Download MATH 225: DIFFERENTIAL EQUATIONS AND LINEAR ALGEBRA

MATH 225: DIFFERENTIAL EQUATIONS AND LINEAR
ALGEBRA: Final Exam: May 24, 2006
Final Exam Solutions
1.a) Let A be an n × n matrix and k is a (constant) scalar. Show that
the set of all vectors ~v such that A ~v = k~v is a subspace of Rn . (This means
that eigenspace associated with an eigenvalue is a subspace of Rn ).
Solution: Let W be the subset of Rn such that any vector ~v in W is
an eigenvector of A associated with the eigenvalue k,i.e., A ~v = k~v . W is a
subspace of Rn because (i) Let ~v1 and ~v2 be in W the their sum ~v1 + ~v2 is
also in W .
A (~v1 + ~v2 ) = A~v1 + A~v2 = k~v1 + k~v2 = k(~v1 + ~v2 )
(ii) a scalar multiple of ~v (c~v ) is also in W , i.e., A(c~v ) = cA~v = ck~v = k(c~v )
1.b) Prove that the similar matrices have the same eigenvalues.
Solution: Let A and B be n × n similar matrices, i.e., there exists an
invertible matrix P such that B = P AP −1 . Let the characteristic polynomial
of A is given by det(A − λI) = f (λ). Eigenvalues of A are the zeros of
f (λ) = 0. Since A = P −1 BP then characteristic polynomial of B can be
found as
det(A−λI) = det(P −1 BP −λI) = det[P −1 (B −λI)P ] = det(B −λI) = f (λ)
Then B has the same characteristic polynomial. Hence A and B have the
same eigenvalues.
2.a) Using the Eigenvalue Method for Linear Systems find the solution
of the following initial value problem
x˙1 = x1 + 9 x2 ,
x˙2 = −x1 − 5 x2
with x1 (0) = 1 and x2 (0) = −1.
Solution: One finds repeated eigenvalue of A (λ = −2)
x1 (t) = 3e−2t − 2(3t + 1)e−2t
x2 (t) = −e−2t + 2te−2t
1
(1)
2.b) Using the Eigenvalue Method for Linear Systems find the solution
of the following initial value problem
1
x˙1 = − x1 + x2 ,
2
1
x˙2 = −x1 − x2
2
with x1 (0) = 1 and x2 (0) = 2.
Solution: One finds a complex eigenvalue of A (λ = −1/2 + i).
x1 (t) = e−t/2 cos t + 2e−t/2 sin t
x2 (t) = −e−t/2 sin t + 2e−t/2 cos t
3.a) Find the general solution of following linear of differential equation
y 00 + 9y = 2 sec 3x
Solution:
yc = A sin 3x + B cos 3x
where A and B are arbitrary constants. Hence y1 (x) = sin 3x and y2 (x) =
cos 3x. The using the variation of parameters we have
yp (x) = C1 (x)y1 (x) + C2 (x)y2 (x) = C1 (x) sin 3x + C2 (x) cos 3x
Letting
C10 sin 3x + C20 cos 3x = 0
and using yp in the equation we get
3C10 cos 3x − 3C20 sin 3x = 2 sec 3x
We the solve C1 and C2 as
C10 = 2/3,
C20 = −
2
2 sin 3x
3 cos 3x
Then the general solution is given by
y(x) = yc + yp = A sin 3x + B cos 3x +
2
2
x sin 3x + ln | cos 3x| cos 3x
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3.b) Diagonalize the matrix A given by


3
0
0
2
A = −4 6
16 −15 −5
Solution: Page 375 Example 3. Eigenvalues and their eigenvectors are
found as
λ1 = 3 :
λ2 = 1 :
λ3 = 0 :
~v1 = (1, 0, 2),
~v2 = (0, 2, −5),
~v3 = (0, 1, −3),
Hence the eigenvector matrix is A = P DP −1


1 0
0
1
P = 0 2
2 −5 −3
which leads to the diagonal matrix


3 0 0
D = 0 1 0
0 0 0
4.a). Find the general solution of the following differential equation
y 0000 + y 000 − 3 y 00 − 5y 0 − 2y = 0
Solution: One can write the characteristic equation
r4 + r3 − 3r2 − 5r − 2 = (r + 1)3 (r − 2)
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Hence we have r = −1 with multiplicity 3 and r = 2. Then the solution is
simply given by
y(x) = (A + Bx + Cx2 )e−x + De2x
where A, B, C, D are arbitrary constants
4.b) Find the general solution of the following differential equation
y 000 − 4y 0 = x + 3 cos x + e−2x
Solution: Complementary solution is found easily as
yc (x) = C1 + C2 e2x + C3 e−2x
where C1 , C2 , C3 are arbitrary constants. In f (x) we have the terms x and
e−2x whose derivatives are contained in yc hence
yp (x) = (A + Bx)x + C cos x + D sin x + Exe−2x
Inserting this into the differential equation we get
A = 0, B = −1/8, C = 0, D = −3/5, E = 1/8
Hence the general solution is
1
3
1
y(x) = C1 + C2 e2x + C3 e−2tx − x2 − sin x + txe−2x
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5
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5.a) Suppose that the sequence {~x0 , ~x1 , · · · , ~xn , · · · } vectors, where ~xk =
(ak , bk ),is defined by its initial vector ~x0 = (a0 , b0 ) and the transition matrix
A in the following way ~xk+1 = A ~xk for all k = 0, 1, 2, · · · . Find ak and bk in
terms of the initial values a0 and b0 when the transition matrix is given by
0.4 0.3
A=
,
−0.4 1.2
and 2a0 − 3b0 = 1. Give an approximate solution for ak and bk when k is
large.
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Solution: (Page 235 Example 4) One can show that
~xk = Ak ~x0 ,
k = 0, 1, 2, · · ·
Hence we have to find the kth power A. For this purpose we should diagonalize it. Let P be the matrix diagolizing the matrix A then
A = P DP −1
where
λ1 0
D=
,
0 λ2
where λ1 and λ2 are the eigenvalues of A. For A we have λ1 = 1 and
λ2 = 3/5 = 0.6 and
1 2 −3
1 3
−1
P =
, P =−
,
2 2
4 −2 1
then
Ak = P Dk P −1 ,
For large k, (0.6)k is closer to zero hence we find
1 2 −3 a0
~xk = −
4 4 −6 b0
Hence ak = − 14 and bk = − 12 for large k.
5.b) Solve the following differential equation
(6xy − y 3 ) dx + (4y + 3x2 − 3xy 2 ) dy = 0
Solution: Page 68 Example 9
M (x, y) = 6xy − y 3 ,
one can show that
N (x, y) = 4y + 3x2 − 3xy 2
∂M
∂N
=
= 6x − 3y 2
∂y
∂x
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hence the equation is exact. Solution can be found from Fx = M = 6xy − y 3
which leads to
F (x, y) = 3x2 y − xy 3 + f (y)
We find f (y) from Fy = N = 4y + 3x2 − 3xy 2 which leads to f (y) = 2y 2
hence the solution is
F (x, y) = 3x2 y − xy 3 + 2y 2 = C = constant
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