An ordinary equation asks for a number. A differential equation asks for a whole function, and all it tells you is how that function changes: its slope, its rate of growth, its acceleration. From that rule you rebuild the function. That is how physics predicts a falling stone, how banks grow money continuously, and how doctors know when a medicine has worn off.
A differential equation tells you how an unknown function changes. Your job is to find the function itself.
An equation with derivatives of a dependent variable y with respect to an independent variable x is a differential equation. With only one independent variable it is ordinary: the only kind in this chapter.
Then y‴ for the third derivative, and yₙ for the nth derivative.
A first-order equation says: at the point (x, y) the unknown curve must have slope F(x, y). Draw that slope as a little dash at many points: a slope field. Solving means finding curves that follow it.
Two labels describe a differential equation: its order (the highest derivative in it) and its degree (the power of that derivative).
The order is the order of the highest derivative present. Here that is y‴, so the order is 3.
The degree is the highest power of that highest-order derivative. Order and degree, when defined, are positive integers.
The degree exists only when the equation is a polynomial in y′, y″, … Functions of x or of y itself are allowed, but not sin(y′), ey′, … So this one has order 3 and no degree.
Order and degree of y′ − cos x = 0; xy y″ + x(y′)² − y y′ = 0; y‴ + y² + e^{y′} = 0.
A solution is a function that makes the equation true. Usually a whole family of functions works, one for each value of the constants.
A solution makes the equation true for every x when it and its derivatives are substituted. y = a sin(x + b) solves this one for all values of a and b.
A solution with arbitrary constants is the general solution. It has as many constants as the order: here two, for a second-order equation.
Choosing values for the constants gives a particular solution. A condition such as “y = 1 when x = 0” picks it out of the family.
y = e^{−3x} solves y″ + y′ − 6y = 0.
y = a cos x + b sin x solves y″ + y = 0.
When the slope is an x part times a y part, pull them apart: all the y’s on one side, all the x’s on the other. Then integrate.
The slope is a function of x alone times a function of y alone. Then the variables are separable.
Continuous growth: the rate is proportional to the amount.
How to use it
dy/dx = (x + 1)/(2 − y).
dy/dx = (1 + y²)/(1 + x²).
dy/dx = −4xy², y(0) = 1.
x dy = (2x² + 1) dx through (1, 1).
Slope 2x/y² through (−2, 3).
Principal grows continuously at 5%.
In a homogeneous equation the slope depends only on the ratio y/x. Put y = vx and the equation becomes separable.
Scale x and y by the same λ and F scales by λⁿ. Then F is a homogeneous function of degree n.
The equation is homogeneous when F has degree 0. Then the slope is the same all along each line y = vx through the origin.
If the equation has this form, put x = vy instead.
How to use it
(x − y) dy/dx = x + 2y.
x cos(y/x) dy/dx = y cos(y/x) + x.
2y e^{x/y} dx + (y − 2x e^{x/y}) dy = 0, x = 0 when y = 1.
Slope (x² + y²)/(2xy).
Multiply a linear equation by the right factor and its left side becomes one single derivative. One integration then solves it.
A first-order linear equation, with P and Q constants or functions of x only.
Multiplying by the integrating factor turns the left side into the derivative of y · IF.
Swap the roles of x and y: then x · IF = ∫(Q₁ × IF) dy + C.
How to use it
dy/dx − y = cos x.
x dy/dx + 2y = x².
y dx − (x + 2y²) dy = 0.
dy/dx + y cot x = 2x + x² cot x, y(π/2) = 0.
Slope x + xy through (0, 1).
y = e^{ax}(c₁ cos bx + c₂ sin bx) solves y″ − 2ay′ + (a² + b²)y = 0.
log(dy/dx) = 3x + 4y, y(0) = 0.
(x dy − y dx) y sin(y/x) = (y dx + x dy) x cos(y/x).
(tan⁻¹y − x) dy = (1 + y²) dx.
Fifteen quick questions to check yourself. Tap an option to see whether it is right, and why.
Every MCQ from Exercises 9.1–9.5 and the miscellaneous exercise, plus three concept checks.
A differential equation involves derivatives of the dependent variable. Order = highest derivative; degree = its power, defined only for polynomial equations in the derivatives.
A solution satisfies the equation identically. The general solution has as many arbitrary constants as the order; a particular solution has none.
Separable: dy/h(y) = g(x) dx, integrate both sides.
Homogeneous (dy/dx = f(x, y) with f of degree 0): put y = vx (or x = vy) and separate.
Linear dy/dx + Py = Q: IF = e^{∫P dx}, y · IF = ∫Q · IF dx + C.
Wrote ∫y dy = ½y², introducing the symbols ∫ and dy; this is often called the birthday of differential equations.
Found the methods of separating variables, solving homogeneous first-order equations, and solving linear first-order equations.
James Bernoulli called the answer the "integral" of the equation; John Bernoulli named separation of variables.
Championed the word "solution" and started the qualitative study of equations whose solutions cannot be written down.
A differential equation tells you the slope of the unknown curve at every point. Each walkthrough draws that as a slope field of little dashes, solves the equation step by step, then draws the whole family of solution curves (one for each value of C) and, when a condition is given, the one particular curve through the given point. The last step checks the answer on the picture: the tangent to the solution has exactly the slope the equation demands. Order-and-degree examples light up a derivative ladder; "verify" examples plot LHS − RHS and show it is flat at 0. Press Next step (or Play); each step is explained in detail under the controls, with links to the basics it uses.
Order is the highest derivative present; degree is its power, when the equation is a polynomial in the derivatives.
Differentiate the function, substitute, and check that LHS = RHS for every x.
Put all the y with dy and all the x with dx, integrate both sides, and use a condition to find C.
When dy/dx depends only on y/x, the substitution y = vx makes the equation separable.
dy/dx + Py = Q: multiply by e^{∫P dx} and the left side becomes the derivative of y × IF.
A verification, a separable equation in disguise, a trigonometric homogeneous equation and a linear equation in x.
A cooling cup of tea, money in a bank, a growing culture of bacteria, medicine leaving the blood and salt mixing into a tank: in each one we only know how fast the quantity changes. A differential equation turns that into a formula for every moment.
Tea at 90 °C cools in a 20 °C room. Newton's law says it loses heat at a rate proportional to the temperature difference: dT/dt = −k(T − 20). Separating gives T = 20 + 70e^{−kt}. It is also linear: dT/dt + kT = 20k, settling at 20 °C.
With continuous compounding dP/dt = rP, so P = P₀e^{rt}. Compounding once a year gives P₀(1 + r)ᵗ, slightly less. Compare them.
dN/dt = kN with e^{2k} = 1.1. The count N = 1,00,000 e^{kt} doubles after 2 log 2/log 1.1 ≈ 14.5 hours, and doubles again every 14.5 hours after that.
The body removes a medicine at a rate proportional to the amount present: dA/dt = −kA, so A = 500e^{−kt} mg for a 500 mg dose. The half-life log 2/k is how long it takes to halve.
A 100 L tank of fresh water receives brine (2 g of salt per litre) at 5 L/min, and the mixture drains at 5 L/min. The salt S grams changes at dS/dt = 10 − S/20: in minus out. It is linear, IF = e^{t/20}, and with S(0) = 0 the solution is S = 200(1 − e^{−t/20}).
Decide first, then tap each card.
Every question that asks you to find, verify, show or solve has an animated walkthrough here; the 12 multiple-choice questions are in the Lesson’s self-check. Pick a question from the chips above each player, then press Play or step through with Next. Solving questions end with the slope check: the solution curve's tangent has exactly the slope the equation demands. Verifying questions end with LHS − RHS plotted flat at 0.
Light up every derivative, read off the highest, then check whether the equation is a polynomial in the derivatives. (Q11–Q12 are in the quiz.)
Differentiate, substitute, and show LHS = RHS. For implicit functions, differentiate implicitly. (Q11–Q12 are in the quiz.)
General solutions, particular solutions through a point, curves from a slope rule, and growth problems. (Q23 is in the quiz.)
Check that dy/dx depends only on y/x (or dx/dy only on x/y), substitute, separate. (Q16–Q17 are in the quiz.)
Write in the form dy/dx + Py = Q (or dx/dy + P₁x = Q₁), find IF = e^{∫P dx}, integrate Q × IF. (Q18–Q19 are in the quiz.)
Order and degree, verification, a proof, and every method of the chapter. (Q13–Q15 are in the quiz.)
Solving differential equations uses nearly everything from the calculus chapters. Here are the eight ideas it rests on, each from its roots: derivatives and their notation, the integration toolkit, why solutions come with arbitrary constants, reading dy/dx as a slope field, treating dy and dx as separate pieces, homogeneous functions, the product rule run backwards, and exponential growth. Every example and exercise links here exactly where you need it, and a button takes you straight back to the problem.
For a curve y = f(x), the first derivative is its slope and the second derivative is how that slope changes.
The first derivative is the slope of the tangent at x: the rate at which y changes.
The second derivative is the derivative of y′: how fast the slope itself changes (it tells whether the curve bends up or down). Then y‴, y⁗, and yₙ for the nth.
When y is only known implicitly, like this, differentiate both sides and remember that every y brings a factor y′ (the chain rule).
Rules you will use
Plus the product, quotient and chain rules. They all come from Chapter 5.
Solving a differential equation is mostly integration. You need the standard integrals, substitution, partial fractions and integration by parts, all from Chapter 7. Two you will meet constantly: ∫dx/x = log|x| (which turns into exponentials when you undo the log), and ∫dx/(1 + x²) = tan⁻¹x.
xⁿ⁺¹/(n + 1) for n ≠ −1; log|x| for n = −1. ∫g′/g dx = log|g|.
e^{kx}/k, −cos x, sin x, tan x; ∫tan x dx = log|sec x|, ∫cot x dx = log|sin x|.
tan⁻¹x and sin⁻¹x; with a²: (1/a)tan⁻¹(x/a) and sin⁻¹(x/a).
t = g(x); partial fractions for rational functions; ∫u v = u∫v − ∫(u′∫v); ∫eˣ[f + f′] = eˣf.
y″ = 0 says the slope never changes. Integrate once: y′ = a. Integrate again: y = ax + b. Two integrations, two arbitrary constants: every straight line solves y″ = 0. That is why a general solution of an nth-order equation has n constants, and why you need n conditions (a point, a slope, …) to pick one particular solution. A particular solution has no arbitrary constants left.
The derivative of a curve at a point is the slope of its tangent there. A first-order equation dy/dx = F(x, y) tells you that slope at every point in the plane, before you know the curve. Draw a short dash with slope F(x, y) at many points: a slope field. A solution is a curve that is tangent to the dashes everywhere, like a leaf floating along a stream. Through each starting point there is one such curve.
Leibniz's notation lets us treat dy and dx like numbers. Take a tiny step dx along the tangent; the tangent rises dy = y′ dx.
The first line can be rewritten as the second, a statement about tiny steps. Integrating adds up the steps on both sides.
This is exactly the substitution rule from Chapter 7, written compactly.
Scale x and y by the same λ. If the value only gets multiplied by a power of λ, the function is homogeneous.
It is homogeneous of degree 2, because every term has total power 2.
Scaling does nothing to it, so it depends only on the ratio y/x. That is the root of the substitution y = vx: along any line through the origin v = y/x is fixed, so F is fixed.
This one is not homogeneous at all.
The integrating factor comes straight from the product rule, read backwards.
The first line is the product rule. The two lines are equal when g′ = Pg.
That g is the integrating factor: multiplying dy/dx + Py = Q by it turns the left side into a single derivative (y · g)′, which one integration undoes.
If a quantity grows at a rate proportional to its size, it grows exponentially. Negative k gives decay (cooling, medicine leaving the blood).
Separate, integrate, then undo the log.
log and e undo each other
Useful facts
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