Download MAP 2302: Solutions to Test #2

MAP 2302:
Solutions to Test #2 (Version A)
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1) We compute the Wronskian:
2 7x 3x2 − 1
7 6x
2
6x W (2, 7x, 3x − 1) = 0 7
= 2 0 6 = 42(2) = 84 6= 0,
0 0
6
so the functions are linearly independent.
2) The theorem only applies to second order equations with initial conditions of the form
y(x0 ) = b0 and y 0 (x0 ) = b1 . Since this initial value problem also has a condition on y 00 (0),
the theorem does not apply to it.
3) The auxiliary equation is r2 − 6r + 9 = 0 which we factor to get (r − 3)2 = 0, so
r = 3 is the solution with multiplicity two. Hence, the general solution of the equation is
y = Ae3x + Bxe3x with y 0 = 3Ae3x + Be3x + 3Bxe3x . We substitute in the initial conditions
and get 1 = y(0) = A and 0 = y 0 (0) = 3A + B = 3 + B, so B = −3. Hence, the solution is
y = e3x − 3xe3x .
4) The auxiliary equation is r3 − 3r2 + 4r − 12 = 0. We solve this by factoring, r2 (r −
3) + 4(r − 3) = 0, so (r2 + 4)(r − 3) = 0, so r = 3, ±2i. Thus, the general solution is
y = Ae3x + B sin(2x) + C cos(2x).
5) We first solve the associated homogeneous equation which has auxiliary equation r2 −
6r + 5 = 0 or (r − 1)(r − 5) = 0. The complementary solution is y = C1 ex + C2 e5x . The UC
function e5x has UC set {e5x } which is a solution to the associated homogeneous equation.
We observe that xe5x does not solve the associated homogeneous equation, so we substitute
in
y = Axe5x
y 0 = Ae5x + 5Axe5x
y 00 = 5Ae5x + 5Ae5x + 25Axe5x = 10Ae5x + 25Axe5x
and get
10Ae5x + 25Axe5x − 6 Ae5x + 5Axe5x + 5Axe5x = 4e4x
(10A − 6A) e5x + (25A − 30A + 5A) xe5x = 4e4x
4Ae5x = 4e4x
so 4A = 4 or A = 1. Hence, the general solution is
y = xe5x + C1 ex + C2 e5x .
1
2
6) We first solve the associated homogeneous equation which has auxiliary equation r2 +4 =
0 and thus general solution
y = C1 cos(2x) + C2 sin(2x).
We use the method of variation of parameters with y1 = cos(2x) and y2 = sin(2x). Hence,
cos(2x)
sin(2x) = 2 cos2 (2x) + 2 sin2 (2x) = 2.
W (y1 , y2 ) = −2 sin(2x) 2 cos(2x)
The variation of parameter formula gives a solution y = v1 y1 + v2 y2 , where
Z
sec(2x) tan(2x) sin(2x)
v1 = −
dx
1(2)
Z
1
=−
tan2 (2x) dx
2
Z
1
=
1 − sec2 (2x) dx
2
x 1
= − tan(2x) + C.
2 4
and
Z
sec(2x) tan(2x) cos(2x)
v2 =
dx
1(2)
Z
1
=
tan(2x) dx
2
1
= ln | sec(2x)| + C.
2
Hence, the general solution is
1
x 1
− tan(2x) + C1 cos(2x) +
ln | sec(2x)| + C2 sin(2x).
y=
2 4
2
7) We make the substitution y = xr and the resulting indicial equation is
r(r − 1)(r − 2)(r − 3) + 6r(r − 1)(r − 2) + 11r(r − 1) + 5r + 4 = 0
(r2 − r)(r2 − 5r + 6) + 6r(r2 − 3r + 2) + 11r2 − 11r + 5r + 4 = 0
r4 − 5r3 + 6r2 − r3 + 5r2 − 6r + 6r3 − 18r2 + 12r + 11r2 − 11r + 5r + 4 = 0
r4 + 4r2 + 4 = 0
(r2 + 2)2 = 0
√
Thus r = ± 2i are solutions with multiplicity two. Hence, the general solution is:
√
√
√
√
y = C1 sin( 2 ln(x)) + C2 cos( 2 ln(x)) + C3 ln(x) sin( 2 ln(x)) + C4 ln(x) cos( 2 ln(x)).
3
8) We set y = vex , so y 0 = v 0 ex + vex , and y 00 = v 00 ex + 2v 0 ex + vex . Substituting in yields:
x(v 00 ex + 2v 0 ex + vex ) − (x + 1)(v 0 ex + vex ) + vex = 0
ex v 00 (x) + v 0 (2x − x − 1) + v(x − x − 1 + 1)) = 0
Set w = v 0 so w0 = v 00 and the equation becomes separable:
xw0 + (x − 1)w = 0
w0
1−x
=
Z 0 w
Z x
w
1
dx =
− 1 dx
w
x
Z
1
dw = ln(x) − x + C
w
ln |w| = ln(x) − x + C
w = C1 xe−x
Then
Z
v=
w dx
Z
=
C1 xe−x dx
We compute the preceding integral using integration by parts with u = x and dv = e−x dx
so du = dx and v = −e−x and we get
Z
−x
v = −C1 xe + C1 e−x dx
= −C1 xe−x − C1 e−x + C2 .
Hence, the general solution is
y = ex v = C1 (x + 1) + C2 ex .