Introduction to Differential Equations: Types, Examples

What Is A Differential Equation?

In any introduction to differential equations, the first definition is this: a differential equation is an equation that relates an unknown function to one or more of its derivatives. Where an algebra equation asks "which number works?", a differential equation asks "which function works?". The answer you are hunting for is a whole function.

The general form of a differential equation in one unknown function $y = y(x)$ is

$$F\left(x,\ y,\ \frac{dy}{dx},\ \frac{d^2y}{dx^2},\ \ldots,\ \frac{d^ny}{dx^n}\right) = 0$$

where:

• $x$ is the independent variable (often time $t$ in applied problems)

• $y$ is the unknown function, also called the dependent variable

• $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, up to $\frac{d^ny}{dx^n}$ are its first, second and $n$th derivatives

• $F$ is any expression that combines them, and the equation must hold for every $x$ in some interval, not just at one point

Take the simplest case, $\frac{dy}{dx} = 2x$. It says "the slope of $y$ at every point equals twice the $x$-coordinate". Integrating gives $y = x^2 + C$, where $C$ is any constant.

Compare that with the algebra equation $x^2 = 4$, whose answers are the two numbers $2$ and $-2$. If the word "derivative" still feels shaky, a quick refresher on the derivative of a function makes the rest of this page much easier.

Why is it called a differential equation? The name comes from Gottfried Leibniz's notation. He wrote small changes as $dx$ and $dy$ and called them differentials, so an equation built from $\frac{dy}{dx}$ and its relatives became a "differential" equation. The derivative is the differential quotient, and the name stuck.

How Do You Read A Differential Equation?

Read a differential equation as a sentence: the left side is a rate of change, and the right side says what drives that rate. Once you can say an equation out loud in plain words, you know what kind of behaviour its solutions will have before you solve anything.

Two habits make reading easier. First, find the unknown function and the variable it depends on; in $\frac{dT}{dt}$, the unknown is temperature $T$ and the variable is time $t$. Second, look at the sign of the right side, because a positive right side means the quantity is increasing at that moment and a negative one means it is falling.

Here is the cooling equation checked against a real solution. With a starting temperature of 90 degrees, the solution is $T = 20 + 70e^{-0.1t}$, and after 10 minutes

$$T(10) = 20 + 70e^{-1} \approx 45.7516$$

so the tea has fallen to about 45.75 degrees, still above the room and still sliding toward it, exactly as the sentence predicted.

What Is The Order Of A Differential Equation?

The order of a differential equation is the order of the highest derivative that appears in it. The order tells you how many arbitrary constants the general solution carries, which is why it is the first thing to check.

The degree is the power of that highest derivative, but only when the equation is a polynomial in its derivatives. For $\left(\frac{d^2y}{dx^2}\right)^3 + \frac{dy}{dx} = x$, the order is 2 and the degree is 3. For $\sin\left(\frac{dy}{dx}\right) + y = 0$, the degree is not defined at all, because the derivative sits inside a sine. Order and degree have their own detailed page on order and degree of a differential equation, with a longer set of drills.

What Are The Types Of Differential Equations?

Differential equations are sorted along a few independent lines: ordinary or partial, linear or nonlinear, and (for linear equations) homogeneous or non-homogeneous. Each label tells you which solution methods are available, so classifying comes before solving. The table gives two examples of each type, the request that turns up again and again on homework forums.

A linear equation can still have messy coefficients in $x$: the test is only about how $y$ appears. That is why $x^2\frac{d^2y}{dx^2} + y = \cos x$ is linear while the innocent-looking $\frac{dy}{dx} = y^2$ is not.

One warning about vocabulary. The word "homogeneous" means two different things in this subject: the "no forcing term" sense in the table, and a separate first-order sense where $\frac{dy}{dx}$ depends only on $\frac{y}{x}$. Even textbooks switch between the two without saying so, and it is a genuinely confusing pair of meanings. The deeper classification of ODEs (autonomous equations, systems, existence of solutions) lives on the ordinary differential equations page.

What Does It Mean To Solve A Differential Equation?

To solve a differential equation is to find a function that makes the equation true for every value of the variable in some interval. You check a candidate by substituting it, together with its derivatives, into both sides; if the two sides agree identically, it is a solution.

Solutions come in two kinds:

• General solution: the whole family, written with arbitrary constants. A first-order equation usually carries one constant, a second-order equation two, and so on. For $\frac{dy}{dx} = 2x$ the general solution is $y = x^2 + C$, and the $+ C$ is the constant of integration you met in integral calculus.

• Particular solution: one member of that family, picked out by extra information. Given $y(1) = 4$, the family $y = x^2 + C$ forces $C = 3$, so the particular solution is $y = x^2 + 3$.

A differential equation together with a starting value is called an initial value problem. Geometrically, the general solution is a stack of curves and the initial value is a point; the particular solution is the one curve passing through that point. The full treatment, including when exactly one curve passes through each point, is on the page on initial value problems. (The guarantee behind "exactly one" is the Picard-Lindelöf theorem, a name you will meet again in a later course.)

Back to the 1961 weather model for a moment. The equations there were the general rule; the measured weather on the first day was the initial value; and changing that initial value in the fourth decimal place picked out a different curve. Same family, different member.

Which Methods Solve Differential Equations?

No single method solves every differential equation; you pick a method from the equation's type. The table is the roadmap for a first course, with each method's own lesson linked rather than repeated here.

The last row matters more than it looks. Most differential equations that model the real world, the 1961 weather equations included, have no solution formula at all, and scientists solve them numerically on computers. A slope field gives you the shape of every solution before you know a single formula.

Examples Of Differential Equations

Example 1

Find the order of each equation and decide whether it is linear: (a) $\frac{dy}{dx} + 3y = x^2$, (b) $\frac{d^2y}{dx^2} + y\frac{dy}{dx} = 0$, (c) $\frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2}$.

(a) The highest derivative is $\frac{dy}{dx}$, so the order is 1. The function $y$ appears only to the first power, so the equation is linear.

(b) The highest derivative is $\frac{d^2y}{dx^2}$, so the order is 2. The term $y\frac{dy}{dx}$ multiplies $y$ by its own derivative, so the equation is nonlinear.

(c) This is a partial differential equation (two variables, $x$ and $t$). The highest derivative is the second partial in $x$, so the order is 2, and it is linear.

Final answer: (a) order 1, linear; (b) order 2, nonlinear; (c) PDE, order 2, linear.

Example 2

Is $y = 1 + x$ a solution of $\frac{dy}{dx} = y^2$?

A tempting shortcut is to test a single point. At $x = 0$:

$$\frac{dy}{dx} = 1$$

$$y^2 = (1 + 0)^2 = 1$$

Both sides equal 1, so it looks like a solution. Now test $x = 1$:

$$\frac{dy}{dx} = 1$$

$$y^2 = (1 + 1)^2 = 4$$

The two sides disagree, so the one-point check was misleading. A solution must work for every $x$ in an interval, so substitute symbolically instead:

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

$$= -x^2 - 2x$$

This is zero only at $x = 0$ and $x = -2$, not on any interval. So $y = 1 + x$ is not a solution. The function that does work through the point $(0, 1)$ is $y = \frac{1}{1 - x}$ for $x < 1$, since

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

Final answer: No. $y = 1 + x$ satisfies the equation only at isolated points; $y = \frac{1}{1 - x}$ is the solution through $(0, 1)$.

Example 3

Show that $y = C_1\cos x + C_2\sin x$ solves $\frac{d^2y}{dx^2} + y = 0$, then find the solution with $y(0) = 2$ and $\frac{dy}{dx}(0) = 3$.

Differentiate twice:

$$\frac{dy}{dx} = -C_1\sin x + C_2\cos x$$

$$\frac{d^2y}{dx^2} = -C_1\cos x - C_2\sin x$$

Add $y$:

$$\frac{d^2y}{dx^2} + y = 0$$

so the family works for every $C_1$ and $C_2$. Apply the two conditions:

$$y(0) = C_1 = 2$$

$$\frac{dy}{dx}(0) = C_2 = 3$$

A second-order equation needed two conditions to fix two constants, exactly as its order predicted.

Final answer: $y = 2\cos x + 3\sin x$.

Example 4

Solve the initial value problem $\frac{dy}{dx} = 6x^2 - 4x$ with $y(1) = 5$.

Integrate both sides directly:

$$y = \int (6x^2 - 4x),dx$$

$$y = 2x^3 - 2x^2 + C$$

Use the initial condition:

$$5 = 2(1)^3 - 2(1)^2 + C$$

$$C = 5$$

Final answer: $y = 2x^3 - 2x^2 + 5$.

Example 5

Find the differential equation whose general solution is $y = Ae^{2x} + Be^{-x}$, where $A$ and $B$ are arbitrary constants.

Two constants mean the equation will be second order. Differentiate twice:

$$\frac{dy}{dx} = 2Ae^{2x} - Be^{-x}$$

$$\frac{d^2y}{dx^2} = 4Ae^{2x} + Be^{-x}$$

Now look for a combination that cancels both constants. Subtract the first derivative and twice $y$:

$$\frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = (4A - 2A - 2A)e^{2x} + (B + B - 2B)e^{-x}$$

$$= 0$$

Final answer: $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = 0$.

Example 6

A drip delivers a medicine into the bloodstream at 3 mg per hour, and the body clears it at a rate of 0.5 times the amount present. Write the differential equation, check that $M = 6(1 - e^{-0.5t})$ solves it with $M(0) = 0$, and find the amount after 2 hours.

Rate of change equals rate in minus rate out:

$$\frac{dM}{dt} = 3 - 0.5M$$

Check the candidate:

$$\frac{dM}{dt} = 3e^{-0.5t}$$

$$3 - 0.5M = 3 - 3(1 - e^{-0.5t}) = 3e^{-0.5t}$$

The two sides match, and $M(0) = 6(1 - 1) = 0$. After 2 hours:

$$M(2) = 6(1 - e^{-1}) \approx 3.7927$$

As $t$ grows, $e^{-0.5t}$ shrinks to 0 and $M$ levels off at 6 mg, the amount where the drip and the clearance balance.

Final answer: $\frac{dM}{dt} = 3 - 0.5M$; about 3.7927 mg after 2 hours, levelling off at 6 mg.

Why Do Differential Equations Exist?

Differential equations exist because nature hands us rules about change, not formulas for amounts. "Nature tells us how things change, not where they end." Newton's second law says how velocity changes; Newton's law of cooling says how temperature changes; a growth law says how a population changes. Turning those rules into predictions is exactly the job of a differential equation, and Isaac Newton was already sorting such equations into kinds in 1671, as the history of the differential equation records.

That one idea gives differential equations three jobs:

• Predict: start from today's state and the rule of change, and compute tomorrow's state (weather, orbits, drug levels).

• Explain: a law such as "rate proportional to amount" explains why bacteria, savings and radioactive samples all follow the same exponential curve.

• Design: engineers choose a spring stiffness or a circuit resistance so that the solution of the equation behaves the way they want.

Algebra could not do this. An algebra equation pins down a value; a differential equation pins down a whole history from a rule and a starting point.

Where Are Differential Equations Used In Real Life?

Differential equations are used wherever a quantity changes at a rate you can describe. A few of the places they run quietly in the background:

• Physics: Newton's second law, $F = m\frac{d^2x}{dt^2}$, is a second-order differential equation for position.

• Heat: Joseph Fourier's 1822 heat equation describes how temperature spreads through a solid; it is a PDE.

• Medicine: drug-dosing schedules come from rate-in, rate-out equations like Example 6.

• Epidemiology: models of infection spread track how the numbers of susceptible, infected and recovered people change.

• Electronics: the current in a circuit with a resistor and an inductor follows a first-order linear equation.

• Weather: forecasts solve huge systems of differential equations numerically, the descendants of the 1961 model.

Where Do Students Trip Up On Differential Equations?

Most of the trouble in a first course comes from carrying algebra habits into a subject where the answer is a function. These are the four errors that cost the most.

Mistake 1: Expecting A Number As The Answer

Where it slips in: The first time a student sees $\frac{dy}{dx} = 2x$, especially when it looks like an algebra exercise.

Don't do this: Do not "solve for $x$" and report $x = 0$ or any other number.

The correct way: Look for a function: $y = x^2 + C$ is the answer, and every value of $C$ gives a valid solution.

The memorizer is the student most likely to fall here, because years of "solve for $x$" drills have trained the reflex to isolate a variable. The fix is to say out loud, before starting, "I am looking for a function."

Mistake 2: Calling An Equation Linear Because It Looks Simple

Where it slips in: Classifying equations such as $\frac{dy}{dx} = y^2$ or $\frac{d^2y}{dx^2} + y\frac{dy}{dx} = 0$.

Don't do this: Do not judge linearity by how short the equation is or by whether $x$ appears with powers.

The correct way: Check only how $y$ appears: first power, never multiplied by itself or its derivatives, never inside $\sin$, $e$ or a root. Powers of $x$ in the coefficients are allowed.

Mistake 3: Assigning A Degree That Does Not Exist

Where it slips in: Order-and-degree questions where a derivative sits inside a function, such as $\sin\left(\frac{dy}{dx}\right) + y = 0$.

Don't do this: Do not write "degree 1" just because the highest derivative is not raised to a visible power.

The correct way: Say the degree is not defined, because the equation is not a polynomial in its derivatives. The order (here 1) is still defined.

Mistake 4: Dropping The Constant, Then Ignoring The Initial Condition

Where it slips in: Integrating to get a general solution, then answering an initial value problem.

Don't do this: Do not write $y = x^2$ for $\frac{dy}{dx} = 2x$ and stop; without $+ C$ there is no way to meet a condition such as $y(1) = 4$.

The correct way: Keep the constant, then use the condition to fix it: $y = x^2 + 3$.

Students who arrive straight from a long run of integration drills tend to drop the $+ C$ out of habit, and the problem only surfaces later when the initial condition cannot be satisfied. The rusher meets a related version: plugging in the condition before integrating, which fixes the wrong constant. Write the general solution first, every time.

The real-world version of Mistake 4 is the story this page opened with. In 1961, meteorologist Edward Lorenz restarted a weather simulation from printed values that differed from the stored ones in the fourth decimal place, and the forecast diverged completely; his 1963 paper "Deterministic Nonperiodic Flow" launched the study of chaos. The equations were right. Treating the initial condition as a detail was the mistake.

Practice Problems On Introduction To Differential Equations

1. State the order of $\frac{d^3y}{dx^3} - 4\frac{dy}{dx} = \sin x$ and whether it is linear.

2. Is $y = e^{4x}$ a solution of $\frac{dy}{dx} = 4y$?

3. Is $y = x^3$ a solution of $x\frac{dy}{dx} = 2y$?

4. Solve $\frac{dy}{dx} = 4x^3 + 2$ with $y(0) = -1$.

5. Find the differential equation of the family $y = Cx^3$.

6. A quantity satisfies $\frac{dQ}{dt} = -0.2Q$ with $Q(0) = 50$. Check that $Q = 50e^{-0.2t}$ is the solution and find $Q(5)$ to 4 decimal places.

Answer to Problem 1: order 3, linear.

Answer to Problem 2: yes, since $\frac{dy}{dx} = 4e^{4x} = 4y$.

Answer to Problem 3: no, since $x \cdot 3x^2 = 3x^3$, which is not $2x^3$.

Answer to Problem 4: $y = x^4 + 2x - 1$.

Answer to Problem 5: $x\frac{dy}{dx} = 3y$.

Answer to Problem 6: $\frac{dQ}{dt} = -10e^{-0.2t} = -0.2Q$ and $Q(0) = 50$; $Q(5) = 50e^{-1} \approx 18.3940$.

The Short Version

• A differential equation relates an unknown function to its derivatives, and its answer is a function, not a number.

• Read every differential equation as a sentence: the left side is a rate of change and the right side is what drives it.

• Order (highest derivative), ODE or PDE, and linear or nonlinear are the labels that decide which method applies.

• An introduction to differential equations always comes back to one check: substitute the candidate and its derivatives, and both sides must agree on a whole interval.

• The general solution is a family of curves, and an initial condition picks the single curve you need.

Your Next Move

Work through Practice Problems 1 to 6 in order; if Problem 3 or Problem 6 trips you up, return to "What Does It Mean To Solve A Differential Equation?" and redo the substitution check from Example 2 line by line.

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