Introduction to Differential Equations

A differential equation is an equation that involves a derivative, like $\dfrac{dP}{dt} = 0.03P$. Instead of telling you what a quantity is, it tells you how the quantity changes. That is often exactly what we know about the real world: populations grow faster when there are more animals, and hot soup cools faster when it is much hotter than the room. This page shows how to turn words into a differential equation and how to check whether a function solves one.

Key ideas

What a differential equation is

A differential equation (often shortened to “DE”) relates a function to one or more of its derivatives. Some examples:

$$\frac{dy}{dx} = 2x \qquad\quad \frac{dy}{dx} = x - y \qquad\quad \frac{dP}{dt} = kP \qquad\quad y'' + 9y = 0$$

The unknown is a function, not a number. Solving a DE means finding a function (or a whole family of functions) that makes the equation true for every value of the input.

Writing a differential equation from words

The phrase “the rate of change of $Q$” means the derivative $\dfrac{dQ}{dt}$ (if $Q$ changes with time $t$). “Is proportional to” means “equals a constant $k$ times”.

Words	Differential equation
The rate of change of $P$ is proportional to $P$.	$\dfrac{dP}{dt} = kP$
The rate of change of $y$ with respect to $x$ is proportional to the square of $y$.	$\dfrac{dy}{dx} = ky^2$
The rate of change of $T$ is proportional to the difference between $T$ and $20$.	$\dfrac{dT}{dt} = k(T - 20)$
The rate of change of $h$ is inversely proportional to $h$.	$\dfrac{dh}{dt} = \dfrac{k}{h}$
The rate of change of $y$ is jointly proportional to $x$ and $y$.	$\dfrac{dy}{dx} = kxy$

The sign of $k$ carries meaning: if the quantity is increasing, $k \gt 0$; if it is decreasing, $k \lt 0$. Some books write a decreasing quantity as $-k$ with $k \gt 0$ instead. Either way is fine, as long as you say which.

Verifying a solution

To check whether a function is a solution, substitute it into the differential equation:

1. Find the derivatives the equation needs ($y'$, and $y''$ if it appears).
2. Substitute $y$ and its derivatives into both sides.
3. Simplify. If the two sides are equal for every $x$, the function is a solution.

You don’t need to know how to solve the equation to check a solution. Checking is just differentiating and simplifying.

General and particular solutions

A differential equation usually has infinitely many solutions. For example, every function $y = x^2 + C$ solves $\dfrac{dy}{dx} = 2x$, because the derivative of any constant is $0$.

• The general solution is the whole family, written with an arbitrary constant: $y = x^2 + C$.
• A particular solution is one specific member, picked out by an initial condition such as $y(1) = 3$. Here $3 = 1^2 + C$ gives $C = 2$, so the particular solution is $y = x^2 + 2$.

Every curve $y = x^2 + C$ solves $\frac{dy}{dx} = 2x$. The initial condition $y(1) = 3$ picks out one of them.

A differential equation together with an initial condition is called an initial value problem. You will learn to find general and particular solutions yourself in separation of variables.

Worked examples

Example 1: From words to an equation

Water drains from a tank so that the rate of change of the depth $h$ (in metres) with respect to time $t$ (in minutes) is proportional to the square root of the depth. Write a differential equation for $h$.

Solution. “The rate of change of the depth” is $\dfrac{dh}{dt}$. “Proportional to the square root of the depth” means $k\sqrt{h}$:

$$\frac{dh}{dt} = k\sqrt{h}$$

The tank is draining, so $h$ is decreasing and $k \lt 0$. (Equivalently, $\dfrac{dh}{dt} = -k\sqrt{h}$ with $k \gt 0$.)

Example 2: Verifying a solution

Show that $y = 3e^{2x} - 1$ is a solution of $\dfrac{dy}{dx} = 2y + 2$.

Solution. Find the derivative of the proposed solution:

$$\frac{dy}{dx} = 6e^{2x}$$

Now substitute $y$ into the right side:

$$2y + 2 = 2(3e^{2x} - 1) + 2 = 6e^{2x} - 2 + 2 = 6e^{2x}$$

Both sides equal $6e^{2x}$ for every $x$, so $y = 3e^{2x} - 1$ is a solution.

Example 3: A second-order equation

Show that $y = \sin(3x) + \cos(3x)$ solves $y'' + 9y = 0$. (Angles are in radians, as always in calculus.)

Solution. Differentiate twice using the chain rule:

$$\begin{aligned} y' &= 3\cos(3x) - 3\sin(3x) \\ y'' &= -9\sin(3x) - 9\cos(3x) \end{aligned}$$

Substitute into the left side:

$$y'' + 9y = -9\sin(3x) - 9\cos(3x) + 9\big(\sin(3x) + \cos(3x)\big) = 0$$

The left side equals the right side, $0$, so it is a solution. The same steps show that $y = 2\sin(3x)$ and $y = -5\cos(3x)$ are solutions too: this equation has infinitely many solutions.

Example 4: Finding a constant that works

For which values of $k$ is $y = e^{kx}$ a solution of $y'' - y' - 6y = 0$?

Solution. With $y = e^{kx}$, we have $y' = ke^{kx}$ and $y'' = k^2e^{kx}$. Substitute:

$$k^2e^{kx} - ke^{kx} - 6e^{kx} = e^{kx}(k^2 - k - 6) = 0$$

Since $e^{kx}$ is never $0$, we need

$$k^2 - k - 6 = 0 \quad\Rightarrow\quad (k - 3)(k + 2) = 0 \quad\Rightarrow\quad k = 3 \ \text{or}\ k = -2$$

Check $k = 3$: $9e^{3x} - 3e^{3x} - 6e^{3x} = 0$. ✓

Common mistakes

Checking only one side, or only one point. A function is a solution only if both sides agree for every $x$ in an interval. Plugging in $x = 0$ and getting a match is not enough. Simplify both sides into the same expression.

Forgetting the chain rule. The derivative of $e^{2x}$ is $2e^{2x}$, and the derivative of $\sin(3x)$ is $3\cos(3x)$. Missing the inner derivative makes a true solution look wrong.

Mixing up “proportional to” and “equal to”. “The rate is proportional to $P$” is $\dfrac{dP}{dt} = kP$, not $\dfrac{dP}{dt} = P$. Always include the constant $k$.

Writing the quantity instead of its rate. “The rate of change of $T$ is proportional to $T - 20$” describes $\dfrac{dT}{dt}$, so the equation is $\dfrac{dT}{dt} = k(T - 20)$, not $T = k(T - 20)$.

Thinking a DE has only one answer. Without an initial condition, there is a whole family of solutions. Write the general solution with its constant, and only find the constant when you are given a point.

Practice

1. (Warm-up) The rate of change of the mass $m$ of a radioactive sample with respect to time $t$ is proportional to the mass. Write a differential equation. What is the sign of the constant?

Solution

$$\frac{dm}{dt} = km$$

The sample is decaying, so $m$ is decreasing and $k \lt 0$.

2. (Warm-up) Show that $y = 5e^{-3t}$ is a solution of $\dfrac{dy}{dt} = -3y$.

Solution

Left side: $\dfrac{dy}{dt} = -15e^{-3t}$.

Right side: $-3y = -3(5e^{-3t}) = -15e^{-3t}$.

The sides are equal for every $t$, so it is a solution.

3. (Core) A cup of coffee cools in a room kept at $21^\circ\text{C}$. The rate of change of the coffee’s temperature $T$ (in degrees Celsius) with respect to time $t$ (in minutes) is proportional to the difference between $T$ and the room temperature. Write a differential equation, and explain the sign of the constant while the coffee is hotter than the room.

Solution

$$\frac{dT}{dt} = k(T - 21)$$

While the coffee is hotter than the room, $T - 21 \gt 0$ and the coffee is cooling, so $\dfrac{dT}{dt} \lt 0$. A positive number times $k$ is negative, so $k \lt 0$.

4. (Core) Show that $y = \dfrac{1}{x + C}$ is a solution of $\dfrac{dy}{dx} = -y^2$ for every constant $C$. Then find the particular solution with $y(0) = \dfrac{1}{2}$.

Solution

Write $y = (x + C)^{-1}$. Then

$$\frac{dy}{dx} = -(x + C)^{-2} = -\frac{1}{(x + C)^2} = -\left(\frac{1}{x + C}\right)^2 = -y^2$$

So it is a solution for every $C$. For the initial condition:

$$\frac{1}{0 + C} = \frac{1}{2} \quad\Rightarrow\quad C = 2$$

The particular solution is $y = \dfrac{1}{x + 2}$.

5. (Core) Show that $y = xe^x$ is a solution of $y'' - 2y' + y = 0$.

Solution

By the product rule:

$$\begin{aligned} y' &= e^x + xe^x \\ y'' &= e^x + (e^x + xe^x) = 2e^x + xe^x \end{aligned}$$

Substitute:

$$(2e^x + xe^x) - 2(e^x + xe^x) + xe^x = 2e^x + xe^x - 2e^x - 2xe^x + xe^x = 0$$

The left side is $0$, so $y = xe^x$ is a solution.

6. (Core) Which of these are solutions of $\dfrac{dy}{dx} = \dfrac{y}{x}$ (for $x \ne 0$)?

• (a) $y = 4x$
• (b) $y = x + 1$
• (c) $y = -x$

Solution

(a) $\dfrac{dy}{dx} = 4$ and $\dfrac{y}{x} = \dfrac{4x}{x} = 4$. Yes, a solution.

(b) $\dfrac{dy}{dx} = 1$ but $\dfrac{y}{x} = \dfrac{x + 1}{x} = 1 + \dfrac{1}{x}$, which is not $1$. Not a solution.

(c) $\dfrac{dy}{dx} = -1$ and $\dfrac{y}{x} = \dfrac{-x}{x} = -1$. Yes, a solution.

7. (Core) Show that $y = Ce^x - x - 1$ solves $\dfrac{dy}{dx} = x + y$ for every constant $C$. Then find the particular solution through the point $(0, 2)$.

Solution

Left side: $\dfrac{dy}{dx} = Ce^x - 1$.

Right side: $x + y = x + Ce^x - x - 1 = Ce^x - 1$.

The sides match, so it is a solution for every $C$. At $(0, 2)$:

$$2 = Ce^0 - 0 - 1 = C - 1 \quad\Rightarrow\quad C = 3$$

The particular solution is $y = 3e^x - x - 1$.

8. (Challenge) Find all values of $r$ for which $y = x^r$ is a solution of $x^2y'' - 2xy' - 4y = 0$ (for $x \gt 0$).

Solution

With $y = x^r$: $y' = rx^{r-1}$ and $y'' = r(r - 1)x^{r-2}$. Substitute:

$$\begin{aligned} x^2 \cdot r(r - 1)x^{r-2} - 2x \cdot rx^{r-1} - 4x^r &= 0 \\ x^r\big(r(r - 1) - 2r - 4\big) &= 0 \\ x^r(r^2 - 3r - 4) &= 0 \end{aligned}$$

Since $x^r \gt 0$ for $x \gt 0$, we need $r^2 - 3r - 4 = (r - 4)(r + 1) = 0$, so $r = 4$ or $r = -1$.

Check $r = 4$: $x^2(12x^2) - 2x(4x^3) - 4x^4 = 12x^4 - 8x^4 - 4x^4 = 0$. ✓

9. (Challenge) Show that $y = \tan(x + C)$ solves $\dfrac{dy}{dx} = 1 + y^2$ for every constant $C$ (radians). Find the particular solution with $y(0) = 1$, and give the largest open interval containing $x = 0$ on which it is defined.

Solution

$$\frac{dy}{dx} = \sec^2(x + C) = 1 + \tan^2(x + C) = 1 + y^2$$

using the identity $\sec^2\theta = 1 + \tan^2\theta$. So it is a solution for every $C$.

For $y(0) = 1$: $\tan C = 1$, so $C = \dfrac{\pi}{4}$ and $y = \tan\left(x + \dfrac{\pi}{4}\right)$.

Tangent is defined between consecutive vertical asymptotes, so we need $-\dfrac{\pi}{2} \lt x + \dfrac{\pi}{4} \lt \dfrac{\pi}{2}$, which gives

$$-\frac{3\pi}{4} \lt x \lt \frac{\pi}{4}$$