solving homogeneous des with constant coe cients ...math121/notes/annotated_online/week...week #4 :...

Week #4 : Applications of Linear Higher-Order DEs

Goals:

• Solving Homogeneous DEs with Constant Coefficients - Secondand Higher Order•Applications - Damped Spring/Mass system•Applications - Pendulum

1

Homogeneous Equations - Review and Generalizing to nth Order - 1

Homogeneous Equations - Review

We saw last week how to solve second-order homogeneous linear equa-tions.

Note 1: we can make linear combinations of individual solutions tomake more general solutions.

Lemma. If both y1 and y2 are solutions to

y′′ + p(x)y′ + q(x)y = 0,

then any linear combination y = C1y1 + C2y2 is also a solution.In other words, the solutions form a vector space.


Note 2: If the equation has constant coefficients, an informedguess that solutions will be of the form y = ert will always get usstarted.

If we start with the DE

ay′′ + by′ + cy = 0

where a, b, c ∈ R and a 6= 0,

then assuming that y = ert, any valid r must statisfy the DE’scharacteristic equation:

ar2 + br + c = 0.


Homogeneous DEs - Arbitrary order

A linear equation of order n can be expressed in the form

An(x)y(n) + An−1(x)y(n−1) + · · · + A0(x)y = F (x) .

Assuming that An(x) 6= 0, we can rewrite the equation in the stan-dard form:

y(n) + p1(x)y(n−1) + · · · + pn(x)y = G(x)


y(n) + p1(x)y(n−1) + · · · + pn(x)y = G(x)

In this n− th order DE, the general solution will be the span of nlinearly independent solutions, y1, . . . , yn.It therefore behooves us to develop a test for linear independence forsets of n functions.

The Wronskian - n functions - 1

The Wronskian - n functions

For n differentiable function y1, y2, . . . , yn, the Wronskian is

W [y1, . . . , yn] := det

y1(x) . . . yn(x)y′1(x) . . . y′n(x)

... . . . ...

y(n−1)1 (x) . . . y

(n−1)n (x)

Lemma. If the Wronskian is nonzero at some point, then y1, . . . ynare linearly independent.


Problem. Use the Wronskian to show that the functions x, x2, andx−1 are linearly independent.


Problem. Given that x, x2, and x−1 are solutions to x3y′′′+x2y′′−2xy′ + 2y = 0 where x > 0, find the general solution.


Concept check: why weren’t the solutions x, x2, and x−1 to

x3y′′′ + x2y′′ − 2xy′ + 2y = 0

in the form of erx?

Homogenous DEs with Const Coeffs - Cases - 1

Homogenous DEs with Constant Coefficients - Cases

Proposition. If the characteristic equation has distinct rootsr1, . . . , rn, then the general solution is C1e

r1x + C2er2x + · · · +

Cnernx.

Problem. Solve y(4) + y′′′ − 7y′′ − y′ + 6y = 0.


Proposition. If the characteristic equation has a root r of mul-tiplicity m, then part of the general solution is C1e

rx+C2xerx+

· · · + Cmxm−1erx.

Problem. Solve y(4) − y(3) − 3y′′ + 5y′ − 2y = 0.


Proposition. If the characteristic equation has a pair of con-jugate roots a ± b

√−1, then the part of the general solution is

C1eax cos(bx) + C2e

ax sin(bx).

Problem. Solve y(4) − y = 0.


SummaryForm of r. Form of solution term(s)Unique real root r1 y =Duplicated real roots. E.g. r = r1, r1 y1 = and

y2 =

Unique pair of complex conjugate roots y1 = and

r1,2 = a± b√−1 y2 =

Duplicated complex conjugate pair.E.g. r1,2,3,4 = a± b

√−1, a± b

√−1

(4 roots)

Homogenous DEs with Const Coeffs - Example - 1

Homogenous DEs with Constant Coefficients - Example

Problem. Solve y(4) − 8y(3) + 26y′′ − 40y′ + 25y = 0.

Hint. Expand (r2 − 4r + 5)2.

Mechanical Vibrations - Spring-mass system - 1

Mechanical Vibrations - Spring-mass system

We now consider the spring/mass system seen earlier, but with moredetail.Consider a mass m hanging on the end of a vertical spring of un-stretched length `. Using Newton’s second law, build a differentialequation that governs the system.

k c

m


k c

m


Problem. Consider a mass of 0.5 kg with spring constant k =2 N · m−1 in an undamped unforced system. Assume the mass isdisplaced 0.4 m from equilibrium and released. Describe the long-term behaviour of the system.

Oscillating Solutions: Sine/Cos and Shifted-Cos Forms - 1

Oscillating Solutions: Sine/Cos and Shifted-Cos Forms

Remark. We can always write the general solution of an undampedunforced spring system in either of two equivalent forms:

To convert between them,

A =√C2

1 + C22 and

α (in radians) satisfies cos(α) = C1A and sin(α) = C2

AThe parameter α is called the phase angle.


Problem. Explain how this relationship helps us understand thebehaviour of solutions when r comes in complex conjugate pairs.


Problem. Consider a spring/mass system with- mass of 0.5 kg,- spring constant k = 2 N ·m−1 and- damping constant c = 0.5 N · s ·m−1

and no external force applied. Assume the mass is displaced 0.3 mfrom equilibrium and released. Describe the long-term behaviour ofthe system.


- mass of 0.5 kg,

- spring constant k = 2 N ·m−1 and

- damping constant c = 0.5 N · s ·m−1

Unforced Spring/Mass System - Patterns of Behaviour - 1

Unforced Spring/Mass System - Patterns of Behaviour

k c

m

my′′ = −ky − cy′or

my′′ + cy′ + ky = 0

r =−c±

√c2 − 4km

2m

Damping c2 − 4km r DescriptionNone

c = 0

Light

c2 < 4km

Heavy

c2 > 4km

Unforced Spring/Mass System - Patterns of Behaviour - 2

Problem. Sketch possible solutions for all four spring/mass cases.

t

y

Undamped

t

y

Critically Damped

t

y

Damped

t

y

Overdamped

Homogeneous DEs - Arbitrary Order - 2

dny

dxn+ p1(x)

dn−1y

dxn−1+ · · · + pn(x)y = f (x) .

In this n − th order DE, the general solution will be the span of nlinearly independent solutions y1, . . . , yn.It therefore behooves us to develop a test for linear independence forsets of more than 2 functions.

Demonstration - Spring/Mass - 1

Demonstration - Spring/Mass

We will demonstrate how the solutions to the damped spring/massDE change as the damping is gradually increased.

my′′ + cy′ + ky = 0

In this demonstration, we will use m = 1 kg, and k = 25 N/m.


y′′ + cy′ + 25y = 0

Problem. What damping level will produce critical damping?

What will the form of the solutions be when damping is belowcritical?

What will the form of the solutions be when damping is abovecritical?


During demonstration, try to ask yourself the following questions:

•As damping increases in general, does the graph of the solutionchange gradually or dramatically?What about the mathematical form of the solution?•Near critical damping specifically, does the graph of the solution

change gradually or dramatically?What about the mathematical form of the solution?

Applications - Pendulum - 1

Applications - Pendulum

Problem. Consider the simple pendulum (mass at the end of a rod)shown below.

If we start from the rotational (torque) version of Newton’s SecondLaw,

(moment of inertia) · (angular accel) =∑

torques

we obtain(mL2) · (θ′′) = −mgL sin(θ)or, in (almost) standard form:


θ′′ +g

Lsin(θ) = 0

Use a well-known approximation from calculus to simplify this to alinear DE:

What limitations does this put on our interpretation of the solution?


Find the general solution to the linearized differential equation

θ′′ +g

Lθ = 0

Use your solution to predict the period of the oscillations of a pen-dulum, given g and the length of the rod, L.


Comment: what does this mean about the swinging of a pendulumfor larger angles?

solving homogeneous des with constant coe cients ...math121/notes/annotated_online/week...week #4 :...

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