Math 250B
Test #4
Spring 2010
1. Consider the differential equation
(D2 + 4D + 5)(D2 + 4)y = 5 cos 3x.
(a) Determine the general solution to the homogeneous problem.
SOLUTION: The Auxilliary equation is (r2 + 4r + 5)(r2 + 4) = 0, which has the solutions r = −4, −1, 2i, −2i.
Therefore the general solution to the homogeneous problem is yc (x) = c1 e−4x + c2 e−x + c3 cos(2x) + c4 sin(2x) .
(b) Write the annihilator for F (x) = 5 cos 3x.
SOLUTION:
A(D) = D2 + 9
(c) Determine an appropriate trial solution for the given differential equation.
SOLUTION: Since neither term on the RHS is a solution to the homogeneous problem, a trial solution can be
formed from the solutions to A(D)y = 0. Therefore an appropriate trial solution is yp (x) = A0 cos(3x) + A1 sin(3x) .
2. Solve the initial-value problem
y 00 + 9y = 70xex ,
y(0) = 1,
y 0 (0) = −1.
SOLUTION: The homogeneous problem (D2 + 9)y = 0 has the general solution y(x) = c1 cos(3x) + c2 sin(3x). To
obtain a particular solution we can use the method of undetermined coefficients or variation of parameters.
(a) Undetermined coefficients solution: An annihilator for 70xex is (D − 1)2 . Thus, A(D)P (D) = (D − 1)2 (D2 + 9).
So the only solutions of A(D)P (D)y = 0 that are not in the homogeneous solution are ex and xex . We thus use
the following trial solution:
yp (x) = A0 ex + A1 xex
Substitution into the original equation yields the conditions on A0 and A1 : A0 = −7/5, A1 = 7. Therefore the
7
general solution is y(x) = c1 cos(3x) + c2 sin(3x) + ex (5x − 1) . To obtain the general solution to the IVP we
5
need the derivative of y(x):
7
y 0 (x) = −3c1 sin(3x) + 3c2 cos(3x) + ex (5x + 4).
5
7
12
28
11
= 1 ⇒ c1 =
,
y 0 (0) = 3c2 +
= −1 ⇒ c2 = − .
5
5
5
5
11
7 x
12
Therefore the solution to the IVP is y(x) =
cos(3x) −
sin(3x) + e (5x − 1) .
5
5
5
The initial conditions impose y(0) = c1 −
(b) Variation of parameters solution: We form the particular solution as yp (x) = u1 (x) cos(3x)+u2 (x) sin(3x), impose
the condition u01 cos(3x) + u02 sin(3x) = 0 and substitute into the original equation to obtain the conditions
u01 cos(3x) + u02 sin(3x) = 0, −3u01 sin(3x) + 3u02 cos(3x) = 70xex .
Solving for u01 , u02 yields u02 =
70
x
3 xe
cos(3x), and integrating we obtain
u2 (x) =
7 x
e [(4 + 5x) cos(3x) + 3(−1 + 5x) sin(3x)] ,
5
and similarly for u1 . Upon combining the expressions in yp , the cos(3x) sin(3x) terms cancel, and we use the fact
that cos2 (3x) + sin2 (3x) = 1 to obtain the same solution as in the previous method.
3. Consider the equation y 00 − 2y 0 + y = ex ln x
(a) Find the general solution of the homogeneous problem.
SOLUTION: The auxiliary equation is r2 − 2r + 1 = 0, which can be written (r − 1)2 = 0. Thus the LI solutions
of the homogeneous problem are ex and xex , so the general solution is yc = c1 ex + c2 xex .
(b) Will the method of undetermined coefficients work to find a particular solution? Why or why not?
SOLUTION: No. There is no annihilator for ex ln x.
(c) Find a particular solution, and then the general solution of the original problem.
SOLUTION: We use the variation of parameters method and look for a solution of the form yp (x) = u1 (x)ex +
u2 (x)xex . We impose the condition u01 ex + u02 xex = 0 and take derivatives:
yp0 (x) = u1 (x)ex + u2 (x)ex (x + 1),
and yp00 (x) = u01 (x)ex + u02 (x)ex (x + 1) + u1 (x)ex + u2 (x)ex (x + 2)
Substition into the original equation gives the condition
u01 ex + u02 ex (x + 1) = ex ln x
u01 ex + u02 xex = 0
along with
We can solve these equations for u01 and u02 by substition: u01 = −xu02 and then u02 = ln x. Thus u2 (x) =
2
2
2
2
x ln x − x and u1 (x) = − x2 ln x + x4 . Thus a particular solution is yp (x) = ex − x2 ln x + x4 + x2 ln x − x2 =
x2
3
x2 ex 12 ln x − 43 , and the general solution is y(x) = ex c1 + c2 x +
ln x − x2 .
2
4
4. Use the differential operator method to find the general solution for the system x01 = 3x1 + x2 ,
SOLUTION: The system can be written as
x02 = −x1 + x2 .
(D − 3)x1 − x2
=
0
(1)
(D − 1)x2 + x1
=
0
(2)
Applying (D − 1) to (1), and using (2), we obtain the equation for x1 :
(D − 1)(D − 3)x1 + x1 = (D2 − 4D + 4)x1 = (D − 2)2 x1 = 0
Therefore x1 (t) = c1 e2t + c2 te2t . Since x2 = (D − 3)x1 , we have the following general solution:
x1 (t) = c1 e2t + c2 te2t ,
x2 (t) = (c2 − c1 )e2t − c2 te2t .
5. Determine two LI real-valued solution vectors to the vector differential equation x0 = Ax,
A=
−1
−2
2
and write
−1
the general solution. (Do not use differential operators to solve this problem.)
SOLUTION: The characteristic equation is (λ + 1)2 + 4 = 0, so there are two complex eigenvalues λ12 = −1 ± 2i. To
find the eigenvector corresponding to λ1 we form the equation (A − λ1 I)v = 0, which corresponds to the augmented
matrix
i
−2i
2
0 1 1
i
0 2 1 i 0
1.M1
, 2.A12 (2)
∼
∼
−2 −2i 0
−2 −2i 0
0 0 0
2
Therefore the eigenvectors are v1 = (−i, 1) and v2 = (i, 1). These can be written as v12 = (0, 1) ± i(−1, 0) = r ± is,
−t
−t
e sin(2t)
−e cos(2t)
so two real LI solutions are
, and
.
e−t cos(2t)
e−t sin(2t)
6. Find the general solution to
x0 = Ax + b(t),
A=
−1
−2
2
,
−1
b(t) =
−t e
0
SOLUTION: The two LI solutions
to the homogeneous
problem werefound in the previous
problem, so the funsin(2t)
−
cos(2t)
sin(2t)
cos(2t)
damental matrix is X(t) = e−t
. Thus X −1 = et
. A particular solution
cos(2t)
sin(2t)
− cos(2t) sin(2t)
is
Z t Z t
0
sin(2t) − cos(2t)
sin(2s)
cos(2s) e−s
xp (t) = X(t)
X −1 (s)b(s)ds = e−t
es
ds =
cos(2t)
sin(2t)
− cos(2s) sin(2s)
0
− 21 e−t
Combining this with the result of problem #5, the general solution is x(t) = e
−t
sin(2t)
− cos(2t)
0
c1
+ c2
−
.
cos(2t)
sin(2t)
1/2