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Math 334
Assignment 4 — Solutions
- The Wronskian of three functions is defined in terms of a determinant as follows:
W φ 1 , φ 2 , φ 3 :=
φ 1 (x) φ 2 (x) φ 3 (x)
φ ′ 1 (x) φ ′ 2 (x) φ ′ 3 (x)
φ ′′ 1 (x) φ ′′ 2 (x) φ ′′ 3 (x)
(a) Find the Wronskian of the functions
φ 1 (x) = 1, φ 2 (x) = x, φ 3 (x) = x
2 .
(b) Find the Wronskian of the functions
φ 1 (x) = e
x , φ 2 (x) = e
−x , φ 3 (x) = cosh x.
Solution
(a) For the functions φ 1 (x) = 1, φ 2 (x) = x, φ 3 (x) = x
2 we have
W φ 1 , φ 2 , φ 3 =
φ 1 (x) φ 2 (x) φ 3 (x)
φ
′ 1 (x) φ
′ 2 (x) φ
′ 3 (x)
φ ′′ 1 (x) φ ′′ 2 (x) φ ′′ 3 (x)
1 x x 2
0 1 2 x
0 0 2
(b) For the functions φ 1 (x) = e
x , φ 2 (x) = e
−x , φ 3 (x) = cosh x we have
W φ 1 , φ 2 , φ 3 =
φ 1 (x) φ 2 (x) φ 3 (x)
φ ′ 1 (x) φ ′ 2 (x) φ ′ 3 (x)
φ ′′ 1 (x) φ ′′ 2 (x) φ ′′ 3 (x)
e x e −x cosh x
e x −e −x sinh x
e x e −x cosh x
- Use variation of parameters to find the general solution of
(a) y
′′
(b) x 2 y ′′
Solution
(a) Linearly independent solutions to the homogeneous equation y
′′
- 16y = 0 are φ 1 (x) = cos 4x and
φ 2 (x) = sin 4x. Using variation of parameters to look for a particular solution of the form
φp(x) = v 1 (x)φ 1 (x) + v 2 (x)φ 2 (x)
leads to
v
′ 1 (x) =^
−φ 2 (x) sec 4x
W φ 1 , φ 2
tan 4x, v
′ 2 (x) =^
φ 1 (x) sec 4x
W φ 1 , φ 2
Integrating these we get
v 1 (x) = −
ln| cos 4x|, v 2 (x) =
x.
Therefore the general solution is:
y(x) = c 1 cos 4x + c 2 sin 4x +
cos 4x ln| cos 4x| +
x sin 4x.
(b) The homogeneous equation x
2 y
′′ +xy
′ +9y = 0 is a Cauchy–Euler equation. One looks for solutions
of the form y = x
r to get a characteristic equation r(r − 1) + r + 9 = 0. This equation has solution
r = ± 3 i which leads to two linearly independent solutions φ 1 (x) = cos(3 ln x) and φ 2 (x) =
sin(3 ln x). Using variation of parameters to look for a particular solution of the nonhomogeneous
equation of the form
φp(x) = v 1 (x)φ 1 (x) + v 2 (x)φ 2 (x)
leads to
v
′ 1 (x) =^
φ 2 (x) tan(3 ln x)/x 2
W φ 1 , φ 2
sin(3 ln x) tan(3 ln x)
3 x
, v
′ 2 (x) =^ −^
φ 1 (x) tan(3 ln x)/x 2
W φ 1 , φ 2
sin(3 ln x)
3 x
Integrating these we get
v 1 (x) =
sin(3 ln x) tan(3 ln x)
3 x
dx (ξ = 3 ln x, dξ =
x
dx)
sin ξ tan ξ dξ =
(sec ξ − cos ξ) dξ
(ln| sec ξ + tan ξ| − sin ξ) =
(ln| sec(3 ln x) + tan(3 ln x)| − sin(3 ln x))
v 2 (x) = −
sin(3 ln x)
3 x
dx (ξ = 3 ln x, dξ =
x
dx)
sin ξ dξ =
cos ξ =
cos(3 ln x).
Therefore the general solution is:
y(x) = c 1 cos(3 ln x) + c 2 sin(3 ln x) +
cos(3 ln x) ln| sec(3 ln x) + tan(3 ln x)|.
- Use variation of parameters to show that
y(x) = c 1 cos x + c 2 sin x +
∫ (^) x
0
f (s) sin(x − s) ds
is the general solution to the differential equation
y
′′
Solution
Linearly independent solutions to the homogeneous equation y ′′
- y = 0 are φ 1 (x) = cos x
and φ 2 (x) = sin x. Using variation of parameters to look for a particular solution of the form
φp(x) = v 1 (x)φ 1 (x) + v 2 (x)φ 2 (x)
leads to
v
′ 1 (x) =^
−φ 2 (x)f (x)
W φ 1 , φ 2
= −f (x) sin x, v
′ 2 (x) =^
φ 1 (x)f (x)
W φ 1 , φ 2
= f (x) cos x.
Integrating these we get
v 1 (x) = −
∫ (^) x
0
f (s) sin s ds, v 2 (x) =
∫ (^) x
0
f (s) cos s ds.
The difference between initial value problems (IVPs) and boundary value problems (BVPs) is that the
auxiliary conditions for IVPs are applied at one point only, whereas the auxiliary conditions for BVPs
are applied at more than one point. While we have a theorem that guarantees that there is one and
only one solution for an IVP, the situation for BVPs is quite different. The trivial solution y ≡ 0 is
always a solution to a homogeneous BVP, but there may be other solutions. In fact, there may be
infinitely many solutions.
Determine all the values of ω for which the above BVP has at least one nontrivial solution.
Solution
Linearly independent solutions to the homogeneous equation are φ 1 (x) = cos ωx and φ 2 (x) = sin ωx.
The general solution to the equation is: y(x) = c 1 cos ωx + c 2 sin ωx. Next we apply the boundary
conditions: {
y(0) = 0
y(1) = 0
c 1 cos 0 + c 2 sin 0 = 0
c 1 cos ω + c 2 sin ω = 0
c 1 = 0
c 2 = 0 or sin ω = 0
The solution is
y(x) =
0 if ω 6 = nπ, n = 1, 2 , 3 ,... ,
c 2 sin nπx if ω = nπ, n = 1, 2 , 3 ,....
Hence, if ω is an integer multiple of π, the BVP has infinitely many solutions.
- Suppose φ 1 (x) and φ 2 (x) are linearly independent solutions of y
′′
′
ther that φ 1 (x) has at least two zeros. Show that φ 2 (x) has one and only one zero between consecutive
zeros of φ 1 (x).
Solution
Let φ 1 (x) and φ 2 (x) be linearly independent solutions of y
′′
′
- Q(x)y = 0 on some interval
(a, b) on which P (x) and Q(x) are continuous. Assume that φ 1 has consecutive zeros at α, β ∈ (a, b),
where α < β, i.e.
φ 1 (α) = φ 1 (β) = 0 and φ 1 (x) 6 = 0 for all x ∈ (α, β) ⊂ (a, b). (1)
We wish to show that φ 2 has one and only one zero in the interval (α, β). To do this we first show that
there is at least one zero in this interval, then show that there is at most one zero in the interval.
(At least one zero)
Since φ 1 (x) and φ 2 (x) are linearly independent solutions, it follows from theorems given in class
that
W φ 1 , φ 2 6 = 0 for any x ∈ (a, b). (2)
This means that the Wronskian is either strictly positive or strictly negative in the interval (a, b).
Evaluating the Wronskian at the zeros of φ 1 yields
W φ 1 , φ 2 = φ 1 (α)φ
′ 2 (α)^ −^ φ
′ 1 (α)φ^2 (α) =^ −φ
′ 1 (α)φ^2 (α),
W φ 1 , φ 2 = φ 1 (β)φ
′ 2 (β)^ −^ φ
′ 1 (β)φ^2 (β) =^ −φ
′ 1 (β)φ^2 (β).
Equation (2) implies that φ
′ 1 (α)^6 = 0 and^ φ
′ 1 (β)^6 = 0.^ On the other hand, Equation (1) implies
that φ 1 (x) is either strictly positive or strictly negative in the interval (α, β). Therefore φ
′ 1 (α)
and φ
′ 1 (β) must be of opposite sign.^ Combine this fact with the fact that^ W^ φ^1 , φ^2 and
W φ 1 , φ 2 are of the same sign and we conclude that φ 2 (α) and φ 2 (β) must be of opposite
sign. Since φ 2 is positive at one end of the interval (α, β) and negative at the other end, the
Intermediate Value Theorem from elementary calculus implies that φ 2 is zero somewhere between
α and β. Hence, φ 2 has at least one zero in (α, β).
(At most one zero)
Suppose φ 2 has two zeros in (α, β). Then, employing an argument similar to the one used above,
φ 1 would have at least one zero in (α, β) between the zeros of φ 2. But this contradicts the above
stipulation that α and β are consecutive zeros of φ 1. Hence, φ 2 cannot have two zeros but can
only have at most one zero in (α, β).
