Brian Bowers (TA for Hui Sun)
MATH 20D
Homework Assignment 4
November 1, 2013
20D - Homework Assignment 4
First, I will give a brief overview of how to use variation of parameters.
(1) Ensure that the differential equation takes the form y 00 + p(t)y 0 + q(t)y = g(t) where p, q, and g are
continuous. [If we don’t have this form and we can’t get to this form, then we CAN’T USE THIS
METHOD.]
(2) Find a fundamental set of solutions y1 and y2 to the associated homogeneous differential equation (namely
y 00 + p(t)y 0 + q(t)y = 0).
(3) Calculate W (y1 , y2 ) = y10 y2 − y1 y20 .
(4) Calculate
Z
Y = −y1
y2 g
dt + y2
W (y1 , y2 )
Z
y1 g
dt.
W (y1 , y2 )
(5) The general solution is y = c1 y1 + c2 y2 + Y .
3.6 #7,10 For each problem, find the general solution of the given differential equation.
(7) y 00 + 4y 0 + 4y = t−2 e−2t ,
t>0
(10) y 00 − 2y 0 + y = et /(1 + t2 )
(7) y 00 + 4y 0 + 4y = t−2 e−2t ,
t > 0. I will use the method outlined above:
(1) We have the right form, and we see that p, q, and g are continuous [since t > 0].
(2) The homogeneous equation is y 00 + 4y 0 + 4y = 0. This has characteristic equation r2 + 4r + 4 = 0,
which factors as (r + 2)2 = 0, yielding solutions r = −2, −2. Thus, y1 = e−2t and y2 = te−2t .
(3) W (y1 , y2 ) = y1 y20 − y10 y2 = e−2t (e−2t − 2te−2t ) − (−2e−2t )te−2t = e−4t − 2te−4t + 2te−4t = e−4t .
(4) We calculate:
Z
y1 g
y2 g
dt + y2
dt
Y = −y1
W (y1 , y2 )
W (y1 , y2 )
Z
Z
te−2t (t−2 e−2t )
e−2t (t−2 e−2t )
−2t
= −e−2t
dt
+
te
dt
e−4t
e−4t
Z
Z
= −e−2t t−1 dt + te−2t t−2 dt
Z
= −e−2t ln |t| + te−2t (−t−1 )
= −e−2t ln t − e−2t [since t > 0]
(5) The general solution is
y = c1 y 1 + c2 y 2 + Y
= c1 e−2t + c2 te−2t − e−2t ln t − e−2t
= c1 e−2t + c2 te−2t − e−2t ln t [combining the e−2t terms]
1
(10) y 00 − 2y 0 + y = et /(1 + t2 ). I will once again use the method outlined above:
(1) We have the right form, and we see that p, q, and g are continuous [since 1 + t2 is never equal to
0].
(2) The homogeneous equation is y 00 − 2y 0 + y = 0. This has characteristic equation r2 − 2r + 1 = 0,
which factors as (r − 1)2 = 0, yielding solutions r = 1, 1. Thus, y1 = et and y2 = tet .
(3) W (y1 , y2 ) = y1 y20 − y10 y2 = et (et + tet ) − et (tet ) = e2t + te2t − te2t = e2t .
(4) We calculate:
Z
y1 g
y2 g
dt + y2
dt
Y = −y1
W (y1 , y2 )
W (y1 , y2 )
Z
Z
et (et /(1 + t2 ))
tet (et /(1 + t2 ))
t
dt
+
te
dt
= −et
e2t
e2t
Z
Z
t
1
= −et
dt + tet
dt
1 + t2
1 + t2
et
= − ln(1 + t2 ) + tet tan−1 t
2
Z
(5) Thus, the general solution is
y = c1 y1 + c2 y2 + Y
= c1 et + c2 tet −
et
ln(1 + t2 ) + tet tan−1 (t)
2
3.6 #13,17 In each problem, verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular particular solution of the given nonhomogeneous equation.
(13) t2 y 00 − 2y = 3t2 − 1,
t > 0;
(17) x2 y 00 − 3xy 0 + 4y = x2 ln x,
y2 (t) = t−1
y1 (t) = t2 ,
x > 0;
y1 (x2 ),
y2 (x)x2 ln x
(13) First, we confirm that y1 and y2 satisfy the differential equation by plugging them into the left side of
the differential equation and confirming that it equals 0:
t2 y100 − 2y1 = t2 (t2 )00 − 2(t2 ) = t2 (2t)0 − 2(t2 ) = t2 (2) − 2t2 = 0
t2 y200 − 2y2 = t2 (t−1 )00 − 2(t−1 ) = t2 (−t−2 )0 − 2t−1 = t2 (2t−3 ) − 2t−1 = 0
Next, we follow steps 1 through 4 from above:
(a) We don’t have the right form yet, so we divide both sides by t2 to get
y 00 −
2
1
y =3− 2
t2
t
Now we see that p, q, and g are continuous [since the problem statement tells us t > 0].
(b) This step was completed for us in the problem statement.
(c) W (y1 , y2 ) = y1 y20 − y10 y2 = t2 (−t−2 ) − (2t)t−1 = −3. [Note that since W 6= 0, we know that y1
and y2 form a fundamental set of solutions.]
2
(d) We calculate
Z
y1 g
y2 g
dt + y2
dt
Y = −y1
W (y1 , y2 )
W (y1 , y2 )
Z −1
Z 2
t (3 − t12 )
t (3 − t12 )
= −t2
dt + t−1
dt
−3
−3
Z
Z
= −t2 (t−3 /3 − t−1 )dt + t−1 (1/3 − t2 )dt
−2
t
t3
−t
− ln |t| + t−1
−
= −t2
6
3
3
2
1
1 t
= + t2 ln t + −
6
3
3
2
1
t
= + t2 ln t −
2
3
Z
2
We note that Y (t) is a particular solution. However, in the solution, they have combined the − t3 term
with y1 to yield the particular solution
Y (t) =
(17) x2 y 00 − 3xy 0 + 4y = x2 ln x,
x > 0;
1
+ t2 ln t
2
y1 (x) = x2 ,
y2 (x) = x2 ln x.
First, we confirm that y1 and y2 satisfy the differential equation by plugging them into the left side of
the differential equation and confirming that it equals 0:
x2 y100 − 3xy10 + 4y1 = x2 (x2 )00 − 3x(x2 )0 + 4(x2 ) = x2 (2x)0 − 3x(2x) + 4x2 = x2 (2) − 6x2 + 4x2 = 0
x2 y200 − 3xy20 + 4y2 = x2 (x2 ln x)00 − 3x(x2 ln x)0 + 4(x2 ln x)
= x2 (2x ln x + x)0 − 3x(2x ln x + x) + 4x2 ln x
= x2 (2 ln x + 2 + 1) − 6x2 ln x − 3x2 + 4x2 ln x
= 2x2 ln x + 2x2 + x2 − 6x2 ln x − 3x2 + 4x2 ln x
=0
Next, we follow steps 1 through 4 from above:
(a) We don’t have the right form yet, so we divide both sides by x2 to get
y 00 −
3 0
4
y + 2 y = ln x
x
x
Now we see that p, q, and g are continuous [since the problem statement tells us x > 0].
(b) This step was completed for us in the problem statement.
(c) W (y1 , y2 ) = y1 y20 − y10 y2 = x2 (2x ln x + x) − (2x)x2 ln x = 2x3 ln x + x3 − 2x3 ln x = x3 . [Note that
since W 6= 0, we know that y1 and y2 form a fundamental set of solutions.]
3
(d) We calculate
Z
y1 g
y2 g
dx + y2
dx
Y = −y1
W (y1 , y2 )
W (y1 , y2 )
Z 2
Z 2
x ln x(ln x)
x (ln x)
= −x2
dx + x2 ln x
dx
x3
x3
Z
Z
(ln x)2
ln x
= −x2
dx + x2 ln x
dx
x
x
1
1
= −x2 (ln x)3 + x2 ln x (ln x)2
3
2
1 2
1
= − x (ln x)3 + x2 (ln x)3
3
2
1 2
3
= x (ln x)
6
Z
7.1 #1 Transform the given equation into a system of first order equations:
u00 + 0.5u0 + 2u = 0.
In general, to convert a second order differential equation into a system of first order differential equations,
we choose to set x1 equal to the dependent variable (u in this case) and then we set x2 = x01 . (In this case,
it means that x2 = x01 = u0 . Furthermore, we see that u00 = (u0 )0 = (x2 )0 .) So if we plug u = x1 , u0 = x2 ,
and u00 = x02 into our original differential equation, we get x02 + 0.5x2 + 2x1 = 0. Thus, our system of first
order equations is
(
x02 + 0.5x2 − 2x1 = 0
.
x2 = x01
7.1 #5 Transform the given initial value problem into an initial value problem for two first order equations.
u00 + 0.25u0 + 4u = 2 cos(3t),
u(0) = 1,
u0 (0) = −2.
We use the same substitutions as in the previous exercise (u = x1 , u0 = x2 , and u00 = x02 ) to get the
system
(
x02 + 0.25x2 + 4x1 = 2 cos(3t)
,
x1 (0) = 1,
x2 (0) = −2.
x2 = x01
7.1 #7 Systems of first order equations can sometimes be transformed into a single equation of higher
order. Consider the system
x01 = −2x1 + x2 ,
x02 = x1 − 2x2 .
(a) Solve the first equation for x2 and substitute into the second equation, thereby obtaining a second
order equation for x1 . Solve this equation for x1 and then determine x2 also.
(b) Find the solution of the given system that also satisfies the initial conditions x1 (0) = 2, x2 (0) = 3.
(c) Sketch the curve, for t ≥ 0, given parametrically by the expressions for x1 and x2 obtained in part (b).
(a) Solving the first equation for x2 , we get
x2 = x01 + 2x1 .
4
Substituting this into the second equation ad solving for x1 , we get
(x01 + 2x1 )0 = x1 − 2(x01 + 2x1 )
=⇒ x001 + 2x01 = x1 − 2x01 − 4x1
=⇒ x001 + 4x01 + 3x1 = 0
=⇒ r2 + 4r + 3 = 0 [characteristic equation]
=⇒ (r + 3)(r + 1) = 0 [factoring]
=⇒ r = −3, −1
=⇒ x1 = c1 e−3t + c2 e−t
Next, we plug this into our equation for x2 to get
x2 = (c1 e−3t + c2 e−t )0 + 2(c1 e−3t + c2 e−t )
= −3c1 e−3t − c2 e−t + 2c1 e−3t + 2c2 e−t
= c2 e−t − c1 e−3t
(b) Now we use the initial conditions:
x1 (0) = c1 e−3(0) + c2 e−(0) = c1 + c2 = 2
x2 (0) = c2 e−(0) − c1 e−3(0) = c2 − c1 = 3
The first equation yields c1 = 2 − c2 . Plugging this into the second equation, we get c2 − (2 − c2 ) =
2c2 − 2 = 3, which we can solve to get c2 = 5/2. Plugging this back into the first equation, we get
c1 + 5/2 = 2, which we can solve to get c1 = −1/2. Thus, we plug these constants into our formulae for
x1 and x2 :
−1 −3t 5 −t
e
+ e
2
2
5 −t 1 −3t
x2 = e + e
2
2
x1 =
(c) We can get some intuition about the graph by thinking about what happens at t = 0 and what happens
for very large values of t. At t = 0, we have x1 = x1 (0) = 2 and x2 = x2 (0) = 3, so we graph the point
(2, 3). At large values of t, we see that both x1 and x2 will approach 0. By experimentally plugging in
several values of t, we can see that the graph looks like the following:
5
7.1 #11,12 For each problem, proceed as in problem 7.
(a) Transform the given system into a single equation of second order.
(b) Find x1 and x2 that also satisfy the given initial conditions.
(c) Sketch the graph of the solution in the x1 x2 -plane for t ≥ 0.
(11) x01 = 2x2 ,
x1 (0) = 3,
(12) x01 = −0.5x1 + 2x2 ,
x02 = −2x1 ,
x2 (0) = 4.
x02 = −2x1 − 0.5x2 ,
x1 (0) = −2,
x2 (0) = 2.
(11) We complete the steps in order:
(a) First, we solve the first equation for x2 to get
x2 =
1 0
x .
2 1
Next, we plug this into our second equation to get
0
1 0
x
= −2x1
2 1
1
=⇒ x001 = −2x1
2
1
=⇒ x001 + 2x1 = 0
2
(b) Next, we solve this equation for x1 :
1 00
x + 2x1 = 0
2 1
1
=⇒ r2 + 2 = 0 [characteristic equation]
2
=⇒ r = ±2i [solving using quadratic formula]
=⇒ x1 = c1 cos(2t) + c2 sin(2t)
Plugging this back into our equation for x2 , we see
x2 =
1 0
1
x = (c1 cos(2t) + c2 sin(2t))0 = −c1 sin(2t) + c2 cos(2t).
2 1
2
Now we use our initial conditions:
x1 (0) = c1 cos(2(0)) + c2 sin(2(0)) = c1 = 3
x2 (0) = −c1 sin(2(0)) + c2 cos(2(0)) = c2 = 4
Finally, we plug c1 and c2 back into our formulae for x1 and x2 :
x1 = 3 cos(2t) + 4 sin(2t)
x2 = −3 sin(2t) + 4 cos(2t)
(c) We can find by experimentation that this actually graphs a circle (traversed clockwise) centered
at 0 of radius 5. (We can confirm this by calculating k(x1 , x2 )k = 5.)
6
(12) x01 = −0.5x1 + 2x2 ,
x1 (0) = −2,
x02 = −2x1 − 0.5x2 ,
x2 (0) = 2
We complete the steps in order:
(a) First, we solve the first equation for x2 to get
1
1 0
x + x1 .
2 1 4
Next, we plug this into our second equation to get
0
1
1
1 0
1 0
x + x1 = −2x1 − 0.5
x + x1
2 1 4
2 1 4
1
1
1
1
=⇒ x001 + x01 = −2x1 − x01 − x1
2
4
4
8
1 00 1 0
17
=⇒ x1 + x1 + x1 = 0
2
2
8
x2 =
(b) Next, we solve this equation for x1 :
17
1 00 1 0
x + x + x1 = 0
2 1 2 1
8
1 2 1
17
=⇒ r + r +
= 0 [characteristic equation]
2
2
8
1
=⇒ r = − ± 2i
2
=⇒ x1 = c1 e−t/2 cos(2t) + c2 e−t/2 sin(2t)
Plugging this back into our equation for x2 , we see
1
1 0
x + x1
2 1 4
0 1 1 −t/2
c1 e
cos(2t) + c2 e−t/2 sin(2t) +
c1 e−t/2 cos(2t) + c2 e−t/2 sin(2t)
=
2
4
1 c1 −t/2
c2
−t/2
=
− e
cos(2t) − 2c1 e
sin(2t) − e−t/2 sin(2t) + 2c2 e−t/2 cos(2t)
2
2
2
1 −t/2
−t/2
+
c1 e
cos(2t) + c2 e
sin(2t)
4
c
c1 −t/2
c2
c2 −t/2
1
= − c1 + c2 +
e
cos(2t) + −c1 −
+
e
sin(2t)
4
4
4
4
= c2 e−t/2 cos(2t) − c1 e−t/2 sin(2t)
x2 =
Now we use our initial conditions:
x1 (0) = c1 e−(0)/2 cos(2(0)) + c2 e−(0)/2 sin(2(0)) = c1 = −2
x2 (0) = c2 e−(0)/2 cos(2(0)) − c1 e−(0)/2 sin(2(0)) = c2 = 2
7
Finally, we plug c1 and c2 back into our formulae for x1 and x2 :
x1 = −2e−t/2 cos(2t) + 2e−t/2 sin(2t)
x2 = 2e−t/2 cos(2t) + 2e−t/2 sin(2t)
(c) We can note that x1 and x2 are decreasing over time, so our graph should approach the origin. In
particular, if we experiment with plugging in various values of t, we see that the graph will spiral
inward clockwise, as seen below:
7.1 #13 Transform the following equations for the parallel circuit into a single second order equation:
dI
V
= ,
dt
L
dV
I
V
=− −
,
dt
C
RC
where L is the inductance, C is the capacitance, and R is the resistance. Note that I (the current) and V
(the voltage) are the dependent variables here, whereas the other letters are constants.
We begin by solving the first equation for V :
dI
V
=
dt
L
Next, we plug this into our second equation:
=⇒
V =L
dI
.
dt
dV
I
V
=− −
dt
C
RC
L dI
d
dI
I
dt
=⇒
L
=− −
dt
dt
C
RC
2
d I
I
L dI
=⇒ L 2 = − −
dt
C
RC dt
d2 I
dI
=⇒ LRC 2 = −IR − L
dt
dt
d2 I
dI
+ RI = 0
=⇒ LRC 2 + L
dt
dt
=⇒ LRCI 00 + LI 0 + RI = 0
7.2 #2 If A =
1+i
3 + 2i
−1 + 2i
2−i
and B =
i
2
3
, find
−2i
(a) A − 2B
(b) 3A + B
(c) AB
(d) BA
8
(a)
1 + i −1 + 2i
i
3
−2
3 + 2i
2−i
2 −2i
1 + i −1 + 2i
2i
6
=
−
3 + 2i
2−i
4 −4i
1 + i − 2i −1 + 2i − 6
=
3 + 2i − 4 2 − i + 4i
1−i
−7 + 2i
=
−1 + 2i 2 + 3i
A − 2B =
(b)
1 + i −1 + 2i
i
3
3A + B = 3
+
3 + 2i
2−i
2 −2i
3 + 3i −3 + 6i
i
3
=
+
9 + 6i 6 − 3i
2 −2i
3 + 3i + i −3 + 6i + 3
=
9 + 6i + 2 6 − 3i − 2i
3 + 4i
6i
=
11 + 6i 6 − 5i
(c)
1+i
3 + 2i
(1 + i)(i) + (−1 + 2i)(2)
(3 + 2i)(i) + (2 − i)(2)
AB =
−1 + 2i
i
2−i
2
3
−2i
(1 + i)(3) + (−1 + 2i)(−2i)
=
(3 + 2i)(3) + (2 − i)(−2i)
i − 1 − 2 + 4i 3 + 3i + 2i + 4
=
3i − 2 + 4 − 2i 9 + 6i − 4i − 2
−3 + 5i 7 + 5i
=
2+i
7 + 2i
(d)
i
3
1 + i −1 + 2i
BA =
2 −2i
3 + 2i
2−i
(i)(1 + i) + (3)(3 + 2i)
(i)(−1 + 2i) + (3)(2 − i)
=
(2)(1 + i) + (−2i)(3 + 2i) (2)(−1 + 2i) + (−2i)(2 − i)
i − 1 + 9 + 6i −i − 2 + 6 − 3i
=
2 + 2i − 6i + 4 −2 + 4i − 4i − 2
8 + 7i 4 − 4i
=
6 − 4i
−4
7.2 #10 Either compute the inverse of the matrix or show that it is singular:
1 4
−2 3
9
Call the above matrix A. First, we quickly check whether or not the matrix is singular. We see that
det(A) = 1(3) − 4(−2) = 11 6= 0, so the matrix is invertible. Now we augment the matrix with the identity
and use row operations to convert the left side to the identity matrix.
1 4 1 0
[augmenting with identity matrix]
−2 3 0 1
1 4 1 0
[adding 2*(row 1) to (row 2)]
0 11 2 1
1 4 1
0
[dividing (row 2) by 11]
2
1
0 1 11
11
3
−4
1 0 11
11
[adding -4*(row 2) to (row 1)]
2
1
0 1 11 11
Thus, the inverse is the right side of our augmented matrix, namely
3 −4 11
2
11
11
1
11
7.2 #22,23 In each problem, verify that the given vector satisfies the given differential equation.
3 −2
4 2t
(22) x0 =
x,
x=
e .
2 −2
2
2 −1
1
1 t
1
(23) x0 =
x+
et ,
x=
e +2
tet .
3 −2
−1
0
1
(22) First, we simplify the expression for x as x =
4e2t
. Now we substitute in and confirm that the
2e2t
equation is true:
2t 0 2t 4e
3 −2
4e
=
2e2t
2 −2
2e2t
2t 8e
3(4e2t ) − 2(2e2t )
=
4e2t
2(4e2t ) − 2(2e2t )
2t 2t 8e
8e
=
4e2t
4e2t
Thus, the given x satisfies the differential equation.
t
e + 2tet
(23) First, we simplify the expression for x as x =
. Now we substitute in and confirm that the
2tet
equation is true:
t t
0 t
2 −1
e + 2tet
e
e + 2tet
=
+
2tet
3 −2
2tet
−et
t
t
t e + 2et + 2tet
2(e + 2tet ) − 1(2tet )
e
=
+
2et + 2tet
3(et + 2tet ) − 2(2tet )
−et
t
t
t 3e + 2tet
2e + 2tet
e
=
+
2et + 2tet
3et + 2tet
−et
t
t
3e + 2tet
3e + 2tet
=
2et + 2tet
2et + 2tet
10
Thus, the given x satisfies the differential equation.
11