MA 1506
Mathematics II
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MA 1506 Mathematics II
Tutorial 3
Ngo Quoc Anh
Second order differential equations
Question 1
Groups: B03 & B08
Question 2
February 8, 2012
Question 3
Question 4
Ngo Quoc Anh
Department of Mathematics
National University of Singapore
1/15
Question 1: General solution of homogeneous second
order linear DE with constant coefficients y 00 + ay 0 + by = 0
General solution can be found using the so-called
characteristic equation. Following is the method of solving.
MA 1506
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Method of solving
By solving the characteristic equation λ2 + aλ + b = 0, there
are three cases depending on the sign of a2 − 4b.
. If a2 − 4b > 0, then the characteristic equation has two
distinct real roots, say λ1 and λ2 . The general solution
of the associated ODE is
Question 1
Question 2
Question 3
Question 4
yh = c1 eλ1 x + c2 eλ2 x .
. If a2 − 4b = 0, then the characteristic equation has a
real double root λ1 = λ2 = − a2 . The general solution of
the associated ODE is
a
yh = (c1 + c2 x)e− 2 x .
2/15
Question 1: General solution of homogeneous second
order linear DE with constant coefficients y 00 + ay 0 + by = 0
MA 1506
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Method of solving (cont’)
. If a2 − 4b < 0, then the characteristic equation has two
complex roots, say λ1 and λ2 , where
√
√
a
−a ± a2 − 4b
4b − a2
=− ±
λ1,2 =
i.
2
2 | {z
2 }
w
Question 1
Question 2
Question 3
Question 4
The general solution of the associated ODE is
a
yh = c1 cos(wx) + c2 sin(wx) e− 2 x .
As you may see that general solution y of a 2nd order ODE
basically depends on two parameters (c1 and c2 as shown
above). (What happens to the 1st order ODEs?)
In order to specify these numbers, we need two initial
conditions involving, for example, y(x0 ) and y 0 (x0 ).
3/15
Question 1: General solution of homogeneous second
order linear DE with constant coefficients y 00 + ay 0 + by = 0
(a) The characteristic equation λ2 + 6λ + 9 = 0 admits a
double real root λ = −3. Thus, the general solution of the
associated ODE is given as follows
yh = (c1 + c2 x)e−3x .
−1 = y (0 = −3c1 e
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Question 1
Question 2
Since 1 = y(0) = c1 and
0
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Question 3
−3x
+ c2 e
−3x
− 3c2 xe
−3x Question 4
x=0
= −3c1 + c2 ,
we get that c1 = 1 and c2 = 2, thus, yh = (1 + 2x)e−3x .
(b) The characteristic equation has two complex roots
λ1,2 = 1 ± 2πi. Hence, we find that
yh = c1 cos(2πx) + c2 sin(2πx) ex .
Using y(0) = −2 and y 0 (0) = 2(3π − 1) we obtain c1 = −2
and c2 = 3, thus, yh = − 2 cos(2πx) + 3 sin(2πx) ex .
4/15
Question 2: Finding yp : The method of undetermined
coefficients
This is also known as the lucky guess method. In order to
find the particular integral, we need to guess its form, with
some coefficients left as variables to be solved for.
Below is a table of some typical functions r(x) and the
solution yp to guess for them.
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Question 1
Question 2
Question 3
Form of r(x)
Form for yp
ax
ke
Ceax
n
n
kx , n > 0
Kn x + · · · + K1 x + K0
k cos(ax) or k sin(ax)
K cos(ax) + M sin(ax)
keax cos(bx) or keax sin(bx) eax (K cos(bx) + M sin(bx))
ax
Pn
i
More
with either
i=1 ki x e cos(bx) or
Pn generally,
i
ax
i=1 ki x e sin(bx), the function yp can be found using
Question 4
yp = eax (P cos(bx) + Q sin(bx))
where P and Q are usually two polynomials of order n.
5/15
Question 2: Finding yp : The method of undetermined
coefficients
(a) Consider y 00 + 2y 0 + 10y = 25x2 + 3. Since
r(x) = 25x2 + 3 is a polynomial of order 2, we find
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yp = Ax2 + Bx + C.
Question 1
Since yp satisfies the ODE, there holds
Question 2
(Ax2 + Bx + C)0 + 2(Ax2 + Bx + C)0
2
Question 3
2
+ 10(Ax + Bx + C) = 25x + 3.
Question 4
By equalizing coefficients, we get A = 52 , B = −1, C = 0.
(b) Consider y 00 − 6y 0 + 8y = x2 e3x . In this case, we find
yp = (Ax2 + Bx + C)e3x .
By computing derivatives yp00 , yp0 and yp00 − 6yp0 + 8yp = x2 e3x ,


A = −1;
9A − 18A + 8A = 1,
6A + 6A + 9B − 12A − 18B + 8B = 0,
B = 0;


2A + 3B + 3B + 9C − 6B − 18C + 8C = 0,
C = −2.
6/15
Question 2: Finding yp : The method of undetermined
coefficients
(c) Consider y 00 − y = 2x sin x. Since r(x) = 2x sin x, we
find yp of the form
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yp = (Ax + B) sin x + (Cx + D) cos x.
Question 1
Then, it is easy to find that
Question 2
yp0 = (A − Cx − D) sin x + (Ax + B + C) cos x,
yp00 = −(Ax + B + 2C) sin x + (2A − Cx − D) cos x.
Question 3
Question 4
By using the ODE, we get that
−(Ax + B + C) sin x + (A − Cx − D) cos x = x sin x.
By equalizing coefficients of sin and cos, we arrive at
Ax + B + C = −x,
A − Cx − D = 0.
Again, by equalizing coefficients of polynomials, we have
A = −1, B + C = 0, −C = 0, and A − D = 0. Hence,
A = −1, B = 0, C = 0, and D = −1.
7/15
Question 2: Finding yp : The method of undetermined
coefficients
(d) Consider y 00 + 4y = sin2 x. Since r(x) = sin2 x, we have
to lower the order of r(x). To this purpose, we write
sin2 x = 12 (1 − cos(2x)). Therefore (and why?), we find yp
of the following form
yp = A + x B sin(2x) + C cos(2x) .
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Question 1
Question 2
Question 3
Then, we immediately have
yp00
Question 4
= −4(C + xB) sin(2x) + 4(B − xC) cos(2x).
Using the ODE, we arrive at
1
−4C sin(2x) + 4(B − 2xC) cos(2x) + 4A = (1 − cos(2x))
2
which implies that A = 81 , B = − 81 , C = 0. Thus, a
particular solution is
1 1
yp = − cos(2x).
8 8
8/15
Question 3: Finding yp : The method of variation
parameters
Suppose we have already solved the homogeneous equation
y 00 + py 0 + q = 0 and written the solution as yh = c1 y1 + c2 y2
where y1 and y2 are linearly independent solutions.
In view of the method of variation parameters, we look for
yp of the nonhomogeneous equation y 00 + py 0 + q = r of the
form
yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x).
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Question 1
Question 2
Question 3
Question 4
Method of solving
We first have
yp0 = u01 y1 + u02 y2 + u1 y10 + u2 y20 .
Since we have two arbitrary functions u1 and u2 , we impose
two conditions in order to find u1 and u2 . While one
condition is that yp verifies the ODE, we can choose the
other condition so as to simplify our calculations
9/15
Question 3: Finding yp : The method of variation
parameters
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Method of solving (cont’)
u01 y1 + u02 y2 = 0.
Then we can calculate yp00 as follows
Question 1
Question 2
yp00 = u01 y10 + u02 y20 + u1 y100 + u2 y200 .
Question 3
Question 4
Substituting in the ODE, we get
u1 (y100 + py10 + qy1 ) + u2 (y200 + py20 + qy2 ) + (u01 y10 + u02 y20 ) = r,
that is,
u01 y10 + u02 y20 = r.
The problem is equivalent to solving a 2 × 2 linear system
u01 y1 + u02 y2 = 0,
u01 y10 + u02 y20 = r.
10/15
Question 3: Finding yp : The method of variation
parameters
(a) Consider y 00 + 4y = sin2 x. Since the characteristic
equation λ2 + 4 = 0 admits λ = ±2i as complex roots, the
general solution of y 00 + 4y = 0 is just
yh = c1 cos(2x) + c2 sin(2x) e0·x = c1 cos(2x) + c2 sin(2x).
Having the form of yh , we shall find yp of the following form
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Question 1
Question 2
Question 3
Question 4
yp = u1 (x) cos(2x) + u2 (x) sin(2x).
We then immediately reach to
u01 cos(2x)+u02 sin(2x) = 0, −2 sin(2x)u01 +2 cos(2x)u02 = sin2 x.
By solving, we get that
1
u01 = − sin2 x sin(2x),
2
1
u02 = sin2 x cos(2x).
2
11/15
Question 3: Finding yp : The method of variation
parameters
(b) Consider y 00 + y = sec x. Since the characteristic
equation λ2 + 1 = 0 admits λ = ±i as complex roots, the
general solution of y 00 + y 0 = 0 is just
yh = c1 cos x + c2 sin x.
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Question 1
Question 2
We shall find yp of the following form
Question 3
Question 4
yp = u1 (x) cos x) + u2 (x) sin x.
We then arrive at
u01 cos x + u02 sin x = 0,
−u01 sin x + u02 cos x = sec x.
By solving, we get that u01 = − tan x and u02 = 1. By
integrating, we obtain
Z
Z
Z
sin x
d(cos x)
u1 = −
dx =
= ln | cos x|+C, u2 = dx.
cos x
cos x
12/15
Question 4
Recall that the ODE y 00 = F (y) can be solved using two
steps. In the first step, we write it as
d 1 02
y
= F (y)
dy 2
and then integrate both sides w.r.t. y to have
s Z
y 0 = ± 2 F (y)dy.
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Question 1
Question 2
Question 3
Question 4
In the second step, since the preceding ODE is separable, we
can solve it to get y. It is important to note that generally
R
you have to keep the constant C after calculating F (y)dy.
Why? In addtion, the sign of y 0 is also important.
For simplicity, we denote R = 150 × 109 . We now consider
for a background,
the given ODE, see
r00 = −
GM
.
r2
13/15
Question 4
By solving, we first have
s
s Z
GM
1
0
dr = ± 2GM
+C .
r = ± −2
r2
r
Since r denotes the distance,
r0 is clearly the speed of the
Earth. Therefore, we initially
have r0 (0) = 0. Besides, there
holds r(0) = R.
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Question 1
Question 2
2R/3
R
Question 3
Question 4
Using these conditions, we find that C = − R1 . In order to
clarify the sign of r0 , we observe that r0 < 0. Why? (This is
based on the fact that the speed is increasing and the
distance r is decreasing.) Our separable ODE is now just
dr
q
2GM 1r −
1
R
= −dt.
14/15
Question 4
Thus, we arrive at
Z
dr
t=− q
2GM 1r − R1
3
Z
Z
Rd Rr
dx
R2
1
q
q q
= −√
.
= −√
1
1
R
2GM
2GM
−
1
x
R
r −1
Since x = Rr we find that ”the departure” is x = 1 and ”the
arrival” is x = 23 . Hence, the time we need is
3
t[initial, meeting]
R2
= −√
2GM
Notice that the integral
R
Z
1
2
3
dx
q
qdx
1
−1
x
1
x
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Question 1
Question 2
Question 3
Question 4
≈ 3865692.129 (sec.)
−1
can be calculated, see my
blog for the answer. For a better explanation of the universe,
please follow a movie via this link
from 14:23 to 19:05.
15/15