Differential Equations

Lessons

Senior 6 · Mathematics

Differential Equations

5 lessons

Lesson 1 of 5

Forming differential equations; separable equations

35 min

Equations whose unknown is a function

In Unit 2 you met $\frac{dy}{dt} = ky$ : the rate of change of a quantity is proportional to the quantity itself. Notice what kind of statement that is. It does not give $y$ ; it gives a rule for how $y$ changes. Science is full of such rules. Newton's second law says how velocity changes under a force; Kirchhoff's laws say how current changes in a circuit; Newton's law of cooling says how fast a cup of tea loses heat. Each rule is an equation that contains a derivative, and solving it means finding the function itself. This unit is about such equations.

Definition

Differential equation: an equation that relates an unknown function to one or more of its derivatives. If the unknown depends on one variable only, it is an ordinary differential equation (ODE); all the equations in this unit are ordinary.

Order: the order of the highest derivative in the equation. Degree: the power of that highest derivative, once the equation is written as a polynomial in the derivatives (no roots or fractions of them).

Linear: $y$ and its derivatives occur only to the first power, are not multiplied together and are not inside other functions. A linear equation of order 2 has the form $a(x)\,y'' + b(x)\,y' + c(x)\,y = f(x)$ .

Example: $\frac{d^2y}{dx^2} + 4y = 0$ is an ODE of order 2 and degree 1, and it is linear.

equation	order	degree	linear?
$\frac{dy}{dx} = 3x^2$	1	1	yes
$\frac{d^2y}{dx^2} + 4y = 0$	2	1	yes
$\left(\frac{dy}{dx}\right)^2 + y = x$	1	2	no: $y^{'}$ is squared
$y\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$	2	1	no: $y$ multiplies $y^{''}$
$\left(\frac{d^2y}{dx^2}\right)^3 - x\frac{dy}{dx} = \sin x$	2	3	no
$\frac{dy}{dx} = \sin y$	1	1	no: $y$ is inside $\sin$

Notice the last row: $\sin x$ on the right of an equation is harmless, but $\sin y$ makes it non-linear, because $y$ is the unknown.

Definition

Solution: a function that turns the equation into a true statement for every $x$ in an interval. The general solution of an equation of order $n$ contains $n$ arbitrary constants; a particular solution is one member of that family, picked out by extra conditions such as an initial condition $y (x_{0}) = y_{0}$ .

Example: $y = Ce^{-x^2}$ is the general solution of $\frac{dy}{dx} = -2xy$ ; the condition $y (0) = 3$ gives the particular solution $y = 3e^{-x^2}$ .

Forming a differential equation

Going the other way, a family of curves with $n$ arbitrary constants satisfies a differential equation of order $n$ that contains no constants. To find it, differentiate $n$ times and eliminate the constants between the $n + 1$ equations.

Worked example 1: eliminating constants

Form the differential equation of (a) the family of parabolas $y = A x^{2}$ (b) the family $y = Ae^{2x} + Be^{-x}$ .

Solution

One constant, so one differentiation: $y^{'} = 2 A x$ . From the family, $A = \frac{y}{x^2}$ , so $y' = \frac{2y}{x}$ , that is $x\frac{dy}{dx} - 2y = 0 \quad (x \ne 0).$
Two constants, so differentiate twice: $y' = 2Ae^{2x} - Be^{-x}$ and $y'' = 4Ae^{2x} + Be^{-x}$ . Adding the first two equations removes $B$ : $y + y' = 3Ae^{2x}$ . Adding the last two also removes $B$ : $y' + y'' = 6Ae^{2x} = 2(y + y')$ . Hence $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = 0 .$ Check with $y = e^{2x}$ : $4 - 2 - 2 = 0$ ✓; with $y = e^{-x}$ : $1 + 1 - 2 = 0$ ✓.

Definition

Separable equation: a first order equation that can be written as

\frac{dy}{dx} = g(x)\,h(y),

a function of $x$ times a function of $y$ . Method (separating the variables): where $h(y) \ne 0$ , write $\frac{dy}{h(y)} = g(x)\,dx$ and integrate both sides:

\int \frac{dy}{h(y)} = \int g(x)\,dx + C .

Reason: if $H'(y) = \frac{1}{h(y)}$ , the chain rule gives $\frac{d}{dx}H(y) = \frac{y'}{h(y)} = g(x)$ , so $H(y) = \int g(x)\,dx + C$ . One constant is enough: constants on both sides would merge into one.

Each segment has the slope −2xy that the equation demands at that point. The general solution y = Ce^(−x²) is the whole family of curves that fit the segments; the initial condition y(0) = 3 picks out the red one.

Worked example 2: separating the variables

Solve (a) $\frac{dy}{dx} = -2xy$ with $y (0) = 3$ (b) $(1 + x^2)\frac{dy}{dx} = 1 + y^2$ with $y (0) = 1$ .

Solution

For $y \ne 0$ : $\int \frac{dy}{y} = \int -2x\,dx$ , so $\ln|y| = -x^2 + C$ and $|y| = e^{C}e^{-x^2}$ . Hence $y = Ae^{-x^2}$ , where $A = \pm e^C$ ; and $A = 0$ gives the solution $y = 0$ as well. From $y (0) = 3$ , $A = 3$ : $y = 3e^{-x^2} .$
$\int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2}$ , so $\tan^{-1}y = \tan^{-1}x + C$ . At $x = 0$ : $\tan^{-1}1 = \frac{\pi}{4} = C$ . Then $y = \tan\left(\tan^{-1}x + \frac{\pi}{4}\right)$ , and by the addition formula $y = \frac{x + 1}{1 - x}, \qquad x \lt 1 .$ The solution through $(0, 1)$ lives only on $x \lt 1$ , where it stays finite.

Warning

Dividing by $h (y)$ can lose solutions. Each value with $h (y) = 0$ is a constant solution. In part (a), dividing by $y$ hid $y = 0$ ; we recovered it by allowing $A = 0$ . Always check the zeros of $h$ separately.

Worked example 3: a draining tank (illustrative figures)

A rainwater tank has a tap at the bottom. By Torricelli's law, the depth $h$ metres of water in an upright cylindrical tank falls at a rate proportional to $\sqrt{h}$ : $\frac{dh}{dt} = -k\sqrt{h}$ . The depth is $1.44\,\mathrm{m}$ when the tap is opened and $0.81\,\mathrm{m}$ after 10 minutes. When will the tank be empty?

Solution Separate: $\int h^{-\frac{1}{2}}\,dh = \int -k\,dt$ , so $2\sqrt{h} = -kt + C$ . At $t = 0$ : $C = 2\sqrt{1.44} = 2.4$ , so $\sqrt{h} = 1.2 - \frac{k}{2}t$ . At $t = 10$ : $0.9 = 1.2 - 5 k$ , so $k = 0.06$ and

\sqrt{h} = 1.2 - 0.03t, \qquad h = (1.2 - 0.03t)^2 .

The tank is empty when $1.2 - 0.03 t = 0$ , at $t = 40$ minutes. Note that the level falls fast at first and slowly at the end, because the pressure at the tap drops as the tank empties.

Tip

To test whether an equation is separable, try to factor the right-hand side as (something in $x$ ) × (something in $y$ ). $e^{x - y} = e^x \times e^{-y}$ is separable; $x + y$ is not, and needs the methods of the next lesson.

3Blue1Brown: Differential equations, a tourist's guide (about 27 min). Watch for the pendulum: the equation tells you how the state changes from moment to moment, and the picture of arrows (a slope or vector field) shows every solution at once, like the slope field above.

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Practice

Questions 7 and 8 are harder.

State the order and degree, and whether the equation is linear: (a) $\frac{d^2y}{dx^2} + 3\frac{dy}{dx} - y = e^x$ (b) $\left(\frac{dy}{dx}\right)^3 = x + y$ (c) $y\frac{dy}{dx} = x$ (d) $\left(\frac{d^2y}{dx^2}\right)^2 + y\sin x = 0$ .
Form the differential equation of (a) $y = Ae^{3x}$ (b) $y = A\cos 2x + B\sin 2x$ (c) the circles $x^{2} + y^{2} = r^{2}$ , with $r$ arbitrary.
Show that $y = x e^{x}$ is a solution of $y^{''} - 2 y^{'} + y = 0$ .
Solve $\frac{dy}{dx} = \frac{x^2}{y}$ with $y (0) = 2$ .
Find the general solution of $\frac{dy}{dx} = e^{x - y}$ .
Solve $x\frac{dy}{dx} = y + 1$ , $x \gt 0$ , with $y (1) = 3$ .
Solve $\frac{dy}{dx} = y^2\cos x$ with $y (0) = 1$ . For which $x$ near 0 is the solution defined?
A falling stone, with air resistance, has velocity $v$ satisfying $\frac{dv}{dt} = 10 - 0.5v$ (in $\mathrm{m\,s^{-1}}$ , with $t$ in seconds) and $v (0) = 0$ . Find $v (t)$ and the speed the stone approaches (its terminal velocity).

Show answersHide answers

(a) order 2, degree 1, linear (b) order 1, degree 3, non-linear (c) order 1, degree 1, non-linear (d) order 2, degree 2, non-linear
(a) $y^{'} = 3 y$ (b) $y^{''} + 4 y = 0$ (c) $x + y\frac{dy}{dx} = 0$
$y^{'} = (1 + x) e^{x}$ , $y^{''} = (2 + x) e^{x}$ , and $(2 + x) - 2 (1 + x) + x = 0$ .
$\frac{y^2}{2} = \frac{x^3}{3} + C$ , $C = 2$ : $y = \sqrt{\frac{2}{3}x^3 + 4}$
$e^y\,dy = e^x\,dx$ : $y = \ln(e^x + C)$
$\ln|y + 1| = \ln x + C$ : $y = A x - 1$ , and $y (1) = 3$ gives $y = 4 x - 1$ .
$-\frac{1}{y} = \sin x + C$ , $C = - 1$ : $y = \frac{1}{1 - \sin x}$ , defined while $\sin x \ne 1$ , so for $-\frac{3\pi}{2} \lt x \lt \frac{\pi}{2}$ .
$\int \frac{dv}{10 - 0.5v} = \int dt$ gives $v = 20\left(1 - e^{-0.5t}\right)$ ; as $t \to \infty$ , $v \to 20\,\mathrm{m\,s^{-1}}$ .

Sources and credits

Written by Ganzaa: the explanations, the slope field (drawn from computed values), the worked examples, the practice items and the quiz questions. Every solution was checked by substituting it back into its equation numerically. The tank figures are illustrative.
Video: Differential equations, a tourist's guide | DE1, 3Blue1Brown, embedded from YouTube.

1Equations whose unknown is a function

2Forming a differential equation

Equations whose unknown is a function

Forming a differential equation