LECTURE 2: SEPARABLE DIFFERENTIAL EQUATIONS
MINGFENG ZHAO
September 11, 2015
Example 1. Verify that x(t) = Ce−2t is a solution to x0 = −2x. Find C to solve for the initial condition x(0) = 100.
In fact,
x0 = −2Ce−2t =⇒ x0 = −2x =⇒ x(t) = Ce−2t is a solution to x0 = −2x .
If we want x(0) = 100, that is, 100 = x(0) = Ce(−2)·0 = Ce0 = C, then C = 100 . In this case, we say that

 x0 = −2x
x(t) = 100e−2t is a solution to the initial value problem
.
 x(0) = 100.
Remark 1. Usually, there are many solutions to a differential equation, not just one. But if we put some extra “good”
conditions, the solution may be unique. Roughly speaking, for an nth order differential equation, if we give n “good”
conditions, the solution to this differential equation may be unique.
.
Integrals as solutions
Let’s consider the differential equation
y 0 = f (x).
By the definition of the antiderivative, we know that
Z
y=
f (x) dx.
In summary, we have
Z
0
y = f (x) =⇒ y =
f (x) dx .
Example 2. Find the general solution(i.e., all solution) of y 0 = 3x2 .
1
2
MINGFENG ZHAO
It’s easy to see that
Z
y=
Example 3. Find the general solution of y 0 =
x2
3x2 dx = x3 + C .
1
.
−1
In fact, we have
Z
y
1
dx
x2 − 1
Z
1
1
1
=
−
dx
2 x−1 x+1
=
Use the partial fraction decompositions
1
(ln |x − 1| − ln |x + 1|) + C
2
1 x − 1 + C.
ln
2 x + 1
=
=
So the general solution of y 0 =
1
is
x2 − 1
1 x − 1 +C .
y = ln 2
x + 1
Remark 2. In general, for an nth order differential equation, the general solution to this differential equation will have
n free constants.
2
Example 4. Solve y 0 = e−x , y(0) = 1.
Z
2
2
It’s easy to see that the general solution to y 0 = e−x is y = e−x dx. By the fundamental theorem of calculus, we
Z x
Z x
2
2
know that y =
e−t dt + C. Since y(0) = 1, then y =
e−t dt + 1 .
0
0
dx
Example 5. Solve
= sin(t) + t, x(0) = 20.
dt
dx
= sin(t) + t, then
Since
dt
Z
x=
[sin(t) + t] dt = − cos(t) +
t2
+ C.
2
Since x(0) = 20, then −1 + C = 20, that is, C = 21. Hence we know that
1
x = − cos(t) + t2 + 21 .
2
Example 6. Find the general solution of y 00 = x.
Let v(x) = y 0 (x), then v 0 = x. So
Z
v=
x dx =
1 2
x + C1 .
2
LECTURE 2: SEPARABLE DIFFERENTIAL EQUATIONS
3
That is,
y0 =
1 2
x + C1 .
2
So
y=
1 3
x + C1 x + C2 .
6
t
Example 7. Suppose a car drives at a speed e 2 m/s, where t is time in a second. How far did the car get in 2 seconds
(starting at t = 0)? How far in 10 seconds?
Let x(t) be the distance traveled in t seconds, then
the speed at time t = the derivative of the distance traveled at t = x0 (t).
By the assumption, we have
t
x0 (t) = e 2
t
Since x0 = e 2 , then x =
Z
t
and x(0) = 0.
t
t
e 2 dt = 2e 2 + C. Since x(0) = 0, then 2 + C = 0, that is, C = −2. So x(t) = 2e 2 − 2,
which implies that
x(2) = 2e − 2 ≈ 3.436
and
x(10) = 2e5 − 2 ≈ 294.826 .
Separable differential equations
A separable differential equation has the form of:
y0 =
dy
= f (x)g(y).
dx
There are two cases:
Case I: If g(a) = 0 for some constant a, then y(x) ≡ a is a solution.
Case II: If g(y) 6= 0, then moving g(y) to the left hand side, we get
1
dy
·
= f (x).
g(y) dx
Integrate both sides with respect to x, then
Z
Z
1
dy
·
dx = f (x) dx.
g(y) dx
By the change of variables, then
Z
1
dy =
g(y)
Z
f (x) dx.
4
MINGFENG ZHAO
One can just think the cancellation in
dy
dy
= dy.
dx
dx, that is,
dx
dx
In summary, we have



 1) y(x) ≡ a for some constant a such that g(a) = 0
0
Z
Z
y = f (x)g(y) =⇒
.
1


dy = f (x) dx.
 2)
g(y)
Z
Remark 3. In the formula
1
dy =
g(y)
Z
f (x) dx, we call y is an implicit solution to y 0 = f (x)g(y).
Example 8. Solve y 0 = ky, where k is a constant
Case I: y(x) ≡ 0 is a solution.
1
dy
= ky, then dy = k dx, which implies that ln |y| = kx + C, that is
Case II: If y 6= 0, since
dx
y
|y| = ekx+C) = eC · ekx =⇒ y = ±eC · ekx .
Since C is arbitrary constant, then ±eC can take any nonzero constant. So
y = Cekx ,
for some nonzero constant C.
In summary, the general solution of y 0 = ky is
y = Cekx .
Example 9 (Unlimited Population/Exponential Growth Model). Suppose there are 100 bacteria at time 0 and 200
bacteria 10 seconds later. How many bacteria will there will be 1 minute from time 0?
ASSUMPTION for the unlimited population/exponential growth model:
The rate of growth of the population is proportional to the size of the population.
That is,
The rate of growth of the population
= constant.
The size of the population
Let P (t) be the population of bacteria in t seconds, then
dP
= kP,
dt
for some constant k.
By the assumption, we have
P 0 = kP,
P (0) = 100
and P (10) = 200.
LECTURE 2: SEPARABLE DIFFERENTIAL EQUATIONS
By the result of Example 8, we know that the general solution of P 0 = kP is
P (t) = Cekt .
Since P (0) = 100 and P (10) = 200, that is,
P (0) = C = 100
and P (10) = Ce10k = 100e10k = 200.
So e10k = 2, then 10k = ln 2, that is,
k=
ln 2
.
10
Hence
ln 2
P (t) = 100e 10 t .
So
P (60) = 100e6 ln 2 = 100 · 26 = 6400 .
Figure 1. Bacteria growth in the first 60 seconds
Example 10. Find the general solution of y 0 = y 2 .
Case I: y ≡ 0 is a solution.
Z
Z
dy
1
1
1
2
Case II: If y 6= 0, since
= y , then 2 dy = dx, which implies that
dy =
dx, that is, − = x + C. So
dx
y
y2
y
y=−
1
.
x+C
So the general solution of y 0 = y 2 is
either y = 0, or y = −
1
.
x+C
5
6
MINGFENG ZHAO
Example 11. Solve y 0 = xy.
Case I: y(x) ≡ 0 is a solution.
Z
dy
1
x2
1
Case II: If y 6= 0, since
= xy, then dy = x dx, which implies that ln |y| =
+ C, that is, ln |y| = x2 + C. So
dx
y
2
2
x2
y = ±e 2 +C . Since C is arbitrary constant, then
y = Ce
x2
2
for some nonzero constant.
In summary, we know that the general solution of y 0 = xy is y = Ce
x2
2
.
xy
Example 12. Solve y 0 = 2
.
y +1
y
xy
=x· 2
, then
Note that y 0 = 2
y +1
y +1
Case I: y(x) ≡ 0 is a solution.
Z 2
Z
y
y2 + 1
y +1
dy
= x· 2
, then
dy = x dx, which implies that
dy =
x dx, that is,
Case II: If y 6= 0, since
dxZ
y +1
y
y
Z 1
1
1
y+
dy = x dx. So y 2 + ln |y| = x2 + C.
y
2
2
xy
0
In summary, the general solution of y = 2
is
y +1
either y ≡ 0, or
1 2
1
y + ln |y| = x2 + C .
2
2
Problems you can do:
Lebl’s Book [2]: All exercises on Page 17, and all exercises from Page 25 to Page 26. Read Example 1.1.2 on
Page 14, Example 1.1.5, Example 1.1.6 on Page 16, Example 1.3.2 on Page 24, Example 1.3.3 and Example
1.3.4 on Page 25.
Braun’s Book [1]: All exercises from Page 24 to Page 26. Read all examples from Page 20 to Page 24, and all
materials in Section 1.5 and Section 1.6 from Page 26 to Page 46.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca