19_7

Applications ofDifferential Equations

19.7

IntroductionBlocks 19.2 to 19.6 have introduced several techniques for solving commonly-occurring rst-order and second-order ordinary dierential equations. In this Block we solve a number of theseequations which model engineering systems.

PrerequisitesBefore starting this Block you should . . .

understand what is meant by a dierentialequation; (Block 19.1)

be familiar with the terminology associatedwith dierential equations: order,dependent variable and independentvariable; (Block 19.1)

be able to integrate; (Blocks 14.1-14.8)

have completed Blocks 19.2, 19.4, 19.5 and19.6

Learning OutcomesAfter completing this Block you should be ableto . . .

recognise and solve rst-order ordinarydierential equations, modelling simpleelectrical circuits, projectile motion andNewtons law of cooling

recognise and solve second-order ordinarydierential equations with constant coef-cients modelling free electrical and me-chanical oscillations

recognise and solve second-order ordinarydierential equations with constant co-ecients modelling forced electrical andmechanical oscillations

Learning StyleTo achieve what is expected of you . . .

allocate sucient study time

briey revise the prerequisite material

attempt every guided exercise and mostof the other exercises

1. Modelling with First-order Equations

Applying Newtons law of coolingIn Block 19.1 we introduced Newtons law of cooling. The model equation was

d

dt= k( s) (1)

where = (t) is the temperature of the cooling object at time t, s the temperature of theenvironment (assumed constant) and k is a thermal constant related to the object. Let 0 bethe initial temperature of the liquid, i.e.

= 0 at t = 0.

Try each part of this exercise

Solve this initial value problem.

Part (a) Separate the variables to obtain an equation connecting two integrals

Answer

Part (a) Now integrate both sides of this equation

Answer

Part (a) Apply the initial condition and take exponentials to obtain a formula for

Answer

Hence ln( s) = kt + ln(0 s) so that ln( s) ln(0 0) = ktThus, rearranging and inverting, we nd:

ln

( s0 s

)= kt s

0 s = ekt

and so, nally, = s + (0 s)ekt.The graph of against t is shown in the gure below.

0

s

t

We see that as time increases (t ), then the temperature of the object cools down to thatof the environment, that is: s.Note that we could have solved (1) by the integrating factor method.

Engineering Mathematics: Open Learning Unit Level 119.7: Dierential Equations

2


Part (a) Write the equation as

d

dt+ k = ks (2)

What is the integrating factor for this equation?

Answer

Multiplying (2) by this factor we nd that

ektd

dt+ kekt = kse

kt or, rearranging,d

dt(ekt) = kse

kt

Part (b) Now integrate this equation and apply the initial condition

Answer

Hence = s + C ekt. Then, applying the initial condition: when t = 0, = s + C so that

C = 0 s and, nally, = s + (0 s)ekt,

as before.

Electrical circuitsAnother application of rst-order dierential equations arises in the modelling of electrical cir-cuits.In Block 19.1 the dierential equation for the RL circuit of the gure below was shown to be

Ldi

dt+ Ri = E

in which the initial condition is i = 0 at t = 0.

E

+

R Li


Part (a) Write this equation in standard form {dydx

+ P (x)y = Q(x)} and obtain the integratingfactor.

Answer

Multiplying the equation in standard form by the integrating factor gives

eRt/Ldi

dt+ eRt/L

R

Li =

E

LeRt/L

or, rearranging,d

dt(eRt/L i) =

E

LeRt/L.

3 Engineering Mathematics: Open Learning Unit Level 119.7: Dierential Equations

Part (b) Now integrate both sides and apply the initial condition to obtain the solution

Answer

2. Modelling Free Mechanical OscillationsConsider the following schematic diagram of a shock absorber:

Mass

Spring

Dashpot

The equation of motion can be described in terms of the vertical displacement x of the mass.

Let m be the mass, kdx

dtthe damping force resulting from the dashpot and nx the restoring

force resulting from the spring. Here, k and n are constants. Then the equation of motion is

md2x

dt2= kdx

dt nx.

Suppose that the mass is displaced a distance x0 initially and released from rest. Then at t = 0,

x = x0 anddx

dt= 0. Writing the dierential equation in standard form:

md2x

dt2+ k

dx

dt+ nx = 0.

We shall see that the nature of the oscillations described by this dierential equation, dependscrucially upon the relative values of the mechanical constants m, k and n.

Now do this exercise

Find the auxiliary equation of this dierential equation and solve it.

Answer

The value of k controls the amount of damping in the system. We explore the solution for twoparticular values of k.


4

No damping

If k = 0 then there is no damping. We expect, in this case, that once motion has started it willcontinue for ever. The motion that ensues is called simple harmonic motion. In this case wehave

=4m n

2m,

that is,

=

n

mi where i =

1.

In this case the solution for the displacement x is:

x = A cos

(n

mt

)+ B sin

(n

mt

)

where A, B are arbitrary constants.

Now do this exercise

Now apply the initial conditions to nd the unique solution:

Answer

Light damping

If k2 4mn < 0, i.e. k2 < 4mn then the roots of the auxiliary equation are complex:

1 =k + i4mn k2

2m2 =

k i4mn k22m

Then, after some rearrangement:

x = ekt/2m [A cos pt + B sin pt]

in which p =

4mn k2/2m.


Part (a) If m = 1, n = 1 and k = 1 nd 1 and 2 and then nd the solution for thedisplacement x.

Answer

Part (b) Impose the initial conditions x = x0,dxdt

= 0 at t = 0 to nd the arbitrary constants.

Answer


The graph of x against t is shown below. This is the case of light damping. As the dampingin the system decreases (i.e. k 0 ) the number of oscillations (in a given time interval) willincrease. In many mechanical systems these oscillations are usually unwanted and the designerwould choose a value of k to either reduce them or to eliminate them altogether. For the choicek = 4mn, known as the critical damping case, all the oscillations are absent.

t

x0

x

x0et/2

x = x0et/2

[cos

3

2t +

3

3sin

3

2t

]

Heavy dampingIf k2 4mn > 0, i.e. k2 > 4mn then there are two real roots of the auxiliary equation, 1 and2:

1 =k +k2 4mn

2m2 =

k k2 4mn2m

Thenx = Ae1t + Be2t.


Part (a) If m = 1, n = 1 and k = 2.5 nd 1 and 2 and then nd the solution for thedisplacement x.

Answer

Part (b) Impose the initial conditions x = x0,dxdt

= 0 at t = 0 to nd the arbitrary constants.

Answer

3. Modelling Forced Mechanical OscillationsSuppose now that the mass is subject to a force f(t) after the initial disturbance. Then theequation of motion is

md2x

dt2+ k

dx

dt+ nx = f(t)

Consider the case f(t) = F cos t, that is, an oscillatory force of magnitude F and angularfrequency . Choosing specic values for the constants in the model: m = n = 1, k = 0, and = 2 we nd

d2x

dt2+ x = F cos 2t


6


Part (a) Find the complementary function for this equation.

Answer

Part (b) Now nd a particular integral for the dierential equation.

Answer

Part (c) Finally, apply the initial conditions to nd the solution for the displacement x.

Answer

If the angular frequency of the applied force is nearly equal to that of the free oscillation thephenomenon of beats occurs. If the angular frequencies are equal we get the phenomenon ofresonance. Both these phenomena are dealt with in the Computer Exercises. Note that we caneliminate resonance by introducing damping into the system. See the Computer Exercises.


More exercises for you to try

1. In an RC circuit (a resistor and a capacitor in series) the applied emf is a constant E.Given that dq

dt= i where q is the charge in the capacitor, i the current in the circuit, R

the resistanceand C the capacitance the equation for the circuit is

Ri +q

C= E.

If the initial charge is zero nd the charge subsequently.

2. If the voltage in the RC circuit is E = E0 cos t nd the charge and the current attime t.

3. An object is projected from the Earths surface. What is the least velocity (the escapevelocity) of projection in order to escape the gravitational eld, ignoring air resistance.The equation of motion is

m vdv

dx= m gR

2

x2

where the mass of the object is m, its distance from the centre of the Earth is x and theradius of the Earth is R.

4. The radial stress p at distance r for the axis of the thick cylinder subjected to internalpressure is given by p + r dp

dr= A p where A is a constant. If p = p0 at the inner wall

r = r1 and is negligible (p = 0) at the outer wall r = r2 nd an expression for p.

5. The equation for an LCR circuit with applied voltage E is

Ldi

dt+ Ri +

1

Cq = E.

By dierentiating this equation nd the solution for q(t) and i(t) if L = 1, R = 100,C = 104 and E = 1000 given that q = 0 and i = 0 at t = 0.

6. Consider the free vibration problem when m = 1, n = 1 and k = 2. (Critical damping)Find the solution for x(t).

7. Repeat problem 6 for the casem = 1, n = 1 and k = 1.5 (light damping)

8. Consider the forced vibration problem with m = 1, n = 25, k = 8, E = sin 3t, x0 = 0with an initial velocity of 3.

Answer


8

4. Computer Exercise or ActivityFor this exercise it will be necessary for you to access thecomputer package DERIVE.To solve a dierential equation using DERIVE it is necessaryto load what is called a Utility File. In this case you will eitherneed to load ode1 or ode2. To do this is simple. Proceed asfollows: In DERIVE, choose File:Load:Math and select thele (double click) on the ode1 or ode2 icon. This will loada number of commands which enable you to solve rst-orderand second-order dierential equations. You can use the Helpfacility to learn more about these if you wish.

Also note that many of the dierential equations presented in this Block are linear dierentialequations having the general form

dy

dx+ p(x)y = q(x) y(x0) = y0

Such equations can also be solved in DERIVE using the command Linear1(p,q,x,y,x0,y0) orby using the related command Linear1 Gen(p,q,x,y,c) for a solution to a dierential equationwithout initial conditions but which contains a single arbitrary constant c.

Also use the Help command to nd out about the more general commands Dsolve1(p, q, x, y, x0, y0)and Dsolve1 Gen(p, q, x, y, c) used for solving very general rst-order ordinary dierential equa-tions.For second-order equations use the DERIVE command

Dsolve2(p, q, r, x, c1, c2)

which nds the general solution (containing two arbitrary constants c1, c2) to the second orderdierential equation

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x)

For many of the examples in this Block both p(x) and q(x) are given constants.As an exercise use DERIVE to check the correctness of the solutions requested in the examplesand guided exercises of this Block.

As a more involved exercise use DERIVE to explore the solution to md2x

dt2+ k

dx

dt+ nx =

f(t) with k2 < 4mn (trigonometric solutions with frequency s =

4mn k2

2m) and when

f(t) = sin t. You will nd that after an initial period the system will vibrate with the samefrequency as the forcing function. The initial response is aected by the transient function(the complementary function). If there is damping in the system this will die away, due tothe decaying exponential terms in the complementary function, to leave only that part of thesolution arising from the particular integral. An interesting case arises when s. You willnd, by plotting the solution curve, that the phenomenon of beats occurs. That is, the systenbeats with a frequency neither of the system frequency nor of the frequency of the forcingfunction. As s the magnitude of the peaks and troughs of the solution curve increase.


If there is no damping in the system the response becomes unbounded as time increases in thelimit = s. The system is in resonance with the forcing function. These extreme oscillationscan be reduced by introducing damping into the system. Examine this phenomenon as thedamping (k) is reduced.MAPLE will solve a wide range of ordinary dierential equations including systems of dierentialequations using the command

dsolve(deqns,vars,eqns)where:deqns ordinary dierential equation in vars, or set of equations and/or initial conditions.vars variable or set of variables to be solved foreqns optional equation of the form keyword=valueFor example to solve

d2y

dt2+ 2

dy

dt+ 2y = et y(0) = 0, y(0) = 0

we would key in> dsolve({di(y(t),t$2)+2*di(y(t),t)+2*y(t)=exp(-t),y(0)=0, D(y)(0)=0},y(t),type=exact);MAPLE responds with

1 cos(t)et

If the initial conditions are omitted MAPLE will present the solution with the correct numberof arbitrary constants denoted by C1, C2 . . . . Thus the general solution of

d2y

dt2+ 2

dy

dt+ 2y = et

is obtained by keying in> dsolve({di(y(t),t$2)+2*di(y(t),t)+2*y(t)=exp(-t)},y(t),type=exact);and MAPLE responds with

y(t) = exp(t) + C1 exp(t) cos(t) + C2 exp(t) sin(t)


10

End of Block 19.7


d

s =

k dt

Back to the theory


12

ln( s) = kt + C where C is constant

Back to the theory


ln(0 s) = C.

Back to the theory


14

e

k dt = ekt is the integrating factor.

Back to the theory


= s + (0 s)ekt since integration produces ekt = sekt + C, where C is an arbitraryconstant.

Back to the theory


16

divide the dierential equation through by L to obtain

di

dt+

R

Li =

E

L

which is now in standard form. The integrating factor is e

RL

dt = eRt/L.

Back to the theory


You should obtain i =E

R(1 eRt/L), since integrating the dierential equation gives:

eRt/L i =E

ReRt/L + C

where C is a constant. Then applying the initial condition i = 0 when t = 0 gives

0 =E

R+ C

so that C = ER

and eRt/L i =E

ReRt/L E

R. Finally, i =

E

R(1 eRt/L). Note that as t ,

i ER

.

Back to the theory


18

Put x = et then the auxiliary equation is

m 2 + k + n = 0.

Hence =k k2 4m n

2m.

Back to the theory


You should obtain x = x0 cos(

nm

t). To see this we rst need an expression for the

derivative,dx

dt:

dx

dt=

n

mA sin

(n

mt

)+

n

mB cos

(n

mt

)

When t = 0,dx

dt= 0 so that

n

mB = 0 so that B = 0.

Therefore

x = A cos

(n

mt

).

Imposing the remaining initial condition: when t = 0, x = x0 so that x0 = A and nally:

x = x0 cos

(n

mt

).

Back to the theory


20

= 1+i

412

= 1/2 i3/2. Hence x = et/2[A cos

3

2t + B sin

3

2t].

Back to the theory


Dierentiating, we obtain

dx

dt= 1

2et/2

[A cos

3

2t + B sin

3

2t

]+ et/2

[

3

2A sin

3

2t +

3

2B cos

3

2t

]

At t = 0,

x = x0 = A (i)

dx

dt= 0 = 1

2A +

3

2B (ii)

Solving (i) and (ii) we obtain

A = x0 B =

33

x0

Then

x = x0et/2

[cos

3

2t +

3

3sin

3

2t

].

Back to the theory


22

=2.56.25 4

2= 1.25 0.75

Hence 1, 2 = 0.5,2 and so x = Ae0.5t + Be2t

Back to the theory


Dierentiating, we obtaindx

dt= 0.5Ae0.5t 2Be2t

At t = 0,

x = x0 = A + B (i)

dx

dt= 0 = 0.5A 2B (ii)

Solving (i) and (ii) we obtainA = 4

3x0 B = 13x0

Thenx = 1

3x0(4e

0.5t e2t).The graph of x against t is shown below. This is the case of heavy damping. Other cases aredealt with in the Exercises at the end of the Block.

t

x0

x

Back to the theory


24

The homogeneous equation isd2x

dt2+ x = 0

with auxiliary equation 2 + 1 = 0. Hence the complementary function is

xcf = A cos t + B sin t.

Back to the theory


Sayxp = C cos 2t + D sin 2t

so thatd2xpdt2

= 4C cos 2t 4D sin 2t.Substituting into the dierential equation

(4C + C) cos 2t + (4D + D sin 2t) F cos 2t.

Comparing coecients gives

3C = F and 3D = 0

so thatD = 0 C = 1

3F xp = 13F cos 2t.

The general solution of the dierential equation is therefore

x = xp + xcf = 13F cos 2t + A cos t + B sin t.

Back to the theory


26

You should obtainx = 1

3F cos 2t +

(x0 +

13F

)cos t.

To obtain this we need to determine the derivative and apply the initial conditions:

dx

dt= 2

3F sin 2t A sin t + B cos t.

At t = 0

x = x0 = 13F + A (i)dx

dt= 0 = B (ii)

HenceB = 0 and A = x0 +

13F.

Thenx = 1

3F cos 2t +

(x0 +

13F

)cos t.

The graph of x against t is shown below.

t

x0

x

Back to the theory


1. Use the equation in the form Rdq

dt+

q

C= E or

dq

dt+

1

RCq =

E

R. The integrating factor is

et/RC and the general solution is

q = EC(1 et/RC) and as t q EC.

2. q =E0C

1 + 2R2C2[cos t et/RC + RC sin t]

i =dq

dt=

E0C

1 + 2R2C2

[ sin t + 1

RCet/RC + 2RC cos t

].

3. vmin =

2gR. If R = 6378 km and g = 9.81 m s2 then vmin = 11.2 km s1.

4. p =p0r

21

r22 r21

(1 r

22

r2

)

5. q = 0.1 110

3e50t(sin 50

3t +

3 cos 50

3t) i =

203e50t sin 50

3t.

6. x = x0(1 + t)et

7. x = x0e0.75t(cos

7

4t + 3

7sin

7

4t) 8.

x =1

104

[e4t (3 cos 3t + 106 sin 3t) 3 cos 3t + 2 sin 3t)]

Back to the theory


28

19_7

Documents

order dierential equations

ln s0 s

ln s ln0

kt s0 s

kt ln0 s

answerhence ln s

s c ekt

time t