $+c$.

A Level

## 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 $t=0$, $P=15$

$15=Ae^{0.2\times0}$

$15=Ae^{0}$

$A=15$

$P=15e^{0.2t}$

A Level

## Example Questions

$\dfrac{dy}{dx}=4\cos(x)e^{y}$

$e^{-y}\dfrac{dy}{dx}=4\cos(x)$

$e^{-y}dy=4\cos(x)dx$

\begin{aligned}\int e^{-y}dy=\int4\cos(x)dx\end{aligned}

$-e^{-y}=4\sin(x)+c$

$e^{-y}=-4\sin(x)+c$

$-y=\ln(c-4\sin(x))$

$y=-\ln(c-4\sin(x))$
$\dfrac{dy}{dx}=3y^{3}-4xy^{3}$

$\dfrac{dy}{dx}=y^{3}(3-4x)$

$\dfrac{1}{y^{3}}\dfrac{dy}{dx}=3-4x$

$y^{-3}dy=\left( 3-4x\right) dx$

\begin{aligned}\int y^{-3}dy=\int\left( 3-4x\right) dx\end{aligned}

$-\dfrac{1}{2}y^{-2}=3x-2x^{2}+c$

$\dfrac{1}{2y^{2}}=2x^{2}-3x+c$

$\dfrac{1}{2(2x^{2}-3x+c)}=y^{2}$

$y^{2}=\dfrac{1}{4x^{2}-6x+c}$

$y=\dfrac{1}{\sqrt{4x^{2}-6x+c}}$
$\dfrac{dN}{da}=0.05N$

$\dfrac{1}{N}\dfrac{dN}{da}=0.05$

$\dfrac{1}{N}dN=0.05da$

\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 $a=0$, $N=1000$

$1000=Ae^{0.05\times0}$

$1000=Ae^{0}$

$A=1000$

$N=1000e^{0.05a}$

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 $t=9$, $P=68$

$68=e^{9k}$

$9k=\ln(68)$

$k=\dfrac{\ln(68)}{9}$

$P=e^{\frac{\ln(68)}{9}t}$

A Level

A Level

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