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Math 334
Assignment 3 — Solutions
- Let φ 1 (x) and φ 2 (x) be linearly independent solutions of the equation y
′′
′
(a, b).
(a) Show that they can not both be zero at the same point x 0 ∈ (a, b).
(b) Show that they can not both have an extremum at the same point x 0 ∈ (a, b).
Solution
(a) Suppose that φ 1 (x 0 ) = 0 and φ 2 (x 0 ) = 0 for some x 0 ∈ (a, b). Then
W [φ 1 , φ 2 ](x 0 ) =
φ 1 (x 0 ) φ 2 (x 0 )
φ
′ 1 (x^0 )^ φ
′ 2 (x^0 )
φ
′ 1 (x^0 )^ φ
′ 2 (x^0 )
But this contradicts the theorem in the notes which states that for linearly independent solutions
of the given differential equation we must have W φ 1 , φ 2 6 = 0 for all x ∈ (a, b). Hence, linearly
independent solutions can not both be zero at the same point.
(b) Suppose that both solutions have an extremum at the same point in (a, b), that is that φ
′ 1 (x^0 ) = 0
and φ
′ 2 (x^0 ) = 0 for some^ x^0 ∈^ (a, b). Then
W [φ 1 , φ 2 ](x 0 ) =
φ 1 (x 0 ) φ 2 (x 0 )
φ
′ 1 (x^0 )^ φ
′ 2 (x^0 )
φ 1 (x 0 ) φ 2 (x 0 )
0 0
As in part (a) this contradicts the theorem in the notes. Therefore linearly independent solutions
can not both have an extremum at the same point.
- Consider the following differential equation:
(x − 1)y
′′ − y
′ = 0.
(a) Find the general solution of this equation.
(b) Evaluate W φ 1 , φ 2 where φ 1 and φ 2 are linearly independent solutions of the equation and W
is the Wronskian. How do you reconcile this with the theorem in the notes?
Solution
(a) The general solution to (x − 1)y
′′ − y
′ = 0 is easily found to be y(x) = c 1 + c 2 (x
2 − 2 x).
(b) Two linearly independent solutions are given by φ 1 (x) = 1 and φ 2 (x) = x
2 − 2 x.
W φ 1 , φ 2 =
1 x
2 − 2 x
0 2 x − 2
= 2(x − 1), therefore W φ 1 , φ 2 = 0. This does not contradict the
theorem in the notes since the coefficient function P (x) =
x − 1
is not continuous at x = 1.
- Find the general solution to the following homogeneous equations:
(a) y ′′
(b) y ′′
(c) y ′′
(d) y ′′ − 6 y ′
Solution
(a) y
′′
′
The characteristic equation is: r
2
- 5r + 6 = 0. The roots are: r = − 2 , −3.
Therefore the general solution is y(x) = c 1 e − 2 x
(b) y
′′
′ − y = 0.
The characteristic equation is: r 2
- r − 1 = 0. The roots are: r =
Therefore the general solution is y(x) = c 1 e
(−1+
√ 5)x/ 2
(− 1 −
√ 5)x/ 2 .
(c) y ′′
The characteristic equation is: r
2
- 9 = 0. The roots are: r = ± 3 i.
Therefore the general solution is y(x) = c 1 cos 3x + c 2 sin 3x.
(d) y ′′ − 6 y ′
The characteristic equation is: r
2 − 6 r + 10 = 0. The roots are: r = 3 ± i.
Therefore the general solution is y(x) = e
3 x (c 1 cos x + c 2 sin x).
- Find the solution to the following initial value problems:
(a) y
′′
′
′ (0) = −3;
(b) y ′′ − 2 y ′ − 2 y = 0, y(0) = 0, y ′ (0) = 3;
(c) y
′′ − 2 y
′
π , y
′ (π) = 0.
Solution
(a) y
′′
′
′ (0) = −3.
The characteristic equation is: r
2
- 2r + 1 = 0. The roots are: r = − 1 , −1.
The general solution to the equation is y(x) = (c 1 + c 2 x)e
−x
. From the initial conditions we get:
y(0) = 1, y
′ (0) = − 3 =⇒ c 1 = 1, c 2 = − 2.
Therefore the solution to the initial value problem is y(x) = (1 − 2 x)e −x .
- Consider the differential equation y
′′ − λ
2 y = 0, where λ is a positive constant.
(a) Show that the general solution can be written in the form c 1 e λx
(b) Show that the general solution can also be written in the form a 1 cosh λx + a 2 sinh λx.
(c) Use each of these general solution formats to solve the initial value problem:
y
′′ − λ
2 y = 0, y(0) = α 0 , y
′ (0) = α 1.
Which is more convenient?
Solution
The characteristic equation is: r
2 − λ
2 = 0. The roots are: r = ±λ.
(a) The general solution to the equation is y(x) = c 1 e λx
(b) From the definitions of the hyperbolic functions cosh ξ = (e
ξ
−ξ )/2 and sinh ξ = (e
ξ − e
−ξ )/2,
we get e
ξ = cosh ξ + sinh ξ and e
−ξ = cosh ξ − sinh ξ. Inserting these formulas into the general
solution of part (a) we get:
y(x) = c 1 (cosh λx + sinh λx) + c 2 (cosh λx − sinh λx) = a 1 cosh λx + a 2 sinh λx,
where a 1 = c 1 + c 2 and a 2 = c 1 − c 2.
(c) i. Using the initial conditions y(0) = α 0 and y
′ (0) = α 1 on the general solution of part (a)
results in the pair of equations
{
c 1 + c 2 = α 0
c 1 λ − c 2 λ = α 1
which are easily solved to get
c 1 = (α 0 λ + α 1 )/(2λ)
c 2 = (α 0 λ − α 1 )/(2λ)
The solution is y(x) =
2 λ
[(α 0 λ + α 1 )e λx
ii. Using the initial conditions y(0) = α 0 and y
′ (0) = α 1 on the general solution of part (b)
results in the pair of equations
{
a 1 = α 0
a 2 λ = α 1
which are easily solved to get
a 1 = α 0
a 2 = α 1 /λ
The solution is y(x) = α 0 cosh λx + (α 1 /λ) sinh λx.
These two solutions are, of course, identical since
y(x) =
2 λ
[(α 0 λ + α 1 )e
λx
−λx ]
α 0
[e
λx
−λx ] +
α 1
2 λ
[e
λx − e
−λx ]
= α 0 cosh λx +
α 1
λ
sinh λx,
but the latter solution was by far the easier to obtain.
