Differential Equations

A2 · P3 · 4 min

A differential equation relates a quantity to its rate of change. P3 covers the separable first-order type: rearrange so all the $y$ terms are with $dy$ and all the $x$ terms with $dx$ , integrate both sides, and use a given condition to fix the constant. Many questions start with a sentence in words that you must first turn into the equation.

Forming the equation

Key result

"The rate of change of $P$ is proportional to $P$ " means $\dfrac{dP}{dt} = kP$ .

"The rate at which the water level falls is proportional to the square root of the depth" means $\dfrac{dh}{dt} = -k\sqrt{h}$ with $k > 0$ .

Introduce a constant of proportionality; its sign carries the direction (a decrease is negative).

Solving by separating the variables

Method

Write the equation as $g(y)\,\dfrac{dy}{dx} = f(x)$ .
Integrate both sides with respect to $x$ : $\displaystyle\int g(y) \, dy = \int f(x) \, dx$ .
Add one constant $c$ (to one side only).
Use the initial condition to find $c$ .
Rearrange to the form asked for, often $y = \ldots$

A basic separable equation

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ given that $y = 2$ when $x = 0$ .

Solution

$\displaystyle\int y \, dy = \int x \, dx \Rightarrow \tfrac{1}{2}y^2 = \tfrac{1}{2}x^2 + c$ .

$y = 2, x = 0$ : $2 = c$ . So $y^2 = x^2 + 4$ , and since $y > 0$ initially, $y = \sqrt{x^2 + 4}$ .

Exponential growth

The population $P$ of a colony satisfies $\dfrac{dP}{dt} = 0.05P$ . Initially $P = 400$ . Find $P$ in terms of $t$ and the time for the population to reach $1000$ .

Solution

$\displaystyle\int \frac{1}{P} \, dP = \int 0.05 \, dt \Rightarrow \ln P = 0.05t + c$ . At $t = 0$ : $c = \ln 400$ .

$\ln\dfrac{P}{400} = 0.05t \Rightarrow P = 400e^{0.05t}$ .

$1000 = 400e^{0.05t} \Rightarrow t = 20\ln 2.5 \approx 18.3$ .

Newton's law of cooling

A drink at $80^\circ$ cools in a room at $20^\circ$ so that $\dfrac{d\theta}{dt} = -k(\theta - 20)$ . After 5 minutes the temperature is $60^\circ$ . Find $\theta$ as a function of $t$ .

Solution

$\displaystyle\int \frac{1}{\theta - 20} \, d\theta = -\int k \, dt \Rightarrow \ln(\theta - 20) = -kt + c$ .

At $t = 0$ , $\theta = 80$ : $c = \ln 60$ . So $\theta - 20 = 60e^{-kt}$ .

At $t = 5$ , $\theta = 60$ : $40 = 60e^{-5k} \Rightarrow k = -\tfrac{1}{5}\ln\tfrac{2}{3} = \tfrac{1}{5}\ln 1.5$ .

$\theta = 20 + 60e^{-0.0811t}$ .

Using partial fractions

Solve $\dfrac{dy}{dx} = y(1 - y)$ given $y = \tfrac{1}{2}$ when $x = 0$ , giving $y$ in terms of $x$ .

Solution

$\displaystyle\int \frac{1}{y(1 - y)} \, dy = \int dx$ . Partial fractions: $\dfrac{1}{y(1 - y)} = \dfrac{1}{y} + \dfrac{1}{1 - y}$ .

$\ln y - \ln(1 - y) = x + c \Rightarrow \ln\dfrac{y}{1 - y} = x + c$ . At $(0, \tfrac{1}{2})$ : $c = 0$ .

$\dfrac{y}{1 - y} = e^x \Rightarrow y = e^x(1 - y) \Rightarrow y = \dfrac{e^x}{1 + e^x}$ .

Forming and solving from words

A tank drains so that the depth $h$ metres decreases at a rate proportional to $\sqrt{h}$ . Initially $h = 4$ and after 10 minutes $h = 1$ . Find how long it takes to empty.

Solution

$\dfrac{dh}{dt} = -k\sqrt{h}$ . Separate: $\displaystyle\int h^{-1/2} \, dh = -\int k \, dt \Rightarrow 2\sqrt{h} = -kt + c$ .

$t = 0$ , $h = 4$ : $c = 4$ . $t = 10$ , $h = 1$ : $2 = -10k + 4 \Rightarrow k = 0.2$ .

$2\sqrt{h} = 4 - 0.2t$ ; empty when $h = 0$ : $t = 20$ minutes.

Interpreting the solution

Questions end by asking what happens as $t \to \infty$ (a limiting value), whether a quantity ever reaches a target, or how the model behaves for large $t$ . Answer from the solved equation: exponentials with negative exponent tend to zero; $\dfrac{e^x}{1 + e^x} \to 1$ ; and so on.

Watch out

Separate before integrating: $\dfrac{dy}{dx} = xy$ becomes $\dfrac{1}{y}\,dy = x\,dx$ . Integrating $xy$ with respect to $x$ as if $y$ were constant is wrong. Also, $\ln y = x + c$ gives $y = Ae^x$ with $A = e^c$ , not $y = e^x + c$ .

Exam tip

"Find the general solution" means keep the constant. "Find the particular solution" means use the condition. "Express $y$ in terms of $x$ " means finish with $y =$ ; a solution left as $\ln y = \ldots$ loses the last mark.

Practice

Question

Solve $\dfrac{dy}{dx} = 3x^2 y$ with $y = 2$ at $x = 0$ .
Solve $\dfrac{dy}{dx} = \dfrac{\cos x}{y^2}$ with $y = 1$ at $x = 0$ .
Solve $x\dfrac{dy}{dx} = y + 1$ with $y = 0$ at $x = 1$ , for $x > 0$ .
The mass $m$ of a substance decays so that $\dfrac{dm}{dt} = -km$ . It halves every 8 hours. Find $k$ and the time for $m$ to fall to $10\%$ of its initial value.

Answers

$\ln y = x^3 + \ln 2 \Rightarrow y = 2e^{x^3}$ .
$\tfrac{1}{3}y^3 = \sin x + \tfrac{1}{3} \Rightarrow y = \sqrt[3]{3\sin x + 1}$ .
$\ln(y + 1) = \ln x + c$ , $c = 0$ : $y = x - 1$ .
$m = m_0 e^{-kt}$ ; $\tfrac{1}{2} = e^{-8k} \Rightarrow k = \tfrac{\ln 2}{8} \approx 0.0866$ ; $0.1 = e^{-kt} \Rightarrow t = \dfrac{\ln 10}{k} \approx 26.6$ hours.

Forming the equation

Solving by separating the variables

Interpreting the solution

Practice

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