Download 3.4.1 Distinct Real Roots

Math 334
3.4. CONSTANT COEFFICIENT EQUATIONS
34
Assume that P (x) = p (const.) and Q(x) = q (const.), so that we have
y ′ + py ′ + qy = 0,
or
L[y] = 0 (where L :=
d
d2
+p
+ q).
dx2
dx
(3.7)
Guess a solution of the form y = erx . The equation becomes
erx (r 2 + pr + q) = 0.
But erx 6= 0, which means that y = erx is a solution iff r satisfies
C(r) := r 2 + pr + q = 0.
(3.8)
Definition 3.24. The polynoial C(r) is called the characteristic polynomial of the linear operator
L, and Eq. (3.8) is called the characteristic equation (or auxiliary equation) of the homogeneous
differential equation (3.7). ◭
p
1
The solution to the characteristic equation (3.8) is r =
−q ± p2 − 4q . There are three cases to
2
consider:


> 0
2
p − 4q = 0 .


<0
3.4.1
Distinct Real Roots
If p2 − 4q > 0, we get two distinct real roots:
p
1
−p − p2 − 4q ,
r1 =
2
r2 =
p
1
−p + p2 − 4q .
2
Lemma 3.25. If p2 − 4q > 0, then the general solution to (3.7) is
y = c1 e r 1 x + c2 e r 2 x .
Proof
We already know that er1 x and er2 x are solutions. We have
rx
p
e1
er2 x r1 x r2 x
= (r2 − r1 )e(r1 +r2 )x = p2 − 4q e−px 6= 0.
W[e , e ] = r1 x
r1 e
r2 e r 2 x Therefore er1 x and er2 x are linearly independent and the result follws from theorem 3.22.
Example 3.26. Consider the differential equation y ′′ −5y ′ +6y = 0. Substituting y = erx into the equation
leads to the auxiliary equation r 2 − 5r + 6 = 0. This has distinct real roots r1 = 2 and r2 = 3, so the general
solution is y(x) = c1 e2x + c2 e3x . (compare with example 1.4.)
Math 334
3.4.2
3.4. CONSTANT COEFFICIENT EQUATIONS
35
Equal Real Roots
If p2 − 4q = 0, we get one real root: r = −p/2. One solution is ϕ1 (x) = e−px/2 . We need another linearly
independent solution. To get one we use the “reduction of order” technique of section 3.3.
Let ϕ2 (x) = v(x)ϕ1(x) = v(x) e−px/2. Then the equation for v becomes:
1
v ′′ + (−p + p)v ′ − (p2 − 4q)v = 0,
2
which reduces to
v ′′ = 0.
Therefore v(x) = ax + b, and the second linearly independent solution is ϕ2 (x) = x ϕ1 (x) = x e−px/2 .
Lemma 3.27. If p2 − 4q = 0, then the general solution to (3.7) is
y = (c1 + c2 x) e−px/2.
Example 3.28. Consider the differential equation y ′′ +4y ′ +4y = 0. Substituting y = erx into the equation
leads to the auxiliary equation r 2 + 4r + 4 = 0. This has a double root r = −2. The general solution is
y(x) = (c1 + c2 x)e−2x .
3.4.3
Complex Roots
If p2 − 4q < 0, we get a pair of complex conjugate roots:
r = α ± iβ,
where
p
α=− ,
2
β=
We have complex solutions:
1p
4q − p2 .
2
er1 x = e(α+iβ)x = eαx eiβx = eαx (cos βx + i sin βx),
er2 x = e(α−iβ)x = eαx e−iβx = eαx (cos βx − i sin βx).
We want real solutions, so take linear combinations:
1 r1 x
(e + er2 x ) = eαx cos βx,
2
1 r1 x
(e − er2 x ) = eαx sin βx.
2i
Lemma 3.29. If p2 − 4q < 0, then the general solution to (3.7) is
y = eαx (c1 cos x + c2 sin x),
1p
p
4q − p2 .
where α = − , and β =
2
2
(real function)
(real function)
Math 334
3.5. CAUCHY–EULER EQUATION
36
Proof
One can easily verify by direct substitution that ϕ1 (x) = eαx cos βx and ϕ2 (x) = eαx sin βx are solutiuons
to the differential equation. To show that they are linearly independent, consider the wronskian:
eαx cos βx
eαx sin βx
W[ϕ1 , ϕ2 ] = αx
αx
e (α cos βx − β sin βx) e (α sin βx + β cos βx)
= e2αx α sin βx cos βx + β cos2 βx − α cos βx sin βx + β sin2 βx
1p
= βe2αx =
4q − p2 e−px 6= 0.
2
Thus, ϕ1 and ϕ2 are linearly independent and the result follows.
Example 3.30. Consider the differential equation y ′′ +2y ′ +4y = 0. Substituting y = erx into the equation
√
leads to the auxiliary equation r 2 + 2r + 4 =
0. This has√a pair of complex conjugate roots r = −1 ± 3i,
√
so the general solution is y(x) = e−x (c1 cos 3x + c2 sin 3x).
3.5
Cauchy–Euler Equation
The second order, homogeneous equation
x2 y ′′ + pxy ′ + qy = 0,
(3.9)
where p and q are constant, is called the Cauchy-Euler equation. Since this equation does not have constant
coefficients, the method of the previous section will not work. However, this equation can be transformed to
a constant coefficient equation by means of the transformation
x = et ,
y(x) = u(t).
Then we have
u′ (t) =
dx d
(y(x)) = xy ′ (x),
dt dx
u′′ (t) =
dx d
(xy ′ (x)) = x2 y ′′ + xy ′ ,
dt dx
which, upon rearrangement gives
xy ′ (x) = u′ (t),
x2 y ′′ (x) = u′′ (t) − u′ (t).
The transformed equation becomes
u′′ + (p − 1)u′ + q = 0.
This is a constant coefficient equation whose characteristic equation is
r 2 + (p − 1)r + q = 0.
The solution to the constant coefficient equation (3.10) is

r1 t
r2 t

c1 e + c2 r ,
rt
u(t) = (c1 + c2 t)e ,

 αt
e (c1 cos βt + c2 sin βt),
for distinct real roots;
for equal real roots;
for complex roots.
(3.10)
Math 334
3.6. NONHOMOGENEOUS EQUATIONS
37
Using y(x) = u(t) = u(ln x) leads to the solution of Eq. (3.9):

r1
r2

for distinct real roots;
c1 x + c2 x ,
y(x) = (c1 + c2 ln x)xr ,
for equal real roots;

 α
x (c1 cos(β ln x) + c2 sin(β ln x)), for complex roots.
Example 3.31. Consider the differential equation 3x2 y ′′ + 11xy ′ − 3y = 0. Substituting y = xr into the
equation leads to the auxiliary equation 3r 2 + 8r − 3 = 0. This has distinct real roots r1 = 1/3 and r2 = −3,
so the general solution is y(x) = c1 x1/3 + c2 x−3 .
3.6
Nonhomogeneous Equations
Consider the nonhomogeneous differential equation:
dy
d2 y
+ P (x)
+ Q(x) y = g(x).
dx2
dx
(3.11)
or, more compactly
L[y] = g,
where L is the differential operator defined by (3.2).
Theorem 3.32.
If
(i) ϕp (x) is any particular solution of the nonhomogeneous equation L[y] = g on (a, b); and
(ii) ϕ1 and ϕ2 are linearly independent solutions to the homogeneous equation L[y] = 0 on (a, b),
then the general solution of the nonhomogeneous equation L[y] = g on (a, b) is given by
y(x) = c1 ϕ1 (x) + c2 ϕ2 (x) + ϕp (x).
Proof
Hypothesis (i) implies that L[ϕp ] = g. Let y(x) be any solution of (3.13), so that L[y] = g. Now define
another function v(x) = y(x) − ϕp (x). Letting L act ov v we get
L[v] = L[y − ϕp ] = L[y] − L[ϕp ] = g − g = 0.
Thus v, being a solution of the homogeneous equation, satisfies
v(x) = c1 ϕ1 (x) + c2 ϕ2 (x)
for some constants c1 and c2 . The result now follows.
Example 3.33. Find the general solution of y ′′ − y = 2 − x2 , given that ϕp (x) = x2 is a particular solution.
Solution
One easily checks that ϕp is indeed a solution of the equation, as ϕ′′p − ϕp = 2 − x2 . The corresponding
homogeneous equation is y ′′ − y = 0. Substituting y = erx into the homogeneous equation yields
r2 − 1 = 0
=⇒
r = ±1
=⇒
ϕ1 (x) = ex , ϕ2 (x) = e−x .
Thus, the general solution to the original nonhomogeneous equation is y(x) = c1 ex + c2 e−x + x2 .
Math 334
3.7. UNDETERMINED COEFFICIENTS
38
In order to solve nonhomogeneous equations, we require a method to determine a particular solution to the
nonhomogeneous equation. Methods for determining particular solutions will be developed in the next two
sections.
3.7
Undetermined Coefficients
In this section we develop a method that will only work for constant coefficient equations. The method
involves “guessing” the form of the particular solution and then just determining the value of a constant,
hence the name: method of undetermined coefficients. The method is best illustrated by means of
eamples. We begin with examples in which the nonhomogeneous term is an exponential.
Example 3.34. Find a particular solution to the equation y ′′ + 3y ′ + 2y = e3x .
Solution
Given the nature of the nonhomogeneous term, a reasonable guess for a particular solution is ϕp (x) = Ae3x ,
where the constant A is to be determined. Letting the linear operator L (defined by the left side of the DE)
act on ϕp we get
L[ϕp ] = ϕ′′p + 3ϕ′p + 2ϕp = 9Ae3x + 3(3Ae3x) + 2Ae3x = 20Ae3x.
Thus, from the DE we have
L[ϕp ] = e3x
=⇒
A=
1
20
=⇒
ϕp (x) =
1 3x
e .
20
Example 3.35. Find a particular solution to the equation y ′′ − 5y ′ + 6y = e2x .
Solution
Again, given the nature of the nonhomogeneous term, a reasonable guess for a particular solution would
seem to be ϕp (x) = Ae2x , where the constant A is to be determined. Letting the linear operator L (defined
by the left side of the DE) act on ϕp we get
L[ϕp ] = ϕ′′p − 5ϕ′p + 6ϕp = 4Ae2x − 5(2Ae2x) + 6Ae2x = 0.
Thus L[ϕp ] 6= e2x for any value of the constant A! The guessed form of particular solution doesn’t work.
What went wrong?
Let us examine this situation more closely. Consider the more general equation y ′′ + py ′ + qy = aeλx for
which the linear operator is
d2
d
L :=
+p
+ q.
dx2
dx
If we guess a solution of the form y = Aerx and plug into the equation, we get
L[Aerx ] = aeλx
=⇒
AC(r)erx = aeλx ,
where C(r) is the characteristic polynomial for the operator L. We can see that our guess can only be
a solution if r = λ, in which case we get A = a/C(λ). However, if eλx is a solution to the homogeneous
equation, then C(λ) = 0, so y = Aeλx can’t possibly be a solution to the nonhomogeneous equation. Suppose
we go back to the expression
L[Aerx ] = AC(r)erx
Math 334
3.7. UNDETERMINED COEFFICIENTS
39
and differentiate with respect to r. This leads to
L[Axerx ] = A[C ′ (r) + xC(r)]erx.
(3.12)
Now, setting r = λ into this we get
L[Axeλx ] = AC ′ (λ)eλx ,
Plugging this into the equation we get
L[Axeλx ] = aeλx
=⇒
AC ′(λ)eλx = aeλx
=⇒
A=
a
,
C ′ (λ)
provided C ′ (λ) 6= 0. What if C ′ (λ) = 0, in addition to C(λ) = 0 (i.e. λ is a double root of C(r), i.e. xeλx
is also a solution to the homogeneous equation)? Then return to (3.12), and differentiate with respect to r
again to get
L[Ax2 erx ] = A[C ′ (r) + 2xC ′(r) + x2 C(r)]erx.
Now, setting r = λ into this we get
L[Ax2 eλx ] = AC ′′ (λ)eλx .
Plugging this into the equation we get
L[Ax2 eλx ] = aeλx
=⇒
AC ′′(λ)eλx = aeλx
=⇒
A=
a
,
C ′′ (λ)
provided C ′′ (λ) 6= 0. But, C(r) = r 2 + pr + q, which means that C ′′ (r) = 2 6= 0, so that A = a/2 and Ax2 eλx
is a particular solution.
To summarize, for equations of the form y ′′ + py ′ + qy = aeλx , the following particular solution can be found:

λx

if L[eλx ] 6= 0,
(i.e. if C(λ) 6= 0)
Ae ,
λx
λx
λx
ϕp (x) = Axe , if L[e ] = 0, L[xe ] 6= 0,
(i.e. if C(λ) = 0, C ′ (λ) 6= 0)

 2 λx
Ax e , if L[eλx ] = L[xeλx ] = 0.
(i.e. if C(λ) = C ′ (λ) = 0)
Example 3.36. Find the general solution to y ′′ − 5y ′ + 6y = e2x .
Solution
First look at the homogeneous equation: y ′′ − 5y ′ + 6y = 0. Substituting y = erx yields a characteristic
equation r 2 − 5r + 6 = 0 which has solutions r = 2 and r = 3. The solution to the homogeneous equation is:
ϕh (x) = c1 e2x + c2 e3x
The nonhomogeneous term e2x is a solution to the homogeneous equation (i.e. L[e2x ] = 0), so try a particular
solution of the form ϕp (x) = Axe2x . Now, letting the linear operator L (defined by the left side of the DE)
act on ϕp we get
L[ϕp ] = ϕ′′p − 5ϕ′p + 6ϕp = AC(2)xe2x + AC ′(2)e2x = −Ae2x ,
so, inserting this into the original equation yields:
L[ϕp ] = e2x
=⇒
A = −1
=⇒
ϕp (x) = −xe2x .
Therefore the general solution to the original equation is y(x) = c1 e2x + c2 e3x − xe2x .
Now we look at an equation with a triginometric inhomogeneity.
Math 334
3.7. UNDETERMINED COEFFICIENTS
40
Example 3.37. Find the general solution to y ′′ − y ′ − y = sin x.
Solution
From the nature of the nonhomogeneous term, a reasonable guess for a particular solution to this equation
is ϕp (x) = A cos x + B sin x. Applying the linear operator L (defined by the left side of the DE) to this gives
L[ϕp ] = ϕ′′p − ϕ′p − ϕp = (−2A + B) cos x + (A − 2B) sin x.
Inserting this into the DE yields:
L[ϕp ] = sin x
=⇒
(
−2A + B = 0
A − 2B = 1
Therefore the particular solution is ϕp (x) =
=⇒
A=
1
,
5
2
B=− .
5
1
(cos x − 2 sin x).
5
In general, the nonhomogeneous equation L[y] = a cos λx + b sin λx will have a particular solution of the
form
(
A cos λx + B sin λx,
if L[cos λx] 6= 0,
ϕp (x) =
x(A cos λx + B sin λx), if L[cos λx] = 0.
Now we look at an example with a polynoial inhomogeneity.
Example 3.38. Find a particular solution to the equation y ′′ + 3y ′ + 2y = 2x.
Solution
Try ϕp (x) = Ax. Applying the linear operator L (defined by the left side of the DE) to this gives
L[ϕp ] = 0 + 3A + 2Ax = A(3 + 2x).
Thus L[ϕp ] 6= 2x for any value of the constant A. Instead try ϕp (x) = Ax + B. Then we have
L[ϕp ] = 0 + 3A + 2(Ax + B) = 2Ax + 3A + B,
so that
L[ϕp ] = 2x
=⇒
(
2A = 2
3A + B = 0
=⇒
A = 1,
B = −3.
Therefore ϕp (x) = x − 3.
In general, for a differential equation of the form
y ′′ + py ′ + qy = a0 + a1 x + a2 x2 + · · · + an xn ,
try a particular solution of the form
(
A 0 + A 1 x + A 2 x2 + · · · + A n xn ,
ϕp (x) =
x(A0 + A1 x + A2 x2 + · · · + An xn),
if q 6= 0,
if q = 0.
We have found particular solutions for nonhomogeneous equations where the nonhomogeneous term g(x) is
an exponential, sine or cosine, and a polynoial. We can also find a particular solution if g(x) is a product of
these.
Math 334
3.7. UNDETERMINED COEFFICIENTS
41
Let
rn (x) = a0 + a1 x + a2 x2 + · · · + an xn ,
sm (x) = b0 + b1 x + b2 x2 + · · · + bm xm ,
Rn (x) = A0 + A1 x + A2 x2 + · · · + Anxn ,
Sm (x) = B0 + B1 x + B2 x2 + · · · + Bm xm .
The following table summarizes the form of the particular solution for a given form of g(x):
g(x)
rn (x)
aeλx
a cos βx + b sin βx
rn (x)eλx
rn (x) cos βx + sm (x) sin βx
eλx [a cos βx + b sin βx]
λx
e [rn (x) cos βx + sm (x) sin βx]
ϕp (x)
xσ Rn (x)
xσ Aeλx
σ
x (A cos βx + B sin βx)
xσ Rn (x)eλx
σ
x (RN (x) cos βx + SN (x) sin βx)
xσ eλx [A cos βx + B sin βx]
σ λx
x e [RN (x) cos βx + SN (x) sin βx]
In the right column of the above table σ is the smallest integer which guarantees that no term in ϕp (x) is a
solution to the corresponding homogeneous equation, and N = max{n, m}.
Example 3.39. Find the general solution to the differential equation y ′′ − y = (2 − 4x)e−x + 10 cos 2x.
Solution
We first solve the homogeneous equation
y ′′ − y = 0,
or
L[y] = 0,
where L :=
d2
− 1.
dx2
Setting y = erx , we get
r2 − 1 = 0
=⇒
r = ±1
=⇒
{ex , e−x } are linearly independent solutions.
Therefore the solution to the homogeneous equation is ϕh (x) = c1 ex + c2 e−x .
Now look for a particular solution to the nonhomogeneous equation. Given the form of the nonhomogeneous
term, we try a particular solution of the form
ϕp (x) = xσ1 (A0 + A1 x)e−x + xσ2 (A cos 2x + B sin 2x).
Since A0 e−x is a solution to the homogeneous equation, we take σ1 = 1. We can take σ2 = 0 and the
particular solution takes the form
ϕp (x) = x(A0 + A1 x)e−x + A cos 2x + B sin 2x.
Applying the differential operator to ϕp yields
L[ϕp ] = −4A1 xe−x + (2A1 − 2A2 )e−x − 5A cos 2x − 5B sin 2x.
Inserting this into the equation yields
L[ϕp ] = (2 − 4x)e−x + 10 cos 2x
=⇒

−4A1 = −4



2A − 2A = 2
1
0

−5A
=
10



−5B = 0
=⇒

A0 = 0



A = 1
1

A
= −2



B=0
Therefore a particular solution is ϕp (x) = x2 e−x − 2 cos 2x and the general solution is:
y(x) = c1 ex + (c2 + x2 )e−x − 2 cos 2x.
.
Math 334
3.8
3.8. VARIATION OF PARAMETERS
42
Variation of Parameters
Consider the nonhomogeneous equation
y ′′ + P (x)y ′ + Q(x)y = g(x).
(3.13)
In the previous section the method of undetermined coefficients was used to find particular solutions to
equations for which the linear operator had constant coefficients. Even then, it only works for a certain type
of nonhomogeneous term, one for which the form of the particular solution can be guessed. In this section
we discuss a method more general that the previous one.
The method called variation of parameters, that we will discuss here, has one major advantage over
the method of undetermined coefficients. It always works, even for general coefficients P (x) and Q(x),
and it requires no guessing, provided that two linearly independent solutions to the homogeneous equation
are known. So why did we bother with the method of undetermined coefficients? Because the method of
variation of parameters can sometimes be difficult or tedious to implement, and if one can guess the form of
a particular solution, it is sometimes easier to use the method of undetermined coefficients.
We now derive the procedure known as variation of parameters. First consider the homogeneous counterpart
to (3.13), namely
y ′′ + P (x)y ′ + Q(x)y = 0.
(3.14)
Suppose that {ϕ1 (x), ϕ2 (x)} are linearly independent solutions to (3.14), so that the general solution to
(3.14) is
y(x) = c1 ϕ1 (x) + c2 ϕ2 (x).
The idea behind this method is to look for a particular solution to (3.13) of the form
ϕp (x) = v1 (x)ϕ1 (x) + v2 (x)ϕ2 (x).
(3.15)
This is the homogeneous solution in which the constants, or parameters, c1 and c2 are allowed to vary, hence
the name variation of parameters. By looking for a particular solution of this form, we have introduced two
new unknown functions, v1 and v2 , so we will require two conditions to solve for them. One condition will,
of course, come from requiring that ϕp satisfy (3.13). But what about a second condition? Since there is no
obvious way to get one, we are at liberty to arbitrarily impose one. What we will do is impose a condition
at the appropriate point in the calculation so as to simplify things as much as possible.
Differentiate ϕp to get
ϕ′p (x) = v1′ ϕ1 + v2′ ϕ2 + v1 ϕ′1 + v2 ϕ′2 .
We now impose our second condition. We will require that v1 and v2 satisfy
v1′ ϕ1 + v2′ ϕ2 = 0.
(3.16)
Why did we choose this condition? This condition simplifies the expression for ϕ′p , which now reduces to
ϕ′p = v1 ϕ′1 + v2 ϕ′2 .
Now differentiate again to get
ϕ′′p = v1′ ϕ′1 + v2′ ϕ′2 + v1 ϕ′′1 + v2 ϕ′′2 .
Insert these quantities into (3.13) to get
v1′ ϕ′1 + v2′ ϕ′2 + v1 ϕ′′1 + v2 ϕ′′2 + P (x)(v1ϕ′1 + v2 ϕ′2 ) + Q(x)(v1ϕ1 + v2 ϕ2 ) = g(x).
Math 334
3.8. VARIATION OF PARAMETERS
43
Re-arranging this equation we get
v1′ ϕ′1 + v2′ ϕ′2 + v1 [ϕ′′1 + P (x)ϕ′1 + Q(x)ϕ1] + v2 [ϕ′′2 + P (x)ϕ′2 + Q(x)ϕ2] = g(x).
Notice that the terms in square brackets are zero since ϕ1 and ϕ2 are solutions to the homogeneous equation
(3.14), so that the above equation reduces to
v1′ ϕ′1 + v2′ ϕ′2 = g(x).
(3.17)
Equations (3.16) and (3.17) form a pair of equations for the two unknown functions v1 and v2 , which can be
written in matrix form as follows:
0
ϕ1 ϕ2 v1′
=
.
ϕ′1 ϕ′2 v2′
g
This matrix equation has a solution provided its coefficient matrix is non-singular. But this is in fact true
since the determinant of the coefficient matrix is just the Wronskian W[ϕ1 , ϕ2 ], which by Theorem 3.20, is
non-zero for linearly independent solutions of the homogeneous equation (3.14). Thus, this equation can be
solved by Cramer’s rule yielding
ϕ1 0
0 ϕ2 ϕ′1 g
g ϕ′2 ′
′
,
,
v2 = v1 = ϕ1′ ϕ2′ ϕ1′ ϕ2′ ϕ1 ϕ2 ϕ1 ϕ2 which can be written as
v1′ (x) = −
ϕ2 (x) g(x)
,
W[ϕ1 , ϕ2 ](x)
v2′ (x) =
ϕ1 (x) g(x)
.
W[ϕ1 , ϕ2 ](x)
Thus, v1 and v2 are obtained by integration, and the particular solution (3.15) becomes
Z x
ϕ1 (ξ)ϕ2 (x) − ϕ1 (x)ϕ2 (ξ)
ϕp (x) =
g(ξ) dξ.
W[ϕ1 , ϕ2 ](ξ)
x0
Example 3.40. Find the general solution to the equation y ′′ + y = tan x on the interval (−π/2, π/2).
Solution
The homogeneous equation y ′′ + y = 0 has two linearly independent solutions given by
ϕ1 (x) = cos x,
ϕ2 (x) = sin x.
The Wronskian for these is
cos x sin x = cos2 x + sin2 x = 1.
W[ϕ1 , ϕ2 ] = − sin x cos x
We look for a particular solution of the form
ϕp (x) = v1 (x)ϕ1 (x) + v2 (x)ϕ2(x) = v1 (x) cos x + v2 (x) sin x.
The functions v1 and v2 satisfy
v1′ (x) = −
ϕ2 (x) g(x)
= − sin x tan x,
W[ϕ1 , ϕ2 ](x)
v2′ (x) =
ϕ1 (x) g(x)
= cos x tan x.
W[ϕ1 , ϕ2 ](x)
Math 334
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44
Integration yields
Z
Z
Z
cos2 x − 1
sin2 x
dx =
dx
v1 (x) = − sin x tan x dx = −
cos x
cos x
Z
= (cos x − sec x) dx = sin x − ln(sec x + tan x) + c1 ,
and
v2 (x) =
Z
cos x tan x dx =
Z
sin x dx = − cos x + c2 .
We only need one particular solution, so choose c1 = c2 = 0. The particular solution is
ϕp (x) = v1 (x)ϕ1 (x) + v2 (x)ϕ2 (x) = [sin x − ln(sec x + tan x)] cos x − cos x sin x = − cos x ln(sec x + tan x)
and the general solution is
y(x) = c1 cos x + c2 sin x − cos x ln(sec x + tan x).
3.9
Applications
As an application of the various material covered in this chapter, we study the mechanical vibrations of a
spring-mass system. Consider an block of mass m suspended from the ceiling by a spring as depicted in
Figure 3.1. Let x = 0 be the equilibrium point of the mass–spring combination. The forces acting on the
mass are:
gravity:
Fg = mg;
spring:
restoring force:
Fs = −mg;
Fr = −kx;
dx
Ff = −b ;
dt
Fe = mf (t).
friction:
external forcing:
(Hooke’s Law)
The total force acting on the mass is:
F = Fg + Fs + Fr + Ff + Fe = −kx − b
dx
+ mf (t).
dt
k
m
x
Figure 3.1: Mass – spring combination.
Math 334
3.9. APPLICATIONS
45
Apply Newton’s law F = ma to get
m
dx
d2 x
= −kx − b
+ mf (t),
2
dt
dt
or, divide by m and re-arrange to put the equation into a more convenient form:
x′′ + 2βx′ + ω2 x = f (t),
where
b
,
2m
Some terminology. The equation is said to be:
undamped if
damped
if
free
if
forced
if
β=
ω2 =
and
(3.18)
k
.
m
b=0
(i.e. if β = 0);
b>0
(i.e. if β > 0);
f (t) ≡ 0;
f (t) 6≡ 0.
Case 1. (undamped, free vibrations: b = 0, f (t) ≡ 0)
Eq. (3.18) reduces to x′′ + ω2 x = 0, the general solution of which is:
x(t) = c1 cos ωt + c2 sin ωt.
We can re-write this as follows:
x(t) =
q
c21
+
c22
(
c2
c
p 1
cos ωt + p 2
sin ωt
c21 + c22
c1 + c22
)
= A(sin ϕ cos ωt + cos ϕ sin ωt) = A sin(ωt + ϕ),
where
c1
c2
,
cos ϕ = .
A
A
For undamped, free vibration the motion, called simple harmonic motion, is simply a sine wave.
A :=
q
c21 + c22 ,
sin ϕ =
Case 2. (damped, free vibrations: b > 0, f (t) ≡ 0)
Eq. (3.18) reduces to x′′ + 2βx′ + ω2 x = 0. Letting x = ert , we get
r 2 + 2βt + ω2 = 0
=⇒
r = −β ±
p
β 2 − ω2 .
Case 2(a) (underdamped motion: β < ω).
p
In this case we have r = −β ± iν, where ν = ω2 − β 2 . The solution is:
x(t) = e−βt (c1 cos νt + c2 sin νt)
or
x(t) = Ae−βt sin(νt + ϕ).
Note that
lim x(t) = lim Ae−βt sin(νt + ϕ) = 0.
t→∞
t→∞
(3.19)
Math 334
3.9. APPLICATIONS
46
Case 2(b) (critially damped motion: β = ω).
In this case the characteristic equation has a double root r = −β, so the solution is:
x(t) = (c1 + c2 t)e−βt.
Note that
lim x(t) = lim (c1 + c2 t)e−βt = 0.
t→∞
t→∞
If we differentiate we see that
x′ (t) = (c2 − c1 β − c2 βt)e−βt
which means that
x′ (t) = 0
=⇒
t=
c2 − c1 β
c2 β
so that there is at most one maximum point or one minimum point.
Case 2(c) (overdamped motion: β > ω).
p
p
In this case the characteristic equation has two real roots: r1 = −β + β 2 − ω2 and r2 = −β − β 2 − ω2
and the solution is
x(t) = c1 er1 t + c2 er2 t .
Note that both r1 and r2 are negative so that
lim x(t) = 0.
t→∞
If we differentiate we see that
√
2
2
x′ (t) = er1 t (c1 r1 + c2 r2 e(r2 −r2 )t ) = er1 t (c1 r1 + c2 r2 e−2 β −ω t )
which means that there is at most one maximum point or one minimum point. The graphs look much the
same as they do for Case 2(b).
Case 3. (forced vibrations: f (t) 6≡ 0)
We consider a periodic forcing term of the form f (t) = F0 cos γt, where F0 , γ > 0. Eq. (3.18) becomes
x′′ + 2βx′ + ω2 x = F0 cos γt.
We look for a particular solution of the form
xp (t) = A1 cos γt + B1 sin γt.
Plug this into the equation to get
[(ω2 − γ 2 )A1 + 2ββA2 ] cos γt + [(ω2 − γ 2 )A2 − 2βγA1 ] sin γt = F0 cos γt.
This leads to two simultaneous equations for A1 and A2 :
(ω2 − γ 2 )A1 + 2ββA2 = F0 ,
(ω2 − γ 2 )A2 − 2βγA1 = 0,
which are easily solved yielding
A1 =
F0 (ω2 − γ 2 )
,
(ω2 − γ 2 )2 + 4β 2 γ 2
A2 =
2F0 βγ
,
(ω2 − γ 2 )2 + 4β 2 γ 2
Math 334
3.9. APPLICATIONS
47
provided that either β 6= 0 or ω 6= γ. Thus, a particular solution is
xp (t) =
where
F0
[(ω2 − γ 2 ) cos γt + 2βγ sin γt] = F0 M sin(γt + θ),
(ω2 − γ 2 )2 + 4β 2 γ 2
1
,
M := p
2
2
(ω − γ )2 + 4β 2 γ 2
tan θ :=
ω2 − γ 2
.
2βγ
For the homogeneous solution we will only consider the underdamped case (i.e. β < ω), so that the homogeneous solution is given by (3.19):
xh (t) = Ae−βt sin(νt + ϕ).
Therefore the general solution is:
x(t) = xh (t) + xp (t) = Ae−βt sin(νt + ϕ) + F0 M sin(γt + θ).
Figure 3.2: Forced vibrations with transient.
Some terminology:
xh (t) is called a transient solution, since lim xh (t) = 0;
t→∞
xp (t) is called the steady state solution or the forced response.
Let us examine the amplitude of the steady state solution. Fix β and ω and consider the amplitude as a
function of γ:
1
.
M (γ) = p
2
2
(ω − γ )2 + 4β 2 γ 2
Figure 3.3: Amplitude M (γ).
Math 334
3.9. APPLICATIONS
48
We have M (0) = 1/ω and lim M (γ) = 0. To find the maximum we differentiate:
γ→∞
M ′ (γ) =
2γ(ω2 − γ 2 − 2β 2 )
[(ω2 − γ 2 )2 + 4β 2 γ 2 ]3/2
and see that
M ′ (γ) = 0
=⇒
γ = 0,
γ = γc :=
p
ω2 − 2β 2 .
√
(if β < ω/ 2)
The frequency γc is called the resonance frequency and the maximum amplitude is
M (γc) =
Notice that
√
(for β < ω/ 2).
1
p
,
2β ω2 − β 2
lim M (γc) = +∞,
lim γc = ω.
β→0
β→0
What happens when β = 0? Setting β = 0 yields a solution
x(t) = A sin(ωt + ϕ) +
F0
sin(γt + θ).
|ω2 − γ 2 |
For the case of pure resonance, with β = 0 and γ = ω, we can not simply plug γ = 0 into the above formulas.
We need to go back to the original ODE:
x′′ + ω2 x = F0 cos ωt.
Since the forcing term is a solution of the homogeneous equation, we look for a particular solution of the
form xp (t) = t(A1 cos ωt + A2 sin ωt), which leads to
xp (t) =
F0
t sin ωt.
2ω
We see that the solution is unbounded.
Figure 3.4: Pure resonance.
The conclusion: if β ≪ 1 and γ ≈ ω, the system is subject to large oscillations. These resonance vibrations
have been known to cause airplane wings to snap, bridges to collapse, and wine glasses to shatter.