An ordinary equation asks for a number. A differential equation asks for a function, described by how it changes. Its order is the highest derivative present. To verify a candidate, substitute it and its derivatives and see whether both sides agree for all x. A first-order equation has a family of solutions, one per constant, and an initial condition selects one member.
Step 1: Picture It
Start with something you already know.
A vine with a growth rule
- A gardener knows one rule about a vine.
- At every moment, it grows at a rate of one tenth of its height per week.
- The rule never says how tall the vine is on any day.
- A 10-inch vine and a 30-inch vine both follow it, along different paths.
- Tell the gardener the height at the start, and the whole summer is fixed.
The rule for the rate is the differential equation: the height h, in inches, follows dh/dt = 0.1h, with t in weeks.
The unknown is the vine’s whole record of heights, a function, not one height.
Every vine on the same rule is one member of the family, and the starting height is the initial condition that picks one.
Where the picture breaks: a real vine stops growing, but this equation keeps it growing faster forever.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
A function y has a derivative that equals 3 at every x, and its graph passes through (0, 2). What is y when x = 4?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
Why they are everywhere
You rarely know a quantity directly, but you often know the rule for its rate of change: a population grows in proportion to its size, a cup cools in proportion to how hot it is. That rule is a differential equation.
Order
dy/dx = 2y is first order. y″ + y = 0 is second order: y = sin x works, since y″ = −sin x and −sin x + sin x = 0. The order says how many constants the general solution will carry.
A candidate that fails
Is y = x² a solution of dy/dx = 2y? dy/dx = 2x, but 2y = 2x². Equal only at x = 0 and x = 1, not for every x. Not a solution. Agreement at one point is not enough.
A family
y = Ce^(−3x) + 2 solves y′ + 3y = 6 for every C: y′ = −3Ce^(−3x), and −3Ce^(−3x) + 3Ce^(−3x) + 6 = 6. One equation, infinitely many solutions, one per value of C.
An initial condition selects one
Add y(0) = 5: C + 2 = 5, so C = 3 and y = 3e^(−3x) + 2. An equation with a starting value is an initial value problem, and it typically has exactly one solution.
Read the solution
y = 3e^(−3x) + 2 starts at 5 and settles toward 2 as x grows. The equation y′ = 6 − 3y says the same: the rate is zero exactly at y = 2. Reading the equation predicts the solution before solving.
Step 3: Try It Yourself
Tap and try it out.
Verify that y = Ce^(−3x) + 2 solves y′ + 3y = 6 for every C, state the order, then apply y(0) = 5 and check
- 1y′ = −3Ce^(−3x)differentiate the candidate
Step 0 of 4
- The equationdy/dx = a·y
- Through(0, 1)
Every short line shows the slope the equation demands at that point. The curve simply follows them, and the marked point is the initial condition that picks it out of the family.
Step 4: In Real Life
A skydiver
A skydiver’s speed changes at a rate that depends on the speed itself: air resistance grows with it. The unknown is the whole speed history, not one number. That is a differential equation.
Step 5: Watch an Example
One step at a time.
Watch Ana Check a Second-Order Equation
First, a check of a second-order equation. A fresh candidate and a fresh equation: does y = 4cos(2x) solve y″ + 4y = 0? Ana differentiates twice, substitutes, and then names the order.
- Step 1
The equation contains y″, so Ana needs two derivatives of the candidate. The chain rule, with inner function 2x, gives y′ = −8sin(2x).
Watch Dev Fit a Family to a Starting Value
Now a family and a starting value, with the later steps yours. Dev is offered the family y = Ce^(2x) − 1 for the equation y′ = 2y + 2, with the starting value y(0) = 4. He takes the derivative; whether the family fits, the value of C, and where the rate is zero are yours.
- Step 1
Dev differentiates the candidate: y′ = 2Ce^(2x). The −1 is a constant, so it contributes nothing to the derivative.
Step 6: Your Turn
Practice makes it stick.
The Skydiver
Problem 1 of 17
A skydiver’s downward speed v, in feet per second, follows dv/dt = 32 − 0.2v, with t in seconds. At the moment her speed is 60 feet per second, how fast is her speed increasing?
feet per second, each second
The Savings Account
Problem 2 of 17
A savings account earns 4% a year, and its owner withdraws $300 a year, spread evenly. The balance B, in dollars, follows dB/dt = 0.04B − 300, with t in years. At what balance does the account stay exactly level?
dollars
Read the Equation
An order, a rate, a check of a candidate, and a constant from a starting value. Each one is a short calculation.
Problem 3 of 17
What is the order of y‴ + 2y′ = 7x?
Problem 4 of 17
A quantity y follows dy/dt = 6 − 2y. At the moment y = 1, what is dy/dt?
Problem 5 of 17
y = x³ is offered as a solution of x·y′ = 3y. Simplify x·y′ − 3y for this candidate. What number is left?
Problem 6 of 17
y = 2x is offered as a solution of y′ = y. The two sides agree at exactly one value of x. Which value?
Problem 7 of 17
y = Ce^(5x) + 1 solves y′ = 5y − 5 for every C. Which value of C gives y(0) = 7?
Whose Step?
One line on each page does not hold. Find the first one, then put it right.
Problem 8 of 17
Noor checked whether y = e^(−x) solves y′ + y = 0. One line does not hold. Which is the first?
Noor’s check of y = e^(−x) in y′ + y = 0
- 1y = e^(−x)
- 2y′ = e^(−x)
- 3y′ + y = e^(−x) + e^(−x) = 2e^(−x)
- Answernot 0, so not a solution
Problem 9 of 17
With the derivative of y = e^(−x) found correctly, what does y′ + y simplify to?
Problem 10 of 17
Leo found the order of (y′)³ + y″ = x. One line does not hold. Which is the first?
Leo’s order of (y′)³ + y″ = x
- 1the equation is (y′)³ + y″ = x
- 2the derivatives present are y′ and y″
- 3y′ is raised to the power 3
- Answerso the order is 3
Problem 11 of 17
What is the order of (y′)³ + y″ = x?
Problem 12 of 17
Sam checked whether y = x + 1 solves y′ = y. One line does not hold. Which is the first?
Sam’s check of y = x + 1 in y′ = y
- 1y = x + 1, so y′ = 1
- 2at x = 0: y′ = 1 and y = 0 + 1 = 1
- 3the two sides agree at x = 0
- Answerso y = x + 1 solves y′ = y
Problem 13 of 17
For Sam’s candidate y = x + 1, what is y′ − y at x = 2?
Pick the Move
Decide what to do next. Neither needs the equation solved from scratch.
Problem 14 of 17
Ravi wants to know whether y = 2e^(3x) solves y′ = 3y. Which move settles it for certain?
Problem 15 of 17
The family y = Ce^(2x) solves y′ = 2y. A problem adds y(0) = 3. What is the next move?
Same Equation?
Each item pairs candidates with equations. Check every pairing by substituting.
Problem 16 of 17
Select every function that solves y′ = 3y.
Problem 17 of 17
The constant function y = 5 is offered as a solution. Select every equation it solves.
Step 7: Quick Check
Show what you know.
Question 1 of 2
y″ + 4y′ + 3y = 0. What is the order?
Question 2 of 2
What is the unknown in a differential equation?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
For some numbers k, the function y = e^(kx) solves y″ − 5y′ + 6y = 0. Exactly two values of k work. What is the larger one?
Stretch 2 of 2
A straight line y = mx + b solves y′ = y − x for every x. What is b?
What You Learned
- A differential equation describes an unknown function by how it changes.
- Its order is the highest derivative that appears, and says how many constants to expect.
- Verify a candidate by substituting it and comparing both sides for every x.
- A first-order equation has a family of solutions, and an initial condition selects one.
For the grown-up
The misconception this lesson targets is accepting a candidate because the two sides agree at one point. It comes from ordinary equations, where one value making both sides equal is the whole answer; here the unknown is a function, and it has to make the two sides agree for every x. A student who simplifies the difference of the two sides to 0, rather than plugging in a number, will not make it. Two close cousins: reading the order from a power, so that (y′)³ is called third order, when order counts only how many times y has been differentiated; and dropping the minus sign from the chain rule on e^(−x), which can turn a real solution into an apparent failure.