1
Review 1, Math 311, Fall 2014
• Use the method of variation of parameters to find all solutions of
y 00 − 3y 0 + 2y = sin(e−x ).
Solution: y = c1 ex + c2 e2x − e2x sin(e−x ).
• Solve the initial value problem
y 00 + 2y 0 + y = 1,
y(0) = 2, y 0 (0) = −2.
Solution: y = 1 + e−x − xe−x .
• Explain the general method for solving linear differential equation of order one
y 0 + a(x)y = b(x).
Now solve the equation for a(x) = 2x and b(x) = x3 .
Sketch of solution: Choose A(x) such that A0 (x) = a(x). ThenRCe−A(x) is the homogeneous solution, while the general solution is Ce−A(x) + e−A(x) b(x)eA(x) dx. By using
2
this formula we get y = ce−x + 21 x2 − 12 .
• Use Euler’s formula ea+ib = ea (cos(b) + isin(b)), where x and y are real, to compute
the integral
Z
e5x sin(3x)dx.
(Hint: Use a = 5x and b = 3x in Euler).
Solution: By Euler’s formula the integral in question is the imaginary part of
1
ex(5+3i)
+ C. Now, 5+3i
= 5−3i
, therefore
5+3i
34
Z
3
5
e5x sin(3x)dx = − e5x cos(3x) + e5x sin(3x) + C
34
34
R
e5x+3ix dx =
• Solve the equation
y (4) + 81y = 0
Solution: We have to find the roots of r4 + 81 = 0. This is the same as fourth roots of
−81. We represent this number in the trigonometric form −81 = 81(cos(π) + isin(π).
So the fourth roots are (see p.25): 3(cos(π/4)+isin(π/4)), 3(cos(π/4+π/2)+isin(π/4+
π/2)), 3(cos(π/4 + π) + isin(π/4
+ π))
and 3(cos(π/4 + 3π
) + isin(π/4 + 3π
)). To
2
2
√
√
2
2
summarize we get r1,2,3,4 = 3(± 2 ± i 2 ). Now, φ = C1 φ1 (x) + C2 φ2 (x) + C3 φ3 (x) +
C4 φ4 (x), where φi (x) = eri x .
• Problem 1(b) on p.89.
2
• Let eri x , i = 1, ..., n, where ri ’s are all distinct and r1 + r2 + · · · + rn = 0. Show that
W (er1 x , er2 x , ..., ern x ) does not depend on x (thus it’s a number).
Solution: The characteristic equation is (r−r1 ) · · · (r−rn ) = rn −(r1 +· · · rn )rn−1 +· · · =
0. Thus here a1 = 0. Now use the relation W (x) = e−a1 (x−x0 ) W (x0 ), so clearly we
get a number. Alternatively, observe that after cofactorPexpansion every term in the
n
determinant expansion involves products er1 x · · · ern x = e i=1 ri x = 1, so the Wronskian
is just a number.
• Compute the Wronskian in each case and test for linear independence (Warning: in
some examples it is easier to test for independence first and then compute the Wronskian).
(a) φ1 (x) = cos(2x), φ2 (x) = sin2 (x), φ3 (x) = cos2 (x).
(b) φ1 (x) = ex , φ2 (x) = e2x and φ3 (x) = e3x .
(c) φ1 (x) = x2 , φ2 (x) = x3 and φ3 (x) = x|x|.
Solution: We observe the trigonometric identity cos(2x) + sin2 (x) − cos2 (x) = 0. Thus
the set in (a) is linearly dependent and the Wronskian is identically zero. For (b)
we get W (ex , e2x , e3x ) = 2e6x , which is nonzero and thus the set is lin. independent.
For (c) the Wronskian is zero, because both x|x| and x2 have proportional columns,
so a different method must be employed to test for linear independence. So we let
c1 x2 + c2 x3 + c3 x|x| = 0 and plug in x = 1, x = −1 and x = 2. We get c1 + c2 + c3 = 0,
c1 − c2 − c3 = 0 and 4c1 + 8c2 + 4c3 = 0. This system has only the trivial solution
c1 = c2 = c3 = 0. The set is lin. independent.
• Use the method of annihilators (Section 2.11) to solve y 00 + 9y = xe3x .
Solution: We observe that xe3x is a solution of the differential equation whose char.
polynomial is (r − 3)2 = 0. Thus any particular solution of the equation must be of
the form c1 sin(3x) + c2 cos(3x) + c3 e3x + c4 xe3x . Clearly, c1 = c2 = 0. So plugging in
1
1
and c3 = − 54
c3 e3x + c4 xe3x for y gives: 18c3 + 6c4 = 0, 18c4 = 1. We get c4 = 18
1 3x
1
Thus, the general solution is: C1 sin(3x) + C2 cos(3x) − 54
e + 18
xe3x .
• (EC) Consider (undamped) harmonic oscillator governed by
d2 y
+ ω02 y = f (t),
2
dt
(1)
with the input f (t) = Asin(ω0 t).
Experiment shows that this will result in an unstable motion with mechanical breakdown of the system. By analyzing solutions of (1), explain why this is bound to
happen.