Chapter 3: Second Order Linear Equations

Ordinary Differential Equations. Section 3.2

Dr. Marco A Roque Sol

Fall 2020

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Section 3.2: Sols. of Linear Homogeneous Equations; the Wronskian.

Consider the initial value problem

y ′′ + p(t)y ′ + q(t)y = g(t), y(t0) = y0, y ′(t0) = y ′0

Does a solution to this initial value problem exist? If a solutionexists, is the solution unique?

Theorem 3.2.1: (Existence and Uniqueness Theorem)

Consider the initial value problem

y ′′ + p(t)y ′ + q(t)y = g(t), y(t0) = y0, y ′(t0) = y ′0

where p, q, and g are continuous on an open interval I containingt0. Then there is exactly one solution y = φ(t) of this problem,and the solution exists throughout the interval I .

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Example: Determine the longest interval in which the initial valueproblem

(t − 1)y ′′ − 4t2y ′ + 6ty = cos t, y(−2) = 2, y ′(−2) = 1

is certain to have a unique twice-differentiable solution.

(t−1)y ′′−4t2y ′+ 6ty = cos t ⇒ y ′′+(−4t2t−1

)y ′+ 6t

t−1y = cos tt−1 ⇒

the functions p(t), q(t) and g(t) are given by

p(t) = − 4t2

t−1 , q(t) = 6tt−1 , g(t) = cos t

t−1

and they are continuous on −∞ < t < 1, 1 < t <∞.

Therefore, an interval where the solution is certain to exist is−∞ < t < 1 �

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Example: Determine the longest interval in which the initial valueproblem

t(t − 4)y ′′ + 3ty ′ + 4y = 2, y(3) = 0, y ′(3) = −1

is certain to have a unique twice-differentiable solution.

t(t − 4)y ′′ + 3ty ′ + 4y = 2⇒ y ′′ + 3t−4y

′ + 4t(t−4)y = 2

t(t−4) ⇒

the functions p(t), q(t), and g(t) are given by

p(t) = 3t−4 , q(t) = 4

t(t−4) , g(t) = 2t(t−4)

and they are continuous on−∞ < t < 0, 0 < t < 4, 4 < t <∞

therefore, an interval where the solution is certain to exist is0 < t < 4 �

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Theorem 3.2.2: (Superposition Principle)

If y1 and y2 are two solutions of the differential equation

y ′′ + p(t)y ′ + q(t)y = 0,

then the linear combination c1y1 + c2y2 is also a solution for anyvalues of c1 and c2.

Proof:

Let y(t) be the function defined by y(t) = c1y1 + c2y2, then

y ′′ + p(t)y ′ + q(t)y = (c1y1 + c2y2)′′(t) + p(t)(c1y1 + c2y2)′ + ...

q(t)(c1y1 + c2y2) =

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

= c1y′′1 + c2y

′′2 (t) + c1p(t)y ′1 + c2p(t)y ′2 + ...

...+ c1q(t)y1 + c2q(t)y2

= c1y′′1 + c1p(t)y ′1 + c1q(t)y1 + ...

...+ c2y′′2 (t) + c2p(t)y ′2 + c2q(t)y2

= c1(y ′′1 + p(t)y ′1 + q(t)y1) + ...

...+ c2(y ′′2 (t) + c2p(t)y ′2 + q(t)y2)

but y1(t) and y2(t) are solutions of the homogeneus second orderlinear differential equations, therefore

y ′′ + p(t)y ′ + q(t)y = c1(0) + c2(0) = 0 �

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

OBS: The superposition principle states that if we have twosolutions, then we can construct an infinite family of solutionsgiven by

y = c1y1 + c2y2

The next question is whether this general linear combinationincludes all solutions or whether there may be other solutions of adifferent form.

Definition: If y1 and y2 are two solutions of the differentialequation

y ′′ + p(t)y ′ + q(t)y = 0,

then the Wronskian of the solutions is defined by

W (t) = W (y1, y2)(t) =

∣∣∣∣y1 y2y ′1 y ′2

∣∣∣∣ = y1y′2 − y ′1y2

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Example: Find the Wronskian of the given pair of functions.

(a) y1(t) = cos 2t, y2(t) = sin 2t

y1(t) = cos 2t ⇒ y ′1(t) = −2 sin 2t; ⇒y2(t) = sin 2t ⇒ y ′2(t) = 2 cos 2t

W (t) = W (y1, y2)(t) =

∣∣∣∣y1 y2y ′1 y ′2

∣∣∣∣ = y1y′2 − y ′1y2 ⇒

W (t) = W (y1, y2)(t) =

∣∣∣∣ cos 2t sin 2t−2 sin 2t 2 cos 2t

∣∣∣∣ =

(cos 2t)(2 cos 2t)− (sin 2t)(−2 sin 2t) = 2(cos 2t)2 + 2(sin 2t)2 =

2[(cos 2t)2 + (sin 2t)2

]= 2 �

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

(b) y1(t) = e−2t , y2(t) = te−2t

y1(t) = e−2t ⇒ y ′1(t) = −2e−2t ;

y2(t) = te−2t ⇒ y ′2(t) = e−2t + t(−2e−2t) = e−2t(1− 2t)

W (t) = W (y1, y2)(t) =

∣∣∣∣y1 y2y ′1 y ′2

∣∣∣∣ = y1y′2 − y ′1y2 ⇒

W (t) = W (y1, y2)(t) =

∣∣∣∣ e−2t te−2t

−2e−2t e−2t(1− 2t)

∣∣∣∣ =

(e−2t)[e−2t(1− 2t)]− (−2e−2t)(te−2t) =

(e−4t)(1− 2t) + 2te−4t = e−4t

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Theorem 3.2.4: Suppose that y1(t) and y2(t) are two solutions of

y ′′ + p(t)y ′ + q(t) = 0.

Then the family of solutions y(t) = c1y1(t) + c2y2(t) with arbitrarycoefficients c1 and c2 includes every solution if and only if there isa point t0 where the Wronskian of y1(t) and y2(t) is not zero.

OBS: This theorem states that, if and only if the Wronskian ofy1(t) and y2(t) is not everywhere zero, then every solution can bewritten as a linear combination c1y1(t) + c2y2(t) for someconstants c1 and c2. In this case, y1(t) and y2(t) are said to forma fundamental set of solutions.

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Example: Show that y1 =√t and y2 = 1/t form a fundamental

set of solutions of

2t2y ′′ + 3ty ′ − y = 0,

y1 =√t ⇒ y ′1 = 1

2√t⇒ y ′′1 = − 1

4t3/2

2t2y ′′1 + 3ty ′1 − y1 = 2t2(− 1

4t3/2

)+ 3t

(1

2√t

)−√t =

−12

√t + 3

2

√t −√t = −3

2

√t + 3

2

√t = 0

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

y2 = 1t ⇒ y ′2 = − 1

t2⇒ y ′′2 = 2

t3

2t2y ′′2 + 3ty ′2 − y2 = 2t2(2t3

)+ 3t

(− 1

t2

)− 1

t =

4t −

3t −

1t = 0

Thus y1 and y2 are solutions of the ODE. Now, let’s check theWronskian

y1 =√t ⇒ y ′1 = 1

2√t

y2(t) = 1t ⇒ y ′2 = − 1

t2

W (t) = W (y1, y2)(t) =

∣∣∣∣y1 y2y ′1 y ′2

∣∣∣∣ = y1y′2 − y ′1y2 ⇒

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

W (t) = W (y1, y2)(t) =

∣∣∣∣∣√t 1

t1

2√t− 1

t2

∣∣∣∣∣ =

(√t)(− 1

t2)− (1t )( 1

2√t) =

− 1t3/2− 1

2t3/2= − 3

2t3/26= 0

Therefore, y1 and y2 form a fundamnental set of solutions �

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Theorem 3.2.6: If y1 and y2 are two solutions of

y ′′ + p(t)y ′ + q(t)y = 0,

where p and q are continuous on an open interval I , then theWronskian is given by

W (y1, y2)(t) = c exp

[−∫

p(t)dt

]where c is a certain constant that depends on y1 and y2, but noton t. Moreover, W (y1, y2) is either zero for all t ∈ I (if c = 0), oris never zero in I (if c 6= 0 ).

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Example: Find the Wronskian of two solutions of the differentialequation.

t2y ′′ + t(t + 2)y ′ + (t + 2)y = 0,

without solving the equation.From the Last Theorem of this section we have

W (y1, y2)(t) = c exp[−∫p(t)dt

]W (y1, y2)(t) = c exp

[−∫

t+2t dt

]⇒

W (y1, y2)(t) = ce−t−ln(t2)

W (y1, y2)(t) = c e−t

t2�

k

Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2

Chapter 3: Second Order Linear Equations

Chapter 3

Example: If the differential equation.ty ′′ + 2y ′ + teyy = 0,

has a fundamental set of solutions y1 and y2 and W (y1, y2)(1) = 2,find the value of W (y1, y2)(t).

From the Last Theorem of this section we have

W (y1, y2)(t) = c exp

[−∫

p(t)dt

]W (y1, y2)(t) = c exp

[−∫p(t)dt

]⇒

W (y1, y2)(t) = c exp[−∫

2t dt]⇒

W (y1, y2)(t) = ce ln(1/t2) = c

t2

but using the condition W (y1, y2)(1) = 2 we get

W (y1, y2)(1) = c/(12) = 2⇒ c = 2

and the Wronskian is given by W (y1, y2)(t) = 2t2

�Dr. Marco A Roque Sol Ordinary Differential Equations. Section 3.2