Download Math 286 · Exam 2 Solutions · Fall 2012 · Dr. Greene

Math 286 · Exam 2 Solutions · Fall 2012 · Dr. Greene
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1. Solve the initial value problem y 00 + y 0 − 6y = 0, y(0) = 0, y 0 (0) = 1. (10 points)
Solution: The auxillary equation m2 + m − 6 = (m + 3)(m − 2) = 0 has roots
m = −3, 2. The general solution of the differential equation is
y = c1 e−2x + c2 e3 x
which implies
y 0 = −2c1 e−2x + 3c2 e3x .
The initial conditions y(0) = 0 and y 0 (0) = 0 imply
c1 + c2 = 0
−2c1 + 3c2 = 1
The first equation implies c2 = −c1 , which after substituting in the second equation
yields −5c1 = 1. Therefore, c1 = −1/5 and c2 = 1/5. The solution to the initial–value
problem is
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y = − e−2x + e3x .
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2. Use the method of undetermined coefficients to solve y 00 − 4y 0 + 4y = x2 + 2.
(10 points)
Solution: To solve the homogeneous differential equation y 00 − 4y 0 + 4y = 0, observe
that the auxillary equation m2 − 4m + 4 = (m − 2)2 has solution m = 2 with
multiplicity 2. Thus, the homogeneous equation has solution y = c1 e2 x + c2 xe2 x. The
form of the right side of the original nonhomogeneous differential equations suggests
a particular solution of the form
y = Ax2 + Bx + C.
By taking derivatives, we see
y 0 = 2Ax + B
y 00 = 2A.
Substituting these expressions in the differential equation gives
y 00 − 4y 0 + 4y = 2A − 4(2Ax + B) + 4(Ax2 + Bx + C)
= 4Ax2 + (4B − 8A)x + 2A − 4B + 4C
= x2 + 0x + 2
Comparing the coefficients of x2 , we see A = 1/4. Comparing the coefficients of x, we
see that
0 = 4B − 8A = 4B − 8/4 = 4B − 2.
Thus, B = 1/2. Finally, the remaining coefficient gives
2 = 2A − 4B + 4C = 2/4 − 4/2 + 4C = 1/2 − 2 + 4C.
Thus, C = 7/8. Consequently, the general solution of the differential equation is
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y = c1 e2 x + c2 xe2 x + x2 + x + .
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2
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3. Use variation of parameters to solve y 00 + y = sec x. (10 points)
Solution: To solve the homogeneous equation y 00 + y = 0, observe that the auxillary
equation m2 + 1 = 0 has solutions m = 0 ± i. Thus, the complmentary solution is
y = c1 cos x + c2 sin x.
A particular solution to the nonhomogeneous equation is u1 cos x + u2 sin x where u1
and u2 are to be determined.
Solve for u1 :
0
sin
x
sec x cos x
0 · cos x − sin x sec x
0
=
u1 = = − sin x sec x.
2 x − (− sin2 x)
cos
cos
x
sin
x
− sin x cos x
Z
u1 =
Z
− sin x sec xdx =
− sin x
dx =
cos x
Z
du
= ln | cos x| + C
u
Solve for u2 :
cos x
0
− sin x sec x
cos x sec x − 0 · (− sin x)
0
=
u2 = = 1.
cos2 x − (− sin2 x)
cos x sin x − sin x cos x
Z
u2 = 1dx = x + C.
Thus, a particular solution of the nonhomogeneous equation is
yp = u1 y1 + u2 y2 = cos x ln | cos x| + x sin x
and the general solution is
y = c1 cos x + c2 sin x + cos x ln | cos x| + x sin x.
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4. (a) Define what it means for a set of functions {y1 , y2 , y3 } to be a fundamental set of
solutions of a 3rd–order homogeneous differential equation on an interval I.
(10 points)
Solution: {y1 , y2 , y3 } is a set of linearly independent solutions to differential
equation over I.
(b) Determine whether the set of functions {x4 , −x4 } is linearly independent on the
open interval (0, 2). (10 points)
Solution: The functions are linearly dependent since 1 · x4 + 1 · (−x4 ) = 0 for every x
in (0, 2). (Alternatively, the two functions are scalar multiples of each other.)
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5. State a 1st-order homogeneous differential equation for which the functions {x4 , −x4 }
are solutions on the interval (0, 2). Find the general solution of your differential
equation. (5 points)
Solution: Observe that letting y = x4 implies that y 0 = 4x3 . Since 4x4 − 4x4 = 0,
y = x4 is a solution (on the interval (0, 2)) to the differential equation xy 0 − 4y = 0.
Similarly, y = −x4 is also a solution of this differential equation since
xy 0 − 4y = x(−4x3 ) − 4x4 = −4x4 + 4x4 = 0 for all x ∈ (0, 2). Observe that this is a
1st–order homogeneous Cauchy–Euler equation. The general solution (a
one–parameter family) is given by y = cx4 .
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6. State a 2nd-order homogeneous differential equation for which the functions
{x4 , −x4 } are solutions on the interval (0, 2). Find the general solution of your
differential equation. (5 points)
Solution: Letting y = x4 gives y 0 = 4x3 and y 00 = 12x2 . Since 12x4 − 12x4 = 0, y = x4
is a solution (on the interval (0, 2)) to
x2 y 00 − 12y = 0.
Since this is a 2nd–order Cauchy–Euler equation, we solve the auxillary equation
m(m − 1) − 12 = 0. Thus, m2 − m − 12 = (m + 3)(m − 4) = 0 has solutions
m = −3, 4. Therefore, the general solution is
y = c1 x−3 + c2 x4 .
Note: There are infinitely many correct answers to this problem. For example,
y = ±x4 are also solutions to the Cauchy–Euler equation x2 y 00 − xy 0 − 8y = 0 which
has auxillary equation 0 = m(m − 1) − m − 8 = m2 − 2m − 8 = (m + 2)(m − 4) and
general solution y = c1 x−2 + c2 x4 .
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7. Solve the homogeneous Cauchy-Euler equation 2x2 y 00 + 2xy 0 − 8y = 0 (10 points)
Solution: This equation has the same solution set as x2 y 00 + xy 0 − 4y = 0. The
auxillary equation is 0 = m(m − 1) + m − 4 = m2 − 4 = (m − 2)(m + 2). The general
solution is
y = c1 x−2 + c2 x2 .
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8. (Bonus!) Explain why one should not attempt to use the method of undetermined
coefficients to solve y 00 + y = ln x. (+2 points)
Solution: The method of undetermined coefficients does not apply in this case
because the function ln x is not a finite sum of finite products of polynomials,
exponentials emx , and trigonmetric functions sin βx and cos βx..
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