Math 334

Assignment 4 — Solutions

1. The Wronskian of three functions is defined in terms of a determinant as follows:

W [φ1, φ2, φ3](x) :=

∣

∣

∣

∣

∣

∣

φ1(x) φ2(x) φ3(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 functionsφ1(x) = 1, φ2(x) = x, φ3(x) = x2.

(b) Find the Wronskian of the functionsφ1(x) = ex, φ2(x) = e−x, φ3(x) = coshx.

Solution

(a) For the functions φ1(x) = 1, φ2(x) = x, φ3(x) = x2 we have

W [φ1, φ2, φ3](x) =

∣

∣

∣

∣

∣

∣

φ1(x) φ2(x) φ3(x)φ′

1(x) φ′

2(x) φ′

3(x)φ′′

1(x) φ′′

2(x) φ′′

3(x)

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

1 x x2

0 1 2x0 0 2

∣

∣

∣

∣

∣

∣

= 2.

(b) For the functions φ1(x) = ex, φ2(x) = e−x, φ3(x) = coshx we have

W [φ1, φ2, φ3](x) =

∣

∣

∣

∣

∣

∣

φ1(x) φ2(x) φ3(x)φ′

1(x) φ′

2(x) φ′

3(x)

φ′′

1(x) φ′′

2(x) φ′′

3(x)

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

ex e−x coshxex −e−x sinhxex e−x coshx

∣

∣

∣

∣

∣

∣

= 0.

2. Use variation of parameters to find the general solution of

(a) y′′ + 16y = sec 4x;

(b) x2y′′ + xy′ + 9y = − tan(3 lnx).

Solution

(a) Linearly independent solutions to the homogeneous equation y′′ +16y = 0 are φ1(x) = cos 4x andφ2(x) = sin 4x. Using variation of parameters to look for a particular solution of the form

φp(x) = v1(x)φ1(x) + v2(x)φ2(x)

leads to

v′

1(x) =

−φ2(x) sec4x

W [φ1, φ2](x)= −

1

4tan4x, v′

2(x) =

φ1(x) sec4x

W [φ1, φ2](x)=

1

4.

Integrating these we get

v1(x) = −1

16ln| cos 4x|, v2(x) =

1

4x.

Therefore the general solution is:

y(x) = c1 cos 4x + c2 sin 4x +1

16cos 4x ln| cos 4x| +

1

4x sin 4x .

1

Math 334 Assignment 4 — Solutions 2

(b) The homogeneous equation x2y′′+xy′+9y = 0 is a Cauchy–Euler equation. One looks for solutionsof the form y = xr to get a characteristic equation r(r−1)+r+9 = 0. This equation has solutionr = ±3i which leads to two linearly independent solutions φ1(x) = cos(3 ln x) and φ2(x) =sin(3 lnx). Using variation of parameters to look for a particular solution of the nonhomogeneousequation of the form

φp(x) = v1(x)φ1(x) + v2(x)φ2(x)

leads to

v′

1(x) =φ2(x) tan(3 lnx)/x2

W [φ1, φ2](x)=

sin(3 lnx) tan(3 lnx)

3x, v′

2(x) = −φ1(x) tan(3 lnx)/x2

W [φ1, φ2](x)= −

sin(3 lnx)

3x.

Integrating these we get

v1(x) =

∫

sin(3 lnx) tan(3 lnx)

3xdx (ξ = 3 lnx, dξ =

3

xdx)

=1

9

∫

sin ξ tan ξ dξ =1

9

∫

(sec ξ − cos ξ) dξ

=1

9(ln| sec ξ + tan ξ| − sin ξ) =

1

9(ln| sec(3 ln x) + tan(3 lnx)| − sin(3 lnx))

v2(x) = −

∫

sin(3 lnx)

3xdx (ξ = 3 lnx, dξ =

3

xdx)

= −1

9

∫

sin ξ dξ =1

9cos ξ =

1

9cos(3 ln x).

Therefore the general solution is:

y(x) = c1 cos(3 lnx) + c2 sin(3 lnx) +1

9cos(3 ln x) ln| sec(3 ln x) + tan(3 lnx)| .

3. Use variation of parameters to show that

y(x) = c1 cos x + c2 sin x +

∫

x

0

f(s) sin(x − s) ds

is the general solution to the differential equation

y′′ + y = f(x).

SolutionLinearly independent solutions to the homogeneous equation y′′ + y = 0 are φ1(x) = cosxand φ2(x) = sin x. Using variation of parameters to look for a particular solution of the form

φp(x) = v1(x)φ1(x) + v2(x)φ2(x)

leads to

v′

1(x) =−φ2(x)f(x)

W [φ1, φ2](x)= −f(x) sin x, v′

2(x) =φ1(x)f(x)

W [φ1, φ2](x)= f(x) cosx.

Integrating these we get

v1(x) = −

∫

x

0

f(s) sin s ds, v2(x) =

∫

x

0

f(s) coss ds.

Math 334 Assignment 4 — Solutions 3

A particular solution is

φp(x) = v1(x)φ1(x) + v2(x)φ2(x) = − cos x

∫

x

0

f(s) sin s ds + sin x

∫

x

0

f(s) coss ds

=

∫

x

0

f(s)(sinx cos s − cosx sin s) ds =

∫

x

0

f(s) sin(x − s) ds.

Therefore the general solution is:

y(x) = c1 cosx + c2 sin x +

∫

x

0

f(s) sin(x− s) ds .

4. Determine the motion for an undamped system at resonance governed by

d2x

dt2+ 9x = 2 cos 3t, x(0) = 1,

dx

dt(0) = 0,

SolutionLinearly independent solutions to the homogeneous equation are φ1(t) = cos 3t and φ2(t) = sin 3t.Since the nonhomogeneous term is also a solution to the homogeneous equation, we look for a particularsolution of the form φp(t) = a1t cos 3t + a2t sin 3t. Plugging into the equation yields:

6(a2 cos 3t − a1 sin 3t) = 2 cos 3t, from which we get a1 = 0, a2 =1

3.

The general solution to the equation is:

x(t) = c1 cos 3t + c2 sin 3t +1

3t sin 3t.

Apply the initial conditions:

x(0) = 1, x′(0) = 0, =⇒ c1 = 1, c2 = 0.

The final solution is: x(t) = cos 3t +1

3t sin 3t .

5 10 15 20 25

-7.5

-5

-2.5

2.5

5

7.5

x(t) = cos 3t + 13t sin 3t

x

t

5. Consider the following homogeneous boundary value problem on the interval (0, 1) with homogeneousboundary conditions:

y′′ + ω2y = 0, y(0) = 0, y(1) = 0.

Math 334 Assignment 4 — Solutions 4

The difference between initial value problems (IVPs) and boundary value problems (BVPs) is that theauxiliary conditions for IVPs are applied at one point only, whereas the auxiliary conditions for BVPsare applied at more than one point. While we have a theorem that guarantees that there is one andonly one solution for an IVP, the situation for BVPs is quite different. The trivial solution y ≡ 0 isalways a solution to a homogeneous BVP, but there may be other solutions. In fact, there may beinfinitely many solutions.

Determine all the values of ω for which the above BVP has at least one nontrivial solution.

SolutionLinearly independent solutions to the homogeneous equation are φ1(x) = cosωx and φ2(x) = sin ωx.The general solution to the equation is: y(x) = c1 cosωx + c2 sin ωx. Next we apply the boundaryconditions:

{

y(0) = 0

y(1) = 0=⇒

{

c1 cos 0 + c2 sin 0 = 0

c1 cos ω + c2 sin ω = 0=⇒

{

c1 = 0

c2 = 0 or sin ω = 0.

The solution is

y(x) =

{

0 if ω 6= nπ, n = 1, 2, 3, . . . ,

c2 sin nπx if ω = nπ, n = 1, 2, 3, . . . .

Hence, if ω is an integer multiple of π, the BVP has infinitely many solutions.

6. Suppose φ1(x) and φ2(x) are linearly independent solutions of y′′ + P (x)y′ + Q(x)y = 0. Suppose fur-ther that φ1(x) has at least two zeros. Show that φ2(x) has one and only one zero between consecutivezeros of φ1(x).

SolutionLet φ1(x) and φ2(x) be linearly independent solutions of y′′ + P (x)y′ + Q(x)y = 0 on some interval(a, b) on which P (x) and Q(x) are continuous. Assume that φ1 has consecutive zeros at α, β ∈ (a, b),where α < β, i.e.

φ1(α) = φ1(β) = 0 and φ1(x) 6= 0 for all x ∈ (α, β) ⊂ (a, b). (1)

We wish to show that φ2 has one and only one zero in the interval (α, β). To do this we first show thatthere 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 φ1(x) and φ2(x) are linearly independent solutions, it follows from theorems given in classthat

W [φ1, φ2](x) 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 φ1 yields

W [φ1, φ2](α) = φ1(α)φ′

2(α) − φ′

1(α)φ2(α) = −φ′

1(α)φ2(α),

W [φ1, φ2](β) = φ1(β)φ′

2(β) − φ′

1(β)φ2(β) = −φ′

1(β)φ2(β).

Equation (2) implies that φ′

1(α) 6= 0 and φ′

1(β) 6= 0. On the other hand, Equation (1) implies

that φ1(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 [φ1, φ2](α) andW [φ1, φ2](β) are of the same sign and we conclude that φ2(α) and φ2(β) must be of oppositesign. Since φ2 is positive at one end of the interval (α, β) and negative at the other end, theIntermediate Value Theorem from elementary calculus implies that φ2 is zero somewhere betweenα and β. Hence, φ2 has at least one zero in (α, β).

Math 334 Assignment 4 — Solutions 5

(At most one zero)Suppose φ2 has two zeros in (α, β). Then, employing an argument similar to the one used above,φ1 would have at least one zero in (α, β) between the zeros of φ2. But this contradicts the abovestipulation that α and β are consecutive zeros of φ1. Hence, φ2 cannot have two zeros but canonly have at most one zero in (α, β).