A differential equation states how a quantity changes and leaves the quantity itself as the unknown. Many laws of science take this form, because a rate is often easier to describe than an amount.
Differential equations Differential equation
A differential equation is an equation that relates a function of an independent variable to one or more of the function's derivatives. Its order is the order of the highest derivative that appears in it.
Here are three differential equations:
d y d x = cos x − x , d Q d t = Q 2 t 2 + 1 , y ′ ′ + x 2 y = 0. \begin{gathered}
\frac{dy}{dx} = \cos x - x, \\[6pt]
\frac{dQ}{dt} = \frac{Q^2}{t^2 + 1}, \\[6pt]
y'' + x^2 y = 0.
\end{gathered} d x d y = cos x − x , d t d Q = t 2 + 1 Q 2 , y ′′ + x 2 y = 0. The first two are first order and the third is second order. In the first, the right side depends on
x x x alone, so its solutions are the antiderivatives of
cos x − x \cos x - x cos x − x . In the other two, the unknown function itself appears in the equation, so an antiderivative alone won't find it.
Writing a differential equation from words A statement about a rate becomes a differential equation phrase by phrase. These phrases come up again and again.
Words Mathematics the rate of change of Q Q Q t t t d Q d t \dfrac{dQ}{dt} d t d Q proportional to X X X k X kX k X inversely proportional to X X X k X \dfrac{k}{X} X k proportional to the product of X X X Y Y Y k X Y kXY k X Y proportional to the difference between X X X Y Y Y k ( X − Y ) k(X - Y) k ( X − Y ) the acceleration of an object at position s s s d 2 s d t 2 \dfrac{d^2 s}{dt^2} d t 2 d 2 s
Here
k k k is a constant, the constant of proportionality. The context fixes its sign. An increasing quantity has a positive rate, and a decreasing one has a negative rate. Writing
− k -k − k with
k > 0 k > 0 k > 0 makes a decrease visible in the equation.
Example 1
Write a differential equation for each statement, with
k k k a positive constant. Find
k k k when the data fix it.
(a) A silicon wafer sits in an oxidation furnace. The thickness
h h h of its oxide layer, in nanometers, increases at a rate inversely proportional to the thickness, where
t t t is measured in minutes. When the layer is
8 8 8 nanometers thick, it's growing at
1.5 1.5 1.5 nanometers per minute.
(b) A skydiver falls with downward velocity
v ( t ) v(t) v ( t ) meters per second, where
t t t is in seconds. Gravity alone would increase
v v v at
9.8 9.8 9.8 meters per second per second. Air resistance reduces that rate by an amount proportional to the square of the velocity.
(c) A curve
y = f ( x ) y = f(x) y = f ( x ) is shaped so that, as
x x x increases, its slope changes at a rate proportional to the slope and opposite to it in sign.
(a) The rate is
d h d t \frac{dh}{dt} d t d h , and "inversely proportional to the thickness" gives
k h \frac{k}{h} h k . The layer is growing, so the sign is positive:
d h d t = k h . \frac{dh}{dt} = \frac{k}{h}. d t d h = h k . At
h = 8 h = 8 h = 8 the rate is
1.5 1.5 1.5 , so
1.5 = k 8 1.5 = \frac{k}{8} 1.5 = 8 k and
k = 12 k = 12 k = 12 . The model is
d h d t = 12 h \frac{dh}{dt} = \frac{12}{h} d t d h = h 12 , a first-order equation.
(b) Gravity contributes
9.8 9.8 9.8 to
d v d t \frac{dv}{dt} d t d v , and air resistance subtracts
k v 2 kv^2 k v 2 :
d v d t = 9.8 − k v 2 . \frac{dv}{dt} = 9.8 - kv^2. d t d v = 9.8 − k v 2 . This equation is first order, and its right side depends on
v v v alone.
(c) The slope is
d y d x \frac{dy}{dx} d x d y , and the rate at which it changes is the second derivative
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y . The two have opposite signs, so the equation carries a minus sign:
d 2 y d x 2 = − k d y d x . \frac{d^2y}{dx^2} = -k\,\frac{dy}{dx}. d x 2 d 2 y = − k d x d y . The minus sign covers both cases. Where the slope is negative,
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y is positive. This equation is second order. In terms of the slope
p = d y d x p = \frac{dy}{dx} p = d x d y , it's the first-order equation
d p d x = − k p \frac{dp}{dx} = -kp d x d p = − k p .
Solutions To show that a function is a solution, differentiate it. Then evaluate each side of the equation separately and show that they agree. An initial condition
y ( x 0 ) = y 0 y(x_0) = y_0 y ( x 0 ) = y 0 gives one value of the solution, and checking it is a separate step. A free-response answer shows each of these steps.
Example 2
The oxide layer in Example 1(a) is
8 8 8 nanometers thick at time
t = 0 t = 0 t = 0 , so its thickness satisfies
d h d t = 12 h \frac{dh}{dt} = \frac{12}{h} d t d h = h 12 with
h ( 0 ) = 8 h(0) = 8 h ( 0 ) = 8 .
(a) Show that
h ( t ) = 24 t + 64 h(t) = \sqrt{24t + 64} h ( t ) = 24 t + 64 satisfies the differential equation with the initial condition
h ( 0 ) = 8 h(0) = 8 h ( 0 ) = 8 .
(b) Find the thickness of the layer after
1.5 1.5 1.5 minutes.
(c) Find the rate at which the layer is growing at
t = 1.5 t = 1.5 t = 1.5 .
(a) The initial condition holds, since
h ( 0 ) = 64 = 8 h(0) = \sqrt{64} = 8 h ( 0 ) = 64 = 8 . By the chain rule, the derivative is
d h d t = 1 2 ( 24 t + 64 ) − 1 / 2 ⋅ 24 = 12 24 t + 64 . \begin{aligned}
\frac{dh}{dt} &= \frac{1}{2}(24t + 64)^{-1/2} \cdot 24 \\[4pt]
&= \frac{12}{\sqrt{24t + 64}}.
\end{aligned} d t d h = 2 1 ( 24 t + 64 ) − 1/2 ⋅ 24 = 24 t + 64 12 . The right side of the equation is
12 h = 12 24 t + 64 . \frac{12}{h} = \frac{12}{\sqrt{24t + 64}}. h 12 = 24 t + 64 12 . The two sides agree for every
t ≥ 0 t \ge 0 t ≥ 0 , so
h h h satisfies
d h d t = 12 h \frac{dh}{dt} = \frac{12}{h} d t d h = h 12 with
h ( 0 ) = 8 h(0) = 8 h ( 0 ) = 8 .
(b) h ( 1.5 ) = 36 + 64 = 10 h(1.5) = \sqrt{36 + 64} = 10 h ( 1.5 ) = 36 + 64 = 10 nanometers.
(c) The differential equation gives the rate from the thickness alone:
d h d t = 12 10 = 1.2 \frac{dh}{dt} = \frac{12}{10} = 1.2 d t d h = 10 12 = 1.2 nanometers per minute. No derivative of the formula is needed.
The thickness h(t) = √(24t + 64), starting at 8 nm. The tangent line at t = 1.5 has slope 12/10 = 1.2. General and particular solutions For
d y d x = cos x − x \frac{dy}{dx} = \cos x - x d x d y = cos x − x , the general solution is
y = sin x − x 2 2 + C y = \sin x - \frac{x^2}{2} + C y = sin x − 2 x 2 + C . The condition
y ( 0 ) = 4 y(0) = 4 y ( 0 ) = 4 gives
C = 4 C = 4 C = 4 , so the particular solution is
y = sin x − x 2 2 + 4 y = \sin x - \frac{x^2}{2} + 4 y = sin x − 2 x 2 + 4 .
Example 3
Consider the differential equation
d y d x = cos x − sin x − y \frac{dy}{dx} = \cos x - \sin x - y d x d y = cos x − sin x − y .
(a) Show that
y = cos x + C e − x y = \cos x + Ce^{-x} y = cos x + C e − x is a solution for every constant
C C C .
(b) Find the particular solution with
y ( 0 ) = 3 y(0) = 3 y ( 0 ) = 3 .
(c) Use the differential equation to find the slope of that particular solution at
x = 0 x = 0 x = 0 .
(a) Differentiate, treating
C C C as a constant:
d y d x = − sin x − C e − x . \frac{dy}{dx} = -\sin x - Ce^{-x}. d x d y = − sin x − C e − x . The right side of the equation is
cos x − sin x − y = cos x − sin x − ( cos x + C e − x ) = − sin x − C e − x . \begin{aligned}
&\cos x - \sin x - y \\[4pt]
&\quad = \cos x - \sin x - \left(\cos x + Ce^{-x}\right) \\[4pt]
&\quad = -\sin x - Ce^{-x}.
\end{aligned} cos x − sin x − y = cos x − sin x − ( cos x + C e − x ) = − sin x − C e − x . The two sides agree for every
x x x and every value of
C C C , so each member of the family is a solution on
( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ ) .
(b) Substitute
x = 0 x = 0 x = 0 and
y = 3 y = 3 y = 3 :
3 = cos 0 + C e 0 = 1 + C 3 = \cos 0 + Ce^0 = 1 + C 3 = cos 0 + C e 0 = 1 + C , so
C = 2 C = 2 C = 2 . The particular solution is
y = cos x + 2 e − x y = \cos x + 2e^{-x} y = cos x + 2 e − x .
(c) At
( 0 , 3 ) (0, 3) ( 0 , 3 ) the equation gives
d y d x = cos 0 − sin 0 − 3 = − 2 \frac{dy}{dx} = \cos 0 - \sin 0 - 3 = -2 d x d y = cos 0 − sin 0 − 3 = − 2 .
Members of the family y = cos x + Ce⁻ˣ for C = −2, −1, 0, 1, and 2. Each one crosses the y-axis at 1 + C and approaches cos x as x increases. The particular solution, C = 2, passes through (0, 3) with slope −2. Equations of higher order Checking a solution of a second-order equation takes two derivatives. Being a solution is a property of the whole function, so a small change to a solution can break it.
Example 4
Consider the differential equation
x 2 y ′ ′ − 2 x y ′ − 4 y = 0 x^2 y'' - 2xy' - 4y = 0 x 2 y ′′ − 2 x y ′ − 4 y = 0 for
x > 0 x > 0 x > 0 .
(a) Show that
y = x 4 y = x^4 y = x 4 is a solution.
(b) Show that
y = 1 x y = \frac{1}{x} y = x 1 is a solution.
(c) Determine whether
y = x 4 + 1 x y = x^4 + \frac{1}{x} y = x 4 + x 1 and
y = x 4 + 1 y = x^4 + 1 y = x 4 + 1 are solutions.
(a) Here
y ′ = 4 x 3 y' = 4x^3 y ′ = 4 x 3 and
y ′ ′ = 12 x 2 y'' = 12x^2 y ′′ = 12 x 2 . The left side is
x 2 ( 12 x 2 ) − 2 x ( 4 x 3 ) − 4 x 4 = 12 x 4 − 8 x 4 − 4 x 4 = 0. \begin{aligned}
&x^2\left(12x^2\right) - 2x\left(4x^3\right) - 4x^4 \\[4pt]
&\quad = 12x^4 - 8x^4 - 4x^4 = 0.
\end{aligned} x 2 ( 12 x 2 ) − 2 x ( 4 x 3 ) − 4 x 4 = 12 x 4 − 8 x 4 − 4 x 4 = 0. (b) Write
y = x − 1 y = x^{-1} y = x − 1 , so
y ′ = − x − 2 y' = -x^{-2} y ′ = − x − 2 and
y ′ ′ = 2 x − 3 y'' = 2x^{-3} y ′′ = 2 x − 3 . The left side is
x 2 ( 2 x − 3 ) − 2 x ( − x − 2 ) − 4 x − 1 = 2 x + 2 x − 4 x = 0. \begin{aligned}
&x^2\left(2x^{-3}\right) - 2x\left(-x^{-2}\right) - 4x^{-1} \\[4pt]
&\quad = \frac{2}{x} + \frac{2}{x} - \frac{4}{x} = 0.
\end{aligned} x 2 ( 2 x − 3 ) − 2 x ( − x − 2 ) − 4 x − 1 = x 2 + x 2 − x 4 = 0. The function
1 x \frac{1}{x} x 1 isn't defined at
0 0 0 . It's a solution on
x > 0 x > 0 x > 0 and on
x < 0 x < 0 x < 0 , but on no interval that contains
0 0 0 .
(c) For
y = x 4 + 1 x y = x^4 + \frac{1}{x} y = x 4 + x 1 , each derivative is the sum of the derivatives in (a) and (b). Each term of the left side is a power of
x x x times
y y y ,
y ′ y' y ′ , or
y ′ ′ y'' y ′′ . So the left side for the sum is the sum of the two left sides,
0 + 0 = 0 0 + 0 = 0 0 + 0 = 0 , and this function is a solution.
For
y = x 4 + 1 y = x^4 + 1 y = x 4 + 1 , the derivatives are the same as for
x 4 x^4 x 4 , but the last term changes:
x 2 ( 12 x 2 ) − 2 x ( 4 x 3 ) − 4 ( x 4 + 1 ) = − 4 ≠ 0. \begin{aligned}
&x^2\left(12x^2\right) - 2x\left(4x^3\right) - 4\left(x^4 + 1\right) \\[4pt]
&\quad = -4 \ne 0.
\end{aligned} x 2 ( 12 x 2 ) − 2 x ( 4 x 3 ) − 4 ( x 4 + 1 ) = − 4 = 0. So
y = x 4 + 1 y = x^4 + 1 y = x 4 + 1 isn't a solution. The constant survives because the equation contains
y y y itself, in the term
− 4 y -4y − 4 y .