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1. Differential equations: solutions and equilibria

Warm-up

Three quick questions from earlier lessons.

Discover

A differential equation links a function with its derivatives. Is $y = e^{2x}$ a solution of $y' = 2y$ ?

Key idea

A solution is a function, not a number.

A differential equation relates

y

, its derivatives and the variable. A function is a solution if substituting it makes both sides equal for every

x

. A constant solution

y = c

(an equilibrium) has

y' = 0

Worked example

Find the equilibrium of $\dfrac{dy}{dt} = 3(y - 5)$ , and check $y = 5 + 2e^{3t}$ .

3(y - 5) = 0 \Rightarrow y = 5

Constant solution: y' = 0.

y' = 6e^{3t}

Differentiate the proposed solution.

It is a solution

Right side: 3(2e^{3t}) = 6e^{3t}. Equal.

Your turn, guided

Your turn: the equilibrium of $\dfrac{dy}{dt} = 2(4 - y)$ .

y = 4

Set 2(4 − y) = 0.

Practice ★

Which function is a solution of $\dfrac{dy}{dx} = 5y$ ?
1. $y = 3e^{ 5x}$
2. $y = e^{3x}$
3. $y = 5xy$
4. $y = e^{x} + 5$
Which function is a solution of $\dfrac{dy}{dx} = 6y$ ?
1. $y = 3e^{ 6x}$
2. $y = e^{3x}$
3. $y = 6xy$
4. $y = e^{x} + 6$

$\dfrac{dy}{dt} = 5(y - 3)$ . Find the constant (equilibrium) solution $y = c$ .
$\dfrac{dy}{dt} = 8(y - 2)$ . Find the constant (equilibrium) solution $y = c$ .

Watch out!

Practice ★★

For which $k > 0$ is $y = e^{kx}$ a solution of $y'' - 9y = 0$ ?
For which $k > 0$ is $y = e^{kx}$ a solution of $y'' - 36y = 0$ ?

Exit ticket

Three questions to check you have got it.

$\dfrac{dy}{dt} = 5(y - 3)$ . Find the constant (equilibrium) solution $y = c$ .
$\dfrac{dy}{dt} = 8(y - 2)$ . Find the constant (equilibrium) solution $y = c$ .
$\dfrac{dy}{dt} = 8(y - 1)$ . Find the constant (equilibrium) solution $y = c$ .

Summary card

Differential equations

Check a solution: compute its derivatives, substitute, compare.
Equilibrium: set the rate equal to 0.

Watch out: equilibrium means $y' = 0$ , not $y = 0$ .

2. Slope fields

Warm-up

Three quick questions from earlier lessons.

Discover

For $\dfrac{dy}{dx} = x - y$ , a slope field draws a short segment at each grid point with that slope. What is the slope at $(3, 1)$ ?

2
−2

Key idea

A slope field shows the shape of all solutions without solving.

At each point

(x, y)

draw a segment of slope

f(x, y)

. Solution curves follow the segments. If

f

depends only on

x

, segments are parallel along vertical lines; only on

y

, along horizontal lines; where

f = 0

, segments are horizontal.

Worked example

Describe the field of $\dfrac{dy}{dx} = y - 2$ .

Same slope along each horizontal line

Depends only on y.

Horizontal segments along

y = 2

y − 2 = 0.

Rising above

y = 2

, falling below

Signs.

Your turn, guided

Your turn: $\dfrac{dy}{dx} = xy$ at $(2, -3)$ .

slope

-6

2 × (−3).

Practice ★

In the slope field of $\dfrac{dy}{dx} = x + 3y$ , what is the slope of the segment at $(-3, 1)$ ?
In the slope field of $\dfrac{dy}{dx} = x + 2y$ , what is the slope of the segment at $(-1, 3)$ ?
In the slope field of $\dfrac{dy}{dx} = x + 2y$ , what is the slope of the segment at $(0, -1)$ ?

Watch out!

Practice ★★

In the slope field of $\dfrac{dy}{dx} = 9x^2$ , which segments are all parallel?
1. Those along any vertical line $x = c$
2. Those along any horizontal line $y = c$
3. None: every segment has its own slope

Slope field of $\dfrac{dy}{dx} = y(3 - y)$ . Tick all the true statements.
1. Segments are horizontal along $y = 0$ and $y = 3$
2. Slopes are positive for $0 < y < 3$
3. Slopes are negative for $y > 3$
4. Slopes are positive for $y > 3$
5. Segments are horizontal along $x = 3$
Slope field of $\dfrac{dy}{dx} = y(5 - y)$ . Tick all the true statements.
1. Segments are horizontal along $y = 0$ and $y = 5$
2. Slopes are positive for $0 < y < 5$
3. Slopes are negative for $y > 5$
4. Slopes are positive for $y > 5$
5. Segments are horizontal along $x = 5$

Exit ticket

Three questions to check you have got it.

In the slope field of $\dfrac{dy}{dx} = x + 3y$ , what is the slope of the segment at $(-3, 1)$ ?
In the slope field of $\dfrac{dy}{dx} = x + 2y$ , what is the slope of the segment at $(-1, 3)$ ?
In the slope field of $\dfrac{dy}{dx} = x + 2y$ , what is the slope of the segment at $(0, -1)$ ?

Summary card

Slope fields

Slope at $(x, y)$ : substitute into $f(x, y)$ .
Only $x$ : parallel along verticals. Only $y$ : along horizontals.
$f = 0$ : horizontal segments (equilibria if constant in $y$ ).

Watch out: $(x, y)$ in that order.

3. Separation of variables and initial conditions

Warm-up

Three quick questions from earlier lessons.

Discover

To solve $\dfrac{dy}{dx} = 2xy$ , Ana integrates both sides with respect to $x$ and writes $y = x^2 y + C$ . Why is that not a solution method?

$y$ is not a constant: the variables must be separated first
It is fine, it just needs a value for $C$

Key idea

Gather each variable with its own differential.

Separable equation

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

: write

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

, integrate both sides (one constant

C

), solve for

y

, then use the initial condition to find

C

Worked example

Solve $\dfrac{dy}{dx} = 2xy$ with $y(0) = 3$ .

\dfrac{dy}{y} = 2x\,dx

Separate.

\ln|y| = x^2 + C

Integrate.

y = 3e^{x^2}

Exponentiate: y = Ae^{x²}; y(0) = 3 gives A = 3.

Your turn, guided

Your turn: $\dfrac{dy}{dx} = 3y$ , $y(0) = 2$ .

y = 2e^{3x}

ln|y| = 3x + C, so y = Ae^{3x}.

Practice ★

Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 5$ .
Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 9$ .
Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 2$ .

Watch out!

Practice ★★

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 8$ (so $y > 0$ ). Give $y$ in terms of $x$ .
Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 6$ (so $y > 0$ ). Give $y$ in terms of $x$ .

Solve $\dfrac{dy}{dx} = y^2$ with $y(0) = \dfrac{1}{ 3 }$ .

Boss problem. Solve $\dfrac{dy}{dx} = 1x^2 y$ with $y(0) = 4$ .

Exit ticket

Three questions to check you have got it.

Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 5$ .
Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 9$ .
Solve $\dfrac{dy}{dx} = 4xy$ with $y(0) = 2$ .

Summary card

Separation of variables

$\dfrac{dy}{h(y)} = g(x)\,dx$ .
Integrate both sides, one $+C$ .
Solve for $y$ ; use the initial value.

Watch out: $e^{x + C} = Ae^x$ , not $e^x + C$ .

4. Exponential growth and decay

Warm-up

Three quick questions from earlier lessons.

Discover

The blue curve is $y = a\,e^{kt}$ , a solution of $y' = ky$ with $y(0) = a$ . Move $k$ through 0.

Solutions y = a e^(kt) of y' = ky. For k > 0 they grow faster and faster; for k < 0 they decay towards 0; k = 0 gives a constant.

For $k < 0$ , what happens to $y$ ?

It decays towards 0
It falls in a straight line and crosses 0

Key idea

Rate proportional to amount means exponential.

\dfrac{dy}{dt} = ky

with

y(0) = y_0

has the solution

y = y_0 e^{kt}

k > 0

: growth, doubling time

\tfrac{\ln 2}{k}

k < 0

: decay, half-life

\tfrac{\ln 2}{|k|}

Worked example

Bacteria double every 3 hours, starting from 500. How many after 12 hours?

500 \cdot 2^4

12 hours = 4 doubling times.

= 8000

Compute.

y = 500e^{(\ln 2/3)t}

Equivalently k = ln 2 / 3.

Your turn, guided

Your turn: half-life 5 years, find $k$ .

k = -\dfrac{\ln 2}{5}

e^{5k} = ½.

Practice ★

A population grows with $\dfrac{dP}{dt} = kP$ . It starts at 104 and doubles every 9 years. What is it after 27 years?
A population grows with $\dfrac{dP}{dt} = kP$ . It starts at 469 and doubles every 4 years. What is it after 12 years?

Solve $\dfrac{dy}{dt} = -2y$ with $y(0) = 5$ .
Solve $\dfrac{dy}{dt} = 4y$ with $y(0) = 8$ .

Watch out!

Practice ★★

A substance decays with $\dfrac{dm}{dt} = km$ and has a half-life of 27 years. Find $k$ (exact).
A substance decays with $\dfrac{dm}{dt} = km$ and has a half-life of 25 years. Find $k$ (exact).

Exit ticket

Three questions to check you have got it.

Solve $\dfrac{dy}{dt} = -2y$ with $y(0) = 5$ .
Solve $\dfrac{dy}{dt} = 4y$ with $y(0) = 8$ .
Solve $\dfrac{dy}{dt} = 3y$ with $y(0) = 4$ .

Summary card

Exponential models

$y' = ky$ , $y(0) = y_0 \Rightarrow y = y_0e^{kt}$ .
Decay: $k < 0$ ; $k = -\ln 2 / (\text{half-life})$ .

Watch out: equal times multiply by equal factors.

5. Newton's law of cooling

Warm-up

Three quick questions from earlier lessons.

Discover

A cup at temperature $b$ °C cools in a 20 °C room. The violet line is the room temperature. Move $b$ and the cooling constant $k$ .

Newton's law of cooling: T = 20 + (b − 20)e^(−kt). Whatever the start, the temperature approaches the room's 20 °C, faster for larger k.

Which temperature does the cup approach?

20 °C
0 °C

Key idea

The difference with the surroundings decays exponentially.

Newton's law of cooling:

\dfrac{dT}{dt} = -k(T - A)

, with

A

the surrounding temperature and

k > 0

. Solution:

T = A + (T_0 - A)e^{-kt}

Worked example

Tea at 90 °C in a 20 °C room, $k = 0.1$ per minute.

T - 20 = 70e^{-0.1t}

u = T − 20 satisfies u' = −0.1u.

T = 20 + 70e^{-0.1t}

So:

It starts cooling at 7 °C per minute

At t = 0 the rate is −0.1 × 70.

Your turn, guided

Your turn: soup at 80 °C in a 25 °C room, $k = 0.2$ . The rate at $t = 0$ ?

-11

°C/min

−0.2 × (80 − 25).

Practice ★

Soup at 60 °C is left in a 17 °C kitchen and follows Newton's law of cooling. Which temperature does it approach in the long run?

$\dfrac{dT}{dt} = -\tfrac{ 4 }{10}(T - 25)$ and $T(0) = 74$ . How fast is the temperature changing at $t = 0$ , in °C/min?
$\dfrac{dT}{dt} = -\tfrac{ 4 }{10}(T - 22)$ and $T(0) = 92$ . How fast is the temperature changing at $t = 0$ , in °C/min?

Watch out!

Practice ★★

A drink at 65 °C cools in a 18 °C room: $\dfrac{dT}{dt} = -\tfrac{ 3 }{10}(T - 18)$ . Find $T(t)$ .
A drink at 89 °C cools in a 16 °C room: $\dfrac{dT}{dt} = -\tfrac{ 2 }{10}(T - 16)$ . Find $T(t)$ .

Exit ticket

Three questions to check you have got it.

A drink at 65 °C cools in a 18 °C room: $\dfrac{dT}{dt} = -\tfrac{ 3 }{10}(T - 18)$ . Find $T(t)$ .
A drink at 89 °C cools in a 16 °C room: $\dfrac{dT}{dt} = -\tfrac{ 2 }{10}(T - 16)$ . Find $T(t)$ .
A drink at 79 °C cools in a 15 °C room: $\dfrac{dT}{dt} = -\tfrac{ 2 }{10}(T - 15)$ . Find $T(t)$ .

Summary card

Newton's law of cooling

$T' = -k(T - A)$ .
$T = A + (T_0 - A)e^{-kt}$ , approaching $A$ .

Watch out: it is the difference $T - A$ that decays.

Answers: Differential equations

1. Differential equations: solutions and equilibria

$y = 3e^{ 5x}$
$y = 3e^{ 6x}$

2. Slope fields

$0$
$5$
$-2$

Those along any vertical line $x = c$

Segments are horizontal along $y = 0$ and $y = 3$ · Slopes are positive for $0 < y < 3$ · Slopes are negative for $y > 3$
Segments are horizontal along $y = 0$ and $y = 5$ · Slopes are positive for $0 < y < 5$ · Slopes are negative for $y > 5$

$0$
$5$
$-2$

3. Separation of variables and initial conditions

$5 e^{2 x^{2}}$
$9 e^{2 x^{2}}$
$2 e^{2 x^{2}}$

$\sqrt{x^{2} + 64}$
$\sqrt{x^{2} + 36}$

$\frac{1}{3 - x}$

$4 e^{\frac{1}{3}}$

$5 e^{2 x^{2}}$
$9 e^{2 x^{2}}$
$2 e^{2 x^{2}}$

4. Exponential growth and decay

$832$
$3752$

$5 e^{- 2 t}$
$8 e^{4 t}$

$- \frac{\ln{\left(2 \right)}}{27}$
$- \frac{\ln{\left(2 \right)}}{25}$

$5 e^{- 2 t}$
$8 e^{4 t}$
$4 e^{3 t}$

5. Newton's law of cooling

$17$

$- \frac{98}{5}$
$-28$

$18 + 47 e^{- \frac{3 t}{10}}$
$16 + 73 e^{- \frac{t}{5}}$

$18 + 47 e^{- \frac{3 t}{10}}$
$16 + 73 e^{- \frac{t}{5}}$
$15 + 64 e^{- \frac{t}{5}}$