Math 220, Differential Equations
Fall 2013 Exam 2 Solutions - 8 November 2013
Landon Kavlie
1. (20 pts) Find a particular solution to the following differential equation using
the Method of Undetermined Coefficients (any other method will receive zero
points):
y 00 (t) + y(t) = 4t cos(t).
First, we solve the homogeneous problem. That is, we write down the associated
equation
r2 + 1 = 0
which has solutions r = ±i. So, our homogeneous solution is
yh = C1 sin(t) + C2 cos(t).
Now, we write down the form of our particular solution. Since it has a t1 term out
front, we know that the form of are particular solution is given by
yp = ts [(At + B) cos(t) + (Ct + D) sin(t)] .
Since the cosine from our particular solution matches the cosine from the homogeneous solution, we then know that s = 1. Thus, our particular solution takes the
form:
yp = At2 cos(t) + Bt cos(t) + Ct2 sin(t) + Dt sin(t).
Taking two derivatives and making use of the product rule a ton, we find that
yp00 =2A cos(t) − 4At sin(t) − At2 cos(t) − 2B sin(t)
− Bt cos(t) + 2C sin(t) + 4Ct cos(t) − Ct2 sin(t)
+ 2D cos(t) − Dt sin(t).
Thus, after plugging into the equation y 00 + y = 4t cos(t), we match coefficients to
get the following equations.
t2 cos(t) : − A + A = 0
t2 sin(t) : − C + C = 0
t cos(t) : 4C − B + B = 4
t sin(t) : − D − 4A + D = 0
sin(t) : 2C − 2B = 0
cos(t) : 2A + 2D = 0.
Solving these equations gives us that A = D = 0 and B = C = 1. Thus, our
particular solution is given by
yp = t cos(t) + t2 sin(t).
2. (20 pts) Find a general solution to the following differential equation using
the Method of Variation of Parameters (any other method will receive zero
1
2
points):
y 00 (t) − 2y 0 (t) + y(t) = t−1 et .
Again, we start by solving the homogeneous equation. We start the associated
equation
r2 − 2r + 1 = 0
which gives us the double root r = 1. So, our homogeneous solution is given by
yh = C1 et + C2 tet .
Now, the method of variation of parameters tells us that the particular solution is
given by
yp = v1 et + v2 tet
where v1 and v2 are functions satisfying the equaions
v10 et + v20 tet = 0
v10 et + v20 tet + v20 et = t−1 et .
Subtracting the two equations and dividing by et quickly gives us that
v20 = t−1
⇒
v2 = ln(t).
Plugging this value back into the first equation (v20 = t−1 ) and dividing through by
et gives us that
v10 = −1 ⇒ v1 = −t.
So, our particular solution is
yp = −tet + ln(t)tet
and the general solution y = yh + yp is given by
y = C1 et + C2 tet − tet + ln(t)tet .
3. (20 pts) Find a general solution to the Cauchy-Euler equation for t > 0,
t2 y 00 (t) − 3ty 0 (t) + 4y(t) = 0.
This is another gimme. Noticing that it is a Caucy-Euler equation, you write
down the associated equation
r2 + (−3 − 1)r + 4 = 0
⇒
r2 − 4r + 4 = 0.
This has the double root r = 2. Thus, our general solution is given by
y(t) = C1 t2 + C2 t2 ln(t).
4. (20 pts) Consider the system of the first order ODEs:
dx
=y−2
dt
dy
= 2 − x.
dt
3
(1) (12 points) Solve the phase plane equation for the system.
(2) (8 points) Sketch by hand several representative trajectories with their flow
arrows.
For the first part, we do the usual trick. We have that
dy
=
dx
dy
dt
dx
dt
=
2−x
.
y−2
This gives us a separable equation to solve. So, after separating, we have that
(y − 2)dy = (2 − x)dx
which we then integrate. This gives us that
1
1 2
y − 2y = 2x − x2 + C.
2
2
Multiplying by 2 and rearraging gives us that
y 2 − 4y + x2 − 4x = C.
After completing the square, we have that
(y − 2)2 + (x − 2)2 = C.
So, the trajectories are circles in the xy-plane centered at the point (2, 2). After
plugging in a point, we find that the trajectories are travelled clockwise. A simple
sketch completes the problem.
5. (20 pts) Find the Laplace Transform of the following functions:
(1) (10 points)
f (t) = e4t cos(5t) + t sin(t).
(2) (10 points)
f (t) =
e2t , 0 < t < 3
1,
t > 3.
For the first equation, we can use the linearity of the Laplace transform as well
as the given transforms in the table to find that
L {f }(s) = L {e4t cos(5t)} + L {t sin(t)}
s−4
d
− L {sin(t)}
(s − 4)2 + 52
ds
s−4
d
1
=
−
(s − 4)2 + 25 ds s2 + 1
s−4
2s
=
+ 2
.
2
(s − 4) + 25 (s + 1)2
=
For the second one, we need to use the definition of the Laplace transform (since
we do not yet have the technology to deal with a discontinuous or piecewise defined
4
function). Thus, we find that
L {f } =
Z
∞
f (t)e−st dt
0
Z
3
f (t)e
=
−st
Z
0
Z
3
2t −st
e e
=
Z
Z
0
=
3
e(2−s)t dt +
f (t)e−st dt
3
∞
dt +
0
=
∞
dt +
3
Z ∞
e−st dt
e−st dt
3
3
∞
1 (2−s)t 1 −st +
e
e
2−s
−s
0
3
1 3(2−s)
1
e−3s
e
−
+
.
2−s
2−s
s
Note that the final steps are sloppy notation since what is really going on is a limit.
But, whose got the time for such things nowadays?
=
5
Table of Laplace Transforms.
1
L {1} = , s > 0
(1)
s
1
at
(2)
L {e } =
, s>a
s−a
n!
(3)
L {tn } = n+1 , s > 0
s
b
L {sin(bt)} = 2
(4)
, s>0
s + b2
s
(5)
L {cos(bt)} = 2
, s>0
s + b2
n!
(6)
L {eat tn } =
, s>a
(s − a)n+1
b
(7)
L {eat sin(bt)} = 2
, s>a
s + b2
s
L {eat cos(bt)} = 2
(8)
, s>a
s + b2
(9)
L {eat f (t)} = L {f }(s − a)
(10)
L {f 0 }(s) = L {f }(s) − f (0)
(11)
L {f 00 }(s) = s2 L {f }(s) − sf (0) − f 0 (0)
(12)
L {f (n) }(s) = sn L {f }(s) − sn−1 f (0) − · · · − f (n−1) (0)
(13)
(14)
dn
L {f }(s)
dsn
L {f (t − a)u(t − a)}(s) = e−as L {f }(s)
L {tn f (t)}(s) = (−1)n
(16)
e−as
s
L {g(t)u(t − a)}(s) = e−as L {g(t + a)}(s)
(17)
L {δ(t − a)}(s) = e−as
(15)
L {u(t − a)}(s) =