Math 3110 Test 2. Fall 2013
1 Solve the initial value problem
y 00 + 3y 0 + 2y = 0,
y(0) = 2, y 0 (0) = 4.
2 For the initial value problem below, determine an interval in which a solution is certain
to exist (you do not have to find this solution).
1
y 00 + tan t · y 0 + y = 0,
t
y(1) = 3, y 0 (1) = 2.
3 Give the general solution to each of the following homogeneous differential equations.
(a)
y 00 − 4y 0 − 5y = 0
(b)
y 00 − y 0 − y = 0
(c)
y 00 + 4y 0 + 4y = 0
(d)
y 00 + 9y = 0
(e)
y 00 − 6y 0 + 10y = 0
4 Write the proper form for a particular solution to each differential equation using the
method of Undetermined Coefficients. You do
not have to determine the values of those coefficients.
(a)
y 00 − 3y 0 + 2y = 2tet
(b)
y 00 + 4y = 8 sin(2t)
(c)
y 00 + y = 3t3
(d)
y 00 + 2y 0 + y = 5e−t + t2
5 Use the Method of Undetermined Coefficients
to find the general solution to the differential
equation
y 00 − 4y 0 + 3y = 6e2t + 3t.
6 Use Variation of Parameters to find the general solution to the differential equation
y 00 − y = 2et .
(c) The characteristic equation
solutions
1 The characteristic equation is r2 + 3r + 2 =
(r+2)(r+1) which has roots r = −1 and r = 2.
Therefore the general solution to the DE is
r2 + 4r + 4 = (r + 2)2
has a double root 2. The general solution is
y(t) = C1 e−2t + C2 te−2t .
y(t) = C1 e−t + C2 e2t .
Its derivative is y 0 = −C1 e−t − 2C2 e−2t , so the
given initial values give the following system
of equations
(
2 = C1 + C2
4 = −C1 − 2C2
(d) The characteristic equation r2 +9 has roots
±3i. The general solution is
y(t) = C1 cos(3t) + C2 sin(3t).
(e) The characteristic equation
r2 − 6r + 10 = 0
Solving this gives the values C1 = 8 and C2 =
−6. Therefore the solution the initial value
problem is
has roots 3±i (by the quadratic formula). The
general solution is
y(t) = 8e−t − 6e−2t .
y(t) = e3t (C1 cos t + C2 sin t).
2 The function 1/t has a discontinuity at zero,
where it is undefined. The function tan t has
discontinuities at π/2 + nπ (again where it
is undefined). The largest region around the
point t = 1 which avoids all of these discontinuities is the open interval (0, π/2).
4
y(t) = t(At + B)et
y(t) = t[A sin(2t) + B cos(2t)]
y(t) = At3 + Bt2 + Ct + D
y(t) = At2 e−t + Bt2 + Ct + D
3 (a) The characteristic equation
5 The associated homogeneous equation is
r2 − 4r − 5 = (r − 5)(r + 1)
has simple roots at 5 and −1. The corresponding general solution is
y(t) = C1 e5t + C2 e−t .
(b) The characteristic equation
r2 − r − 1 = 0
Y (t) = Ae2t .
Its derivatives are Y 0 = 2Ae2t and Y 00 = 4Ae2t .
Plugging that information into the equation
The corresponding general solution is
√
5)t/2
+ C2 e(1−
√
5)t/2
Its characteristic equation is r2 − 4r + 3 = (r −
3)(r − 1) which has simple roots 1 and 3. The
general solution to the homogeneous equation
is
y(t) = C1 et + C2 e3t .
By the Method of Undetermined Coefficients,
there is a specific solution to the non-hom.
equation L[y] = 6e2t of the form
has roots (by the quadratic formula)
√
1± 5
r=
.
2
y(t) = C1 e(1+
y 00 − 4y 0 + 3y = 0.
.
4Ae2t − 8Ae2t + 3Ae2t = 6e2t =⇒ A = −6
There is a specific solution to the non-hom.
equation L[y] = 3t of the form Y (t) = Bt +
C. Plugging that (and its derivatives into the
equation gives:
0−4B +3(Bt+C) = 3t =⇒ B = 1, C = 4/3.
Therefore, the general solution to the non-hom.
DE is
4
y(t) = C1 et + C2 e3t − 6e2t + t + .
3
6 The associated homogeneous solution has
characteristic equation r2 − 1, with roots ±1,
so the general solution to that is
y(t) = C1 et + C2 e−t .
With the method of variation of parameters,
we look for a specific solution to the non-hom.
DE of the form
Y (t) = u1 et + u2 e−t .
The first derivative of this is
Y 0 (t) = u1 et + u01 et − u2 e−t + u02 e−t .
Make a simplifying assumption, that u01 et +
u02 e−t = 0, and take a second derivative
Y 00 (t) = u1 et + u01 et + u2 e−t − u02 e−t .
Plugging this into the DE and simplifying gives
the equation
u01 et − u02 e−t = 2et
Now we have to solve a system of equations
(
u01 et + u02 e−t = 0
u01 et − u02 e−t = 2et
Add the two together to get that
2u01 et = 2et =⇒ u01 = 1 =⇒ u1 = t.
Subtract the two to get
1
2u02 e−t −2et =⇒ u02 = −e2t =⇒ u2 = − e2t .
2
There is then a specific solution to the nonhom. DE
1 2t −t
1 t
t
e,
Y (t) = te − e · e = t −
2
2
and the general solution to the non-hom DE is
1 t
t
−t
y(t) = C1 e + C2 e + t −
e.
2