Download Quiz 2 Solutions, Version A, MAP 2302, Fall`16 1. Verify that ex is a

Quiz 2 Solutions, Version A, MAP 2302, Fall’16
1. Verify that ex is a solution to the equation
y 000 + y 00 − 6y 0 + 4y = 0
and find its general solution.
Solution: Since (ex)0 = ex , one has y 000 + y 00 − 6y 0 + 4y = (1 + 1 − 6 + 4)ex = 0 for all
x, which means that ex is a solution. Therefore the auxiliary equation has a root r = 1 and
can by factorized by means of synthetic division
r3 + r2 − 6r + 4 = 0
(r − 1)(r2 + 2r − 4) = 0 ⇒ (r − 1)[(r + 1)2 − 5] = 0 .
√
So, the other roots are also real, r = −1 ± 5 = r± . Two linearly independent solutions
associated with them are er+ x and er− x Thus, the general solution reads
⇒
y = C1 e x + C2 e r+ x + C3 e r− x .
2. Find the general solution to the equation
y (4) + 13y 00 + 36y = 0
Solution: The auxiliary equation has the following imaginary roots:
13 2 25
13 ± 5
−
= 0 ⇒ r2 = −
r + 13r + 36 = 0 ⇒
r +
2
4
2
2
2
r = −4 ⇒ r = ±2i or r = −9 ⇒ r = ±3i
4
2
2
Each root is simple. For every simple root ±αi, the equation has two linearly independent
solutions cos(αt) and sin(αt). Therefore, the general solution reads
y = C1 cos(2x) + C2 sin(2x) + C3 cos(3x) + C4 sin(3x) .
3. Consider the equation
y 00 + 2y 0 + 4y = f(x)
Verify that the functions y1 (x) = 14 sin(2x) and y2 (x) = 14 x − 81 are solutions to the above equation if, respectively, f(x) = cos(2x) and f(x) = x. Find a particular solution to the equation if
f(x) = 11x − 12 cos(2x).
Solution: By the superposition principle, if y1 and y2 are solutions to the equation when
f = f1 and f = f2, respectively, then the function ay1 (x) + by2(x) is a particular solution to
the equation when f = af1 + bf2. Thus, a particular solution in question is
yp (x) = 11y2 (x) − 12y1 (x) =
11
11
x−
− 3 sin(2x) .
4
8
4. Let D = d/dx be the differentiation operator, that is, Dy = y 0, D2 y = y 00, etc. Find the
general solution to the equation
(D + 3)(D − 1)2 (D2 + 2D + 5)2 D3 y = 0
Solution: The auxiliary (or characteristic) equation (r + 3)(r − 1)2 [(r + 1)2 + 4]2 r3 = 0 for
the given linear differential equation has the following roots r of multiplicity s which determine
the corresponding sets of linearly independent solutions:
r
r
r
r
= −3
=1
= −1 ± 2i
=0
s=1
s=2
s=2
s=3
:
:
:
:
e−3x
ex , xex
e−x cos(2x) ,
1 , x , x2
e−x sin(2x) ,
xe−x cos(2x) ,
xe−x sin(2x)
The general solution is a linear combination of these 10 functions with arbitrary coefficients:
y = C1 + C2x + C3x2 + (C4 + C5x)e−x cos(2x) + (C6 + C7x)e−x sin(2x) + (C8 + C9x)ex + C10e−3x .
5. Use the method of undermined coefficients to determine the form of a particular solution to
the following equation. DO NOT calculate numerical values of the coefficients!
y 000 + 3y 00 − 4y = e−2x
Solution: The right side of the equation has the form f(x) = xm eαx cos(βx) with m = 0,
α = −2, and β = 0. If the number α±iβ = −2 is a root of the auxiliary equation r3 +3r2 −4 = 0
of multiplicity s, then a particular solution can be found in the form yp = xs Pm (x)e−2x where
Pm (x) = P0 (x) = A is a general polynomial of degree m = 0 (a constant A to be determined
by the substitution of yp into the equation). It follows that r = −2 is indeed a root. So one
has to determine its multiplicity. Using the synthetic division:
r3 + 3r2 − 4 = (r + 2)(r2 + r − 2) = (r + 2)[(r + 2)(r − 1)] = (r + 2)2 (r − 1) = 0
it is concluded that the root r = −2 has multiplicity s = 2 and, hence, a particular solution
can be found in the form yp = Ax2e−2x.
6. Find the general solution to the equation
y 000 + 2y 00 − 9y 0 − 18y = −18x2 − 18x + 22
Hint: e3x is a solution to the corresponding homogeneous equation.
Solution: The general solution is y = yh + yp where yh is the general solution of the associated homogeneous equation and yp is a particular solution. The auxiliary equation has a
root r = 3 (by the hint) so that the other roots are obtained by synthetic division:
r3 + 2y 2 − 9r − 18 = (r − 3)(r2 + 5r + 6) = (r − 3)(r + 2)(r + 3) = 0
All three roots are real and simple and, hence,
yh (x) = C1e3x + C2e−2x + C3 e−3x .
A particular solution can be found by the method of undetermined coefficients. The right side
is a linear combination of power functions xm , m = 0, 1, 2, which have the form xm eαx cos(βx)
with α = β = 0. Since α±iβ = 0 is not a root of the auxiliary equation, for every m a particular
solution is a polynomial of degree m. By the superposition principle, a particular solution in the
case when the right side is a linear combination of power functions is the corresponding linear
combinations of particular solutions for each m = 0, 1, 2 (polynomials of degree m = 0, 1, 2).
Therefore it is a polynomial of degree m = 2 (maximal power):
yp = Ax2 + Bx + C ,
yp0 = 2Ax + B ,
yp00 = 2A ,
yp000 = 0 .
The substitution into the equation yields
0 + 4A − 9(2Ax + B) − 18(Ax2 + Bx + C) = −18x2 − 18x − 22
x2 : −18A = −18 ⇒ A = 1
x : −18A − 18B = −18 ⇒ B = 0
1 : −4A − 18C = 22 ⇒ C − 1
⇒
y = C1 e3x + C2 e−2x + C3 e−3x + x2 − 1
7. Let x = x(t) and y = y(t). Use the elimination method to find the general solution to the
system of equations
(
x0 + 2x + y 0 = 0
x0 − x + y 0 − y = sin(t)
Solution: Let D = d/dt be the differentiation operator. The system can be written in the
form
(
(D + 2)x + Dy = 0
(D − 1)x + (D − 1)y = sin(t)
By acting by D − 1 on the first equation and by D + 2 on the second one and subtracting the
second equation from the first one, the function x is eliminated and resulting equation contains
only y:
[(D − 1)D − (D + 2)(D − 1)]y = −(D + 2) sin(t) ⇒ (−2D + 2)y = − cos(t) − 2 sin(t)
1
⇒ y 0 − y = cos(t) + sin(t)
2
The general solution to this equation has the form y = yh + yp where yh = C1 et is the general
solution to the associated homogeneous equation and yp is a particular solution. The latter
is found by the method of undetermined coefficients. Since ±i is not a root of the auxiliary
equation r − 1 = 0, a particular solution can be found in the form yp = A cos(t) + B sin(t). The
substitution into the equation yields:
1
−A sin(t) + B cos(t) − A cos(t) − B sin(t) = cos(t) + sin(t)
2
⇒
(
−A − B = 1
B − A = 12
Therefore A = − 34 , B = − 41 , and y = C1 et − 43 cos(t) − 14 sin(t). Similarly, by eliminating y from
the system, an equation for x is obtained:
[(D − 1)(D + 2) − D(D1 )]x = D sin(t)
⇒
(2D − 2)x = − cos(t)
⇒
1
x0 − x = − cos(t)
2
Its general solution is also obtained by the method of undetermined coefficients:
x = C2 e t +
1
1
cos(t) − sin(t)
4
4
The obtained functions x(t) and y(t) may not be solutions to the original system for any choice
of integrations constants C1 and C2 . In order to find all C1 and C2 for which the found functions
are solutions, one has to substitute x(t) and y(t) into the system. The second equation is the
sum of the equations obtained for x and y Therefore it is identically satisfied for any choice
of C1 and C2 . The substitution into the first equation yields (by collecting coefficients at et,
cos(t), and sin(t)):
1
1 1
1 1 3
cos(t)+ − − +
sin(t)
x0 + 2x + y 0 = (C2 + 2C2 + C1)et + − + −
4 2 4
4 2 4
C1
= (3C2 + C1 )et = 0 ⇒ C2 = −
3
Alternative solution: Let u = x + y. Then the second equation reads
u0 − u = sin(t)
⇒
u(t) = Cet −
1
1
sin(t) − cos(t)
2
2
This general solution is found by the method of undetermined coefficients (as above). The first
equation reads
u0 + 2x = 0
⇒
1
x(t) = − u0(t)
2
⇒
1
y(t) = u(t) − x(t) = u(t) + u0 (t) .
2
The solution obtained by the elimination method above corresponds to the choice C2 = −C/2
and C1 = 3C/2 (evidently, C2 = −C1/3). Note that in the variables (x, u), x satisfies an
algebraic equation and no integration constant arises when solving it! This explains why the
general solution to the system can contain only one arbitrary constant.
8 Extra credit. The motion of an oscillator under the action of an external force is described by the equation y 00 + ω 2 y = f(t) where y(t) is the amplitude of oscillations and ω is
a positive constant. Suppose that initially y(0) = y 0 (0) = 0 and f(t) = f0 sin2 (ωt/2) with
f0 = 10−2016 . Will the amplitude y(t) exceed 2016, i.e. y(t) > 2016, in due course for some
t > 0? Explain your answer!
Solution: The general solution has the form y(t) = C1 cos(ωt) + C2 sin(ωt) + yp(t) where
the first two terms are the general solution to the associated homogeneous equation because
the auxiliary equation r2 + ω 2 = 0 has the complex roots r = ±iω, and yp is a particular solution. The constants C1 and C2 are determined by the initial conditions. But regardless of their
values, the amplitude of the homogeneous solution remains bounded for all t. Let us investigate
the amplitude of a particular solution. By making use of the identity sin2 α = (1 − cos(2α))/2,
the right side can be written in the form f(t) = f0(1 − cos(ωt))/2. By the superposition
principle yp = A + t(B cos(ωt) + C sin(ωt)). The constant term corresponds to the constant
component of the force 21 f0 , while the term proportional to t corresponds to − 21 f0 cos(ωt). Note
the factor t in yp which is necessary since ±iω is the root of the auxiliary equation (recall the
method of undetermined coefficients)! The constants A, B, and C are found by the substitution
of yp into the equation. Since B and C cannot vanish simultaneously (otherwise yp is not a
particular solution), the amplitude of yp grows linearly with increasing t. Therefore for t large
enough it can exceed any preassigned positive number because a linear combination of cos(ωt)
and sin(ωt) is a periodic function (on each interval of length 2π/ω it repeats its values). Thus,
the amplitude y(t) will exceed 2016 regardless of how small f0 > 0 is.