Engineering Math 2.pdf

8/14/2019 Engineering Math 2.pdf

1/16

Differential equati

50

Second order differential equations of

the formad2

ydx2

+ bdydx

+ cy= 0

50.1 Introduction

An equation of the form ad2y

dx2+b

dy

dx+ cy= 0, where

a,b

andc

are constants, is called alinear second

order differential equation with constant coeffi-cients. When the right-hand side of the differentialequation is zero, it is referred to as a homogeneousdifferential equation. When the right-hand side isnot equal to zero (as in Chapter 51) it is referred toas anon-homogeneous differential equation.

There are numerous engineering examplesof second order differential equations. Threeexamples are:

(i) Ld2q

dt2 +R

dq

dt+

1

Cq= 0, representing an equa-

tion for chargeqin an electrical circuit contain-ing resistanceR, inductanceL and capacitanceCin series.

(ii) md2s

dt2+ a

ds

dt+ ks= 0, defining a mechanical

system, where s is the distance from a fixedpoint after tseconds, m is a mass, a the dampingfactor andkthe spring stiffness.

(iii)d2y

dx2+

P

EIy= 0, representing an equation for

the deflected profile y of a pin-ended uniform

strut of length l subjected to a load P. E isYoungs modulus and I is the second momentof area.

If D representsd

dxand D2 represents

d2

dx2 then the

above equation may be stated as

(aD2 + bD + c)y= 0. This equation is said to be inD-operator form.

Ify=Aemx thendy

dx=Amemx and

d2y

dx2=Am2emx.

Substituting these values intoad2y

dx2+ b

dy

dx+ cy= 0

gives:

a(Am2emx) + b(Amemx) + c(Aemx) = 0

i.e. Aemx(am2 + bm+ c) = 0

Thusy=Aemx is a solution of the given equation pro-

vided that (am2 + bm+ c)= 0.am2 + bm+ c= 0 iscalled the auxiliary equation, and since the equationis a quadratic,mmay be obtained either by factoris-ing or by using the quadratic formula. Since, in theauxiliary equation, a, b and c are real values, thenthe equation may have either

(i) two different real roots (whenb

2>

4ac) or(ii) two equal real roots (whenb2 = 4ac) or

(iii) two complex roots (whenb2< 4ac).

50.2 Procedure to solve differentialequations of the form

ad2y

dx2+b

dy

dx+cy= 0

(a) Rewrite the differential equation

ad2y

dx2+ b

dy

dx+ cy = 0

as (aD2 + bD + c)y = 0

(b) Substitute m for D and solve the auxiliary

equationam2 + bm+ c= 0 form.

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476 DIFFERENTIAL EQUATIONS

(c) If the roots of the auxiliary equation are:

(i) real and different, say m= and m=,then the general solution is

y = Aex + Bex

(ii) real and equal, say m=twice, then the

general solution isy = (Ax + B)ex

(iii) complex, saym=j, then the generalsolution is

y = ex{A cosx + B sin x}

(d) Given boundary conditions, constants A and B,may be determined and theparticular solutionof the differential equation obtained.

The particular solutions obtained in the workedproblems of Section 50.3 may each be verified by

substituting expressions for y, dydx

and d2

ydx2

into the

original equation.

50.3 Worked problems on differentialequations of the form

ad2y

dx2+b

dy

dx+cy= 0

Problem 1. Determine the general solution

of 2d2y

dx2+ 5

dy

dx 3y= 0. Find also the parti-

cular solution given that whenx = 0, y= 4 anddy

dx= 9.

Using the above procedure:

(a) 2d2y

dx2+ 5

dy

dx 3y= 0 in D-operator form is

(2D2 + 5D 3)y= 0, where D d

dx

(b) Substituting m for D gives the auxiliary equation

2m2 + 5m 3 = 0.

Factorising gives: (2m 1)(m+ 3)= 0, from

which,m= 12 orm=3.

(c) Since the roots are real and different the general

solution isy=Ae12x+Be3x.

(d) Whenx= 0,y= 4,

hence 4= A+B (1)

Since y= Ae12x +Be3x

then

dy

dx =

1

2Ae

1

2x

3Be3x

When x= 0,dy

dx= 9

thus 9=1

2A 3B (2)

Solving the simultaneous equations (1) and (2)givesA= 6 andB=2.

Hence the particular solution is

y= 6e12x 2e3x

Problem 2. Find the general solution of

9d2y

dt2 24

dy

dt+ 16y= 0 and also the particular

solution given the boundary conditions that

whent= 0,y=dy

dt= 3.

Using the procedure of Section 50.2:

(a) 9d2y

dt2 24

dy

dt+ 16y= 0 in D-operator form is

(9D2 24D+ 16)y= 0 where D d

dt

(b) Substituting m for D gives the auxiliary equation

9m2 24m+ 16= 0.

Factorising gives: (3m 4)(3m 4)= 0, i.e.

m= 43 twice.

(c) Since the roots are real and equal,the general

solution isy= (At+B)e43t.

(d) When t= 0,y= 3hence3= (0+B)e0,i.e.B= 3.

Sincey= (At+B)e43 t

thendy

dt= (At+B)

4

3e

43 t

+ Ae

43 t, by the

product rule.


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SECOND ORDER DIFFERENTIAL EQUATIONS (HOMOGENEOUS) 477

When t= 0, dydt= 3

thus 3 = (0 +B) 43

e0 +Ae0

i.e. 3= 43B+Afrom which,A=1, since

B= 3.Hence the particular solution is

y = (t + 3)e43t or

y = (3t)e43t

Problem 3. Solve the differential equation

d2y

dx2+ 6 dy

dx+ 13y= 0, given that when x= 0,

y

=3 and

dy

dx =7.


(a)d2y

dx2+ 6 dy

dx+ 13y= 0 in D-operator form is

(D2+ 6D+ 13)y= 0, where D ddx

(b) Substitutingm for D gives the auxiliary equation

m2 + 6m+ 13= 0.Using the quadratic formula:

m = 6

[(6)2 4(1)(13)]2(1)

= 6

(16)2

i.e.m = 6j42

= 3 j2(c) Since the roots are complex,the general solu-

tion is

y = e3x(A cos 2x+B sin 2x)(d) Whenx= 0,y= 3, hence

3= e0(A cos0+B sin 0), i.e.A= 3.Sincey= e3x(A cos2x+B sin 2x)

thendy

dx= e3x(2A sin 2x + 2B cos2x) 3e3x(A cos2x +B sin 2x),

by the product rule,

= e3x[(2B 3A)cos2x (2A+ 3B)si n2x]

Whenx = 0, dydx= 7,

hence7= e0[(2B 3A)cos0 (2A+ 3B)sin0]i.e.7= 2B 3A,fromwhich,B= 8,sinceA= 3.Hence the particular solution is

y= e3x(3cos2x + 8 sin 2x)

Since, from Chapter 18, page 178,a cos t+ b sin t=R sin (t+), whereR=

(a2 + b2) and = tan1 a

bthen

3cos2x + 8si n2x=

(32 + 82)sin(2x+ tan1 38 )

=

73 sin(2x + 20.56)=

73 sin(2x + 0.359)

Thus the particular solution may also beexpressed as

y =

73 e3x sin(2x+ 0.359)

Now try the following exercise.

Exercise 188 Further problems on differen-tial equations of the form

ad

2

ydx2

+bdydx

+cy= 0

In Problems 1 to 3, determine the general solu-tion of the given differential equations.

1. 6d2y

dt2 dy

dt 2y= 0

y = Ae 23 t+Be 12 t

2. 4d2

dt2

+4

d

dt+

=0

= (At+B)e 12 t

3.d2y

dx2+ 2 dy

dx+ 5y= 0

[y= ex(A cos2x+B sin 2x)]

In Problems 4 to 9, find the particular solu-tion of the given differential equations for thestated boundary conditions.


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4. 6d2y

dx2+ 5 dy

dx 6y= 0; whenx= 0,y= 5 and

dy

dx= 1.

y = 3e 23x + 2e 32x

5. 4d2y

dt2

5

dy

dt +y

=0; when t

=0, y

=1 and

dy

dt= 2.

y= 4e 14 t 3et

6. (9D2 + 30D + 25)y= 0, where D ddx

;

whenx = 0,y = 0 and dydx

= 2.y= 2xe 53x

7.d2x

dt2

6

dx

dt +9x

=0; when t

=0, x

=2 and

dx

dt= 0. [x = 2(1 3t)e3t]

8.d2y

dx2+ 6 dy

dx+ 13y= 0; whenx= 0,y= 4 and

dy

dx= 0. [y = 2e3x(2cos2x + 3si n2x)]

9. (4D2 + 20D + 125)= 0, where D ddt

;

whent

=0,

=3 and

d

dt =2.5.

[= e2.5t(3cos5t+ 2si n5t)]

50.4 Further worked problems onpractical differential equations of

the formad2y

dx2+b

dy

dx+cy= 0

Problem 4. The equation of motion of a bodyoscillating on the end of a spring is

d2x

dt2+ 100x = 0,

wherex is the displacement in metres of the bodyfrom its equilibrium position after time t sec-onds. Determine x in terms of t given that at

timet= 0,x = 2mand dxdt

= 0.

An equation of the formd2x

dt2+ m2x = 0 is a differ-

ential equation representingsimple harmonic motion(S.H.M.). Using the procedure of Section 50.2:

(a)d2x

dt2+ 100x= 0 in D-operator form is

(D2 + 100)x = 0.(b) The auxiliary equation is m2 + 100 = 0, i.e.

m2 = 100 andm = (100), i.e.m = j10.(c) Since the roots are complex, the general solution

isx = e0(A cos 10t+B sin 10t),i.e.x= (A cos 10t+Bsin10t) metres

(d) Whent= 0,x = 2, thus 2 =Adx

dt= 10A sin 10t+ 10B cos 10t

Whent= 0, dxdt

= 0

thus 0 = 10A sin 0 + 10B cos 0, i.e.B = 0Hence the particular solution is

x= 2cos10tmetres

Problem 5. Given the differential equation

d2V

dt2=2V, where is a constant, show that

its solution may be expressed as:V= 7 cosh t+ 3 sinh t

given the boundary conditions that when

t= 0,V= 7 and dVdt

= 3.


(a)d2V

dt2=2V, i.e. d

2V

dt2 2V= 0 in D-operator

form is (D2

2

)v= 0, where D d

dx

(b) The auxiliary equation is m2 2 = 0, fromwhich,m2 =2 andm = .

(c) Sincethe roots arerealand different, the generalsolution is

V=Aet +Bet

(d) Whent= 0,V= 7 hence 7 =A+B (1)dV

dt=AetBet


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SECOND ORDER DIFFERENTIAL EQUATIONS (HOMOGENEOUS) 479

When t= 0, dVdt= 3,

thus 3 = AB,i.e. 3 = AB (2)From equations (1) and (2), A

=5 and B

=2

Hencethe particular solution is

V= 5et + 2et

Since sinh t= 12

(et et)and cosh t= 12 (et+ et)then sinht+ cosh t= etand cosh t sinh t= et from Chapter 5.Hence the particular solution may also bewritten as

V= 5(sinh t+ cosh t)+ 2(cosh t sinh t)

i.e. V= (5+ 2) cosht+ (5 2) sinhti.e. V= 7 cosht + 3 sinht

Problem 6. The equation

d2i

dt2+ R

L

di

dt+ 1LC

i= 0represents a current i flowing in an elec-trical circuit containing resistance R, induc-tance L and capacitance C connected inseries. If R= 200 ohms, L= 0.20 henry andC= 20 106 farads, solve the equation for igiven the boundary conditions that when t= 0,i= 0 and di

dt= 100.


(a)d2i

dt2+R

L

di

dt+1

LC

i

=0 in D-operator form is

D2 + RL

D+ 1LC

i = 0 where D d

dt

(b) The auxiliary equation ism2+ RLm+ 1

LC= 0

Hencem =RL

RL

2 4(1)

1

LC

2

When R= 200, L= 0.20 and C= 20 106,then

m= 200

0.20

2000.20

2 4

(0.20)(20 106)

2

= 10000

2= 500

(c) Since the two roots are real and equal (i.e.500twice, since for a second order differential equa-tion there must be two solutions), the general

solution is i= (At+B)e500t.

(d) When t= 0, i= 0, hence B= 0di

dt= (At+B)(500e500t)+ (e500t)(A),

by the product rule

When t= 0, didt= 100, thus 100=500B+A

i.e. A= 100, since B= 0Hence the particular solution is

i= 100te500t

Problem 7. The oscillations of a heav-ily damped pendulum satisfy the differential

equation d2

xdt2+ 6 dx

dt+ 8x= 0, where x cm is

the displacement of the bob at time tseconds.The initial displacement is equal to +4cm andthe initial velocity

i.e.

dx

dt

is 8 cm/s. Solve the

equation for x .


(a)d2x

dt2+6

dx

dt +8x=

0 in D-operator form is

(D2 + 6D + 8)x= 0, where D ddt

(b) The auxiliary equation ism2+ 6m+ 8= 0.Factorising gives: (m+ 2)(m+ 4)= 0, fromwhich, m =2 or m =4.

(c) Sincethe roots arerealand different, the general

solution is x=Ae2t +Be4t.

(d) Initial displacement means that time t= 0. Atthis instant, x = 4.


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Thus 4=A+B (1)Velocity,

dx

dt= 2Ae2t 4Be4t

dx

dt = 8 cm/s whent= 0,

thus 8 = 2A 4B (2)From equations (1) and (2),

A = 12 andB = 8


x = 12e2t 8e4t

i.e.displacement,x= 4(3e2t 2e4t) cm


Exercise 189 Further problems on secondorder differential equations of the form

ad2y

dx2+b

dy

dx+cy= 0

1. The charge,q, on a capacitor in a certain elec-trical circuit satisfies the differential equa-

tion d2q

dt2+ 4 dq

dt+ 5q= 0. Initially (i.e. when

t= 0), q=Q and dqdt= 0. Show that the

charge in the circuit can be expressed as:

q=

5Qe2tsin (t+ 0.464)

2. A body moves in a straight line so that itsdistance s metres from the origin after time

tseconds is given byd2s

dt2+ a2s= 0, where a

is a constant. Solve the equation for s giventhats= cand ds

dt= 0 whent= 2

a[s= c cosat]

3. The motion of the pointer of a galvanometerabout its position of equilibrium is repre-sented by the equation

Id2

dt2+ Kd

dt+F= 0.

IfI, the moment of inertia of the pointer aboutits pivot, is 5 103,K, the resistance due tofriction at unit angular velocity, is 2 102and F, the force on the spring necessary toproduce unit displacement, is 0.20, solve theequation for in terms oftgiven that when

t= 0, = 0.3 and ddt= 0.

[= e2t(0.3cos6t+ 0.1si n6t)]4. Determine an expression for x for a differ-

ential equation

d2x

dt2+ 2ndx

dt+ n2

x= 0 whichrepresents a critically damped oscillator,

given that at timet= 0,x= sand dxdt= u.

[x= {s+ (u+ ns)t}ent]

5. Ld2i

dt2+Rdi

dt+ 1

Ci= 0 is an equation repre-

senting current i in an electric circuit. Ifinductance L is 0.25 henry, capacitance C

is 29.76 106 farads and R is 250 ohms,solve the equation for i given the boundary

conditions that when t= 0, i= 0and didt= 34.

i= 120

e160t e840t

6. The displacement s of a body in a dampedmechanical system, with no external forces,satisfies the following differential equation:

2d2s

dt2+ 6 ds

dt+ 4.5s= 0

where t represents time. If initially, when

t= 0, s= 0 and dsdt= 4, solve the differential

equation fors in terms oft. [s= 4te 32 t]


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Differential equati

51

Second order differential equations of

the formad2

ydx2 + bdydx + cy = f(x)

51.1 Complementary function andparticular integral

If in the differential equation

ad2y

dx2+ bdy

dx+ cy = f(x) (1)

the substitutiony= u+ vis made then:

ad2(u+ v)

dx2 + b

d(u+ v)

dx+ c(u+ v) = f(x)

Rearranging gives:

a

d2u

dx2+ b

du

dx+ cu

+a

d2v

dx2+ b

dv

dx+cv

= f(x)

If we let

ad2v

dx2+ b

dv

dx+ cv = f(x) (2)

then

ad2u

dx2 + bdu

dx+ cu = 0 (3)

The general solution,u, of equation (3) will con-tain two unknown constants, as required for thegeneral solution of equation (1). The method of solu-tion of equation (3) is shown in Chapter 50. Thefunction u is called the complementary function(C.F.).

If the particular solution, v, of equation (2) canbe determined without containing any unknown

constants theny= u+ vwill give the general solu-tion of equation (1). The function v is called the par-ticular integral (P.I.). Hence the general solution ofequation (1) is given by:

y = C.F.+P.I.

51.2 Procedure to solve differentialequations of the form

ad2y

dx2+b

dy

dx+cy = f(x)

(i) Rewrite the given differential equation as

(aD2 + bD+ c)y=f(x).

(ii) Substitute m for D, and solve the auxiliaryequationam2 + bm+ c= 0 form.

(iii) Obtain the complementary function,u, whichis achieved using the same procedure as inSection 50.2(c), page 476.

(iv) To determine the particular integral, v, firstlyassume a particular integral which is sug-gested by f(x), but which contains undeter-mined coefficients. Table 51.1 on page 482gives some suggested substitutions for differentfunctionsf(x).

(v) Substitute the suggested P.I. into the dif-ferential equation (aD2 + bD+ c)v=f(x) andequate relevant coefficients to find the constantsintroduced.

(vi) The general solution is given byy=C.F.+P.I., i.e.y= u+ v.

(vii) Given boundary conditions, arbitrary constantsin the C.F. may be determined and the par-ticular solution of the differential equationobtained.


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Table 51.1 Form of particular integral for different functions

Type Straightforward cases Snag cases SeeTry as particular integral: Try as particular integral: problem

(a) f(x) = a constant v = k v = kx (used when C.F. 1, 2contains a constant)

(b) f(x) = polynomial (i.e. v = a + bx + cx2 + 3f(x) =L +Mx +Nx2 + where any of the coefficientsmay be zero)

(c) f(x) = an exponential function v = keax (i) v = kxeax (used when eax 4, 5(i.e.f(x) =Aeax) appears in the C.F.)

(ii) v = kx2eax (used when eax 6andxeax both appear inthe C.F.)

(d) f(x) = a sine or cosine function v =A sinpx +B cospx v =x(A sinpx +B cospx) 7, 8(i.e.f(x) = a sinpx + b cospx, (used when sinpxand/or

wherea orb may be zero) cospxappears in the C.F.)

(e) f(x) = a sum e.g. 9

(i) f(x) = 4x2 3sin2x (i) v = ax2 + bx + c+ dsin 2x + e c os2x

(ii) f(x) = 2 x + e3x (ii) v = ax + b + ce3x

(f) f(x) = a product e.g. v = ex (A sin 2x +B cos 2x) 10f(x) = 2ex cos2x


ad2y

dx2+b

dy

dx+cy = f(x)where

f(x)is a constant or polynomial


d2y

dx2 +

dy

dx 2y = 4.


(i)d2y

dx2 +

dy

dx 2y = 4 in D-operator form is

(D2 + D 2)y = 4.

(ii) Substituting m for D gives the auxiliary equa-

tion m2 + m 2 = 0. Factorising gives: (m 1)(m + 2) = 0, from whichm = 1 orm = 2.

(iii) Since the roots are real and different, the C.F.,

u = Aex +Be2x.

(iv) Since the term on the right hand side of the given

equation is a constant, i.e. f(x) = 4, let the P.I.also be a constant, sayv = k(see Table 51.1(a)).

(v) Substituting v = k into (D2 + D 2)v = 4

gives (D2 + D 2)k= 4. Since D(k) = 0 and

D2(k) = 0 then 2k= 4, from which, k = 2.Hence the P.I., v=2.

(vi) The general solution is given byy = u + v, i.e.

y=Aex+Be2x 2.

Problem 2. Determine the particular solu-

tion of the equation d2

ydx2

3 dydx

= 9, given the

boundary conditions that whenx = 0,y = 0 anddy

dx= 0.


(i)d2y

dx2 3

dy

dx= 9 in D-operator form is

(D2 3D)y = 9.


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SECOND ORDER DIFFERENTIAL EQUATIONS (NON-HOMOGENEOUS) 483

(ii) Substituting m for D gives the auxil-

iary equationm2 3m= 0. Factorising gives:m(m 3)= 0, from which,m= 0 orm= 3.


u=Ae0 +Be3x, i.e.u =A +Be3x.

(iv) Since the C.F. contains a constant (i.e.A) thenlet the P.I., v= kx(see Table 51.1(a)).

(v) Substituting v= kxinto (D2 3D)v= 9 gives

(D2 3D)kx= 9.D(kx)= kand D2(kx)= 0.Hence (D2 3D)kx= 0 3k= 9, from which,k=3.Hence the P.I., v = 3x.

(vi) The general solution is given byy= u+ v, i.e.

y =A +Be3x 3x.

(vii) When x= 0, y= 0, thus 0=A+Be0 0, i.e.

0=A+B (1)dy

dx= 3Be3x 3;

dy

dx= 0 when x= 0, thus

0= 3Be0 3 from which, B= 1. From equa-tion (1),A=1.Hence the particular solution is

y = 1 + 1e3x 3x,

i.e. y = e3x 3x 1

Problem 3. Solve the differential equation2

d2y

dx2 11

dy

dx+ 12y = 3x 2.


(i) 2d2y

dx2 11

dy

dx+ 12y= 3x 2 in D-operator

form is

(2D2 11D + 12)y = 3x 2.

(ii) Substituting m for D gives the auxiliary equa-tion 2m2 11m+ 12= 0. Factorising gives:

(2m 3)(m 4)= 0, from which, m= 32 orm= 4.


u =Ae32x

+Be4x

(iv) Sincef(x)= 3x 2 is a polynomial, let the P.I.,v= ax+ b(see Table 51.1(b)).

(v) Substituting v= ax+ binto

(2D2 11D+ 12)v= 3x 2 gives:

(2D2 11D + 12)(ax + b) = 3x 2,

i.e. 2D2(ax + b) 11D(ax + b)

+ 12(ax + b) = 3x 2

i.e. 0 11a+ 12ax + 12b = 3x 2

Equating the coefficients ofx gives: 12a= 3,

from which,a= 14

.

Equating the constant terms gives:

11a+ 12b=2.

i.e. 11

14

+ 12b=2 from which,

12b=2+11

4=

3

4i.e.b=

1

16

Hence the P.I., v= ax+ b=1

4x +

1

16

(vi) The general solution is given byy=u+ v

, i.e.

y = Ae32x

+Be4x +1

4x +

1

16



ad2y

dx2 + b

dy

dx+ cy = f(x) where f(x) is a

constant or polynomial.

In Problems 1 and 2, find the general solutionsof the given differential equations.

1. 2d2y

dx2+ 5

dy

dx 3y = 6

y = Ae

12x+Be3x 2

2. 6d2y

dx2+ 4

dy

dx 2y = 3x 2

y = Ae

1

3x

+Bex 2

32x

In Problems 3 and 4 find the particular solutionsof the given differential equations.

3. 3d2y

dx2 +

dy

dx 4y= 8; whenx = 0, y= 0 and

dy

dx= 0.

y = 27

(3e

43x+ 4ex) 2


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4. 9d2y

dx2 12

dy

dx+ 4y = 3x 1; when x = 0,

y = 0 anddy

dx=

4

3y =

2 + 3

4x

e

23x + 2 + 3

4x

5. The chargeq in an electric circuit at time tsat-

isfies the equation Ld2q

dt2 +R

dq

dt+

1

Cq =E,

whereL,R,Cand Eare constants. Solve theequation given L = 2H, C= 200 106 FandE= 250 V, when (a)R = 200 and (b)Ris negligible. Assume that when t= 0, q = 0

anddq

dt= 0.

(a) q =

1

20

5

2

t+1

20 e

50t

(b) q =1

20(1 cos 50t)

6. In a galvanometer the deflectionsatisfies the

differential equationd2

dt2 + 4

d

dt+ 4 = 8.

Solve theequation forgiventhat when t= 0,

=d

dt= 2. [ = 2(te2t+ 1)]


ad2y

dx2+b

dy

dx+ cy = f(x) where

f(x)is an exponential function

Problem 4. Solve the equation

d

2

ydx2

2 dydx

+ y = 3e4x given the boundary

conditions that whenx = 0,y = 23

anddy

dx= 4 1

3


(i)d2y

dx2 2

dy

dx+ y = 3e4x in D-operator form is

(D2 2D + 1)y = 3e4x.

(ii) Substituting m for D gives the auxiliary

equation m2 2m + 1 = 0. Factorising gives:(m 1)(m 1) = 0, from which,m = 1 twice.

(iii) Since the roots are real and equal the C.F.,u= (Ax+B)ex.

(iv) Let the particular integral, v = ke4x (see

Table 51.1(c)).

(v) Substituting v = ke4x into

(D2 2D + 1)v = 3e4x gives:

(D2 2D + 1)ke4x = 3e4x

i.e. D2(ke4x) 2D(ke4x) + 1(ke4x) = 3e4x

i.e. 16ke4x 8ke4x + ke4x = 3e4x

Hence 9ke4x = 3e4x, from which, k= 13Hence the P.I., v=ke4x= 1

3e4x.

(vi) The general solution is given byy = u + v, i.e.

y= (Ax+B)ex+ 13

e4x.

(vii) Whenx = 0,y = 23 thus

23

= (0 +B)e0 + 13

e0, from which,B = 1.dy

dx= (Ax +B)ex + ex(A) + 4

3e4x.

Whenx = 0,dy

dx= 4

1

3, thus

13

3=B +A +

4

3from which,A = 4, sinceB = 1.Hence the particular solution is:

y = (4x 1)ex + 13

e4x


2d2y

dx2

dy

dx 3y = 5e

32x

.


(i) 2d2y

dx2

dy

dx 3y = 5e

32x

in D-operator form is

(2D2 D 3)y = 5e32x

.


equation 2m2 m 3 = 0. Factorising gives:

(2m 3)(m + 1) = 0, from which, m = 32 orm = 1. Since the roots are real and different

then the C.F.,u=Ae32x+Bex.

(iii) Since e32x

appears in the C.F. and in theright hand side of the differential equation, let


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the P.I., v= kxe32x

(see Table 51.1(c), snagcase (i)).

(iv) Substituting v= kxe32x

into (2D2 D 3)v=

5e32x

gives: (2D2 D 3)kxe32x= 5e

32x

.

Dkxe

32x= (kx)

32

e32x+

e

32x

(k),

by the product rule,

= ke32x3

2x + 1

D2

kxe

32x

= D

ke

32x

32x + 1

=

ke

32x

32

+ 3

2x + 1 3

2ke

32x

= ke32x

94x + 3

Hence (2D2 D 3)

kxe

32x

= 2

ke

32x

94x + 3

ke

32x

32x + 1

3

kxe

32x

= 5e

32x

i.e. 92kxe

32x+ 6ke

32x

32xke

32x ke

32x

3kxe32x= 5e

32x

Equating coefficients of e32x

gives: 5k= 5, fromwhich,k= 1.

Hence the P.I., v=kxe32x=xe

32x

.

(v) The general solution is y= u+ v, i.e.

y=Ae32x+Bex +xe

32x

.

Problem 6. Solved2y

dx2 4

dy

dx+ 4y = 3e2x.


(i)d2y

dx2 4

dy

dx+ 4y= 3e2x in D-operator form is

(D2 4D+ 4)y= 3e2x.


equation m2 4m+ 4= 0. Factorising gives:(m 2)(m 2)= 0, from which,m= 2 twice.

(iii) Since the roots are real and equal, the C.F.,

u= (Ax+B)e2x.

(iv) Since e2x and xe2x both appear in the C.F.

let the P.I., v= kx2e2x (see Table 51.1(c), snagcase (ii)).

(v) Substituting v= kx2e2x into (D2 4D+ 4)v =

3e2x gives: (D2 4D+ 4)(kx2e2x)= 3e2x

D(kx2e2x) = (kx2)(2e2x) + (e2x)(2kx)

= 2ke2x(x2 +x)

D2(kx2e2x) = D[2ke2x(x2 +x)]

= (2ke2x)(2x + 1) + (x2 +x)(4ke2x)

= 2ke2x

(4x + 1 + 2x2)

Hence (D24D + 4)(kx2e2x)

= [2ke2x(4x + 1 + 2x2)]

4[2ke2x(x2 +x)] + 4[kx2e2x]

= 3e2x

from which, 2ke2x= 3e2x andk= 32Hence the P.I., v=kx2e2x= 3

2x2e2x.

(vi) The general solution,y= u+ v, i.e.

y = (Ax+B)e2x + 32x2e2x



ad2y

dx2+b

dy

dx+cy=f(x)wheref(x) is an expo-

nential function

In Problems 1 to 4, find the general solutions ofthe given differential equations.

1.d2y

dx2

dy

dx 6y = 2ex

y = Ae3x +Be2x 1

3ex

2.d2y

dx2 3

dy

dx 4y = 3ex

y = Ae4x +Bex 35xe

x


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(i)d2y

dx2+ 16y= 10cos4xin D-operator form is

(D2 + 16)y = 10cos4x(ii) The auxiliary equation is m2

+16

=0, from

whichm= 16 = j4.(iii) Since the roots are complex the C.F.,

u= e0(A cos4x +B sin 4x)i.e.u=Acos 4x+B sin 4x

(iv) Since sin 4x occurs in the C.F. and in theright hand side of the given differential equa-tion, let the P.I., v =x(Csin 4x+D cos4x) (seeTable 51.1(d), snag caseconstants C and Dare used sinceA and B have already been usedin the C.F.).

(v) Substituting v =x(Csin 4x+D cos4x) into(D2 + 16)v = 10cos4xgives:

(D2 + 16)[x(Csin 4x +D cos4x)]= 10cos4x

D[x(Csin 4x +D cos4x)]= x(4Ccos4x 4D sin 4x)

+ (Csin 4x +D cos4x)(1),by the product rule

D2[x(Csin 4x +D cos4x)]= x(16Csin 4x 16D cos4x)

+ (4Ccos4x 4D sin 4x)+ (4Ccos4x 4D sin 4x)

Hence (D2 + 16)[x(Csin 4x+D cos4x)]= 16Cx sin 4x16Dx cos4x+ 4Ccos4x

4D sin 4x + 4Ccos4x 4D sin 4x+ 16Cx sin 4x + 16Dx cos4x

= 10cos4x,i.e. 8D sin 4x+ 8Ccos4x= 10cos4xEquating coefficients of cos 4xgives:

8C= 10, from which,C= 108

= 54

Equating coefficients of sin 4xgives:8D= 0, from which,D= 0.Hence the P.I., v=x

54

sin 4x

.

(vi) The general solution,y= u+ v, i.e.y = Acos 4x+Bsin 4x+ 5

4xsin 4x

(vii) Whenx = 0,y= 3, thus3 =A cos0 +B sin 0 + 0, i.e.A= 3.

dy

dx = 4A sin 4x

+4B cos4x

+ 54x(4cos4x) + 5

4sin 4x

Whenx = 0, dydx

= 4, thus

4 = 4A sin 0 + 4B cos0 + 0 + 54

sin 0

i.e. 4 = 4B, from which,B= 1Hence the particular solution is

y = 3 cos 4x+ sin 4x+ 54xsin 4x



ad2y

dx2 + b

dy

dx+ cy = f(x)wheref(x) is a sine

or cosine function


1. 2d2y

dx2 dy

dx 3y = 25sin2xy = Ae 32x +Bex

15

(11 sin 2x 2cos2x)

2.d2y

dx2 4 dy

dx+ 4y= 5cosx

y= (Ax+B)e2x 45sinx+ 35cosx

3.d2y

dx2+ y = 4cosx

[y

=A cosx

+B sinx

+2x sinx]

4. Find the particular solution of the differential

equationd2y

dx2 3 dy

dx 4y= 3 sinx; when

x= 0,y= 0 and dydx

= 0.y =

1

170(6e4x 51ex)

134

(15 sinx 9cosx)


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5. A differential equation representing the

motion of a body isd2y

dt2 + n2y = ksinpt,

where k, n and p are constants. Solve the

equation (given n = 0 andp2 = n2) given that

whent= 0,y =dy

dt

= 0.

y =

k

n2 p2

sinpt

p

nsin nt

6. The motion of a vibrating mass is given by

d2y

dt2 + 8

dy

dt+ 20y = 300 sin 4t. Show that the

general solution of the differential equation isgiven by:

y = e4t(A cos2t+B sin 2t)

+15

13

(si n4t 8cos4t)

7. Ld2q

dt2 +R

dq

dt+

1

Cq =V0sin t represents

the variation of capacitor charge in anelectric circuit. Determine an expressionfor q at time tseconds given that R= 40 ,

L = 0.02H, C= 50 106 F, V0 = 540.8 Vand = 200 rad/s and given the boundaryconditions that when t= 0, q = 0 anddq

dt= 4.8

q = (10t+ 0

.01)e

1000t

+ 0.024 sin 200t 0.010 cos 200t


ad2y

dx2+b

dy

dx+cy = f(x)where

f(x)is a sum or a product

Problem 9. Solved2y

dx2 +

dy

dx 6y = 12x 50 sinx.


(i)d2y

dx2+

dy

dx 6y = 12x 50 sinx in D-operator

form is

(D2 + D 6)y = 12x 50 sinx

(ii) The auxiliary equation is (m2 +m 6) = 0,from which,

(m 2)(m + 3) = 0,

i.e. m = 2 orm = 3


u=Ae2x+Be3x.

(iv) Since the right hand side of the given differen-tial equation is the sum of a polynomial and asine function let the P.I. v = ax + b+ c sinx +dcosx(see Table 51.1(e)).

(v) Substituting v into

(D2 + D 6)v = 12x 50 sinxgives:

(D2 + D 6)(ax + b + c sinx + dcosx)

= 12x 50 sinx

D(ax + b + c sinx + dcosx)

= a + c cosx dsinx

D2(ax + b + c sinx + dcosx)

= c sinx dcosx

Hence (D2 + D 6)(v)

= (c sinx dcosx) + (a + c cosx

dsinx) 6(ax + b + c sinx + dcosx)

= 12x 50 sinx

Equating constant terms gives:

a 6b = 0 (1)

Equating coefficients of x gives: 6a= 12,from which,a = 2.

Hence, from (1),b = 13

Equating the coefficients of cosxgives:

d+ c 6d = 0

i.e.

c 7d = 0

(2)

Equating the coefficients of sinxgives:

c d 6c = 50

i.e. 7c d = 50(3)

Solving equations (2) and (3) gives: c = 7 andd= 1.Hence the P.I.,

= 2x 13+ 7 sinx + cosx


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(vi) The general solution,y= u+ v,i.e. y = Ae2x +Be3x 2x

13+ 7 sinx + cosx

Problem 10. Solve the differential equationd2y

dx2 2 dy

dx+ 2y= 3ex cos2x, given that when

x= 0,y= 2 and dydx= 3.


(i)d2y

dx2 2 dy

dx+ 2y= 3ex cos2x in D-operator

form is

(D2 2D + 2)y = 3ex cos2x(ii) The auxiliary equation ism2 2m+ 2= 0

Using the quadratic formula,

m = 2

[4 4(1)(2)]2

= 2 42

= 2j22

i.e. m = 1j1.(iii) Since the roots are complex, the C.F.,

u = ex(Acos x +Bsinx).

(iv) Since the right hand side of the given dif-ferential equation is a product of an expo-nential and a cosine function, let the P.I.,v= ex(Csin 2x+D cos2x) (see Table 51.1(f) again, constants C and D are used since AandB have already been used for the C.F.).

(v) Substitutingv into (D2 2D+ 2)v= 3ex cos2xgives:

(D2 2D + 2)[ex(Csin 2x +D cos2x)]= 3ex cos2x

D(v

) = ex

(2Ccos2x 2D sin 2x)+ ex(Csin 2x +D cos2x)(ex{(2C+D)cos2x

+ (C 2D)si n2x})D2(v) = ex(4Csin 2x 4D cos2x)

+ ex(2Ccos2x 2D sin 2x)+ ex(2Ccos2x 2D sin 2x)+ ex(Csin 2x +D cos2x)

ex{( 3C 4D)si n2x+ (4C 3D)cos2x}

Hence (D2 2D+ 2)v= ex{(3C 4D)si n2x

+(4C

3D)cos2x

} 2ex{(2C+D)cos2x+ (C 2D)si n2x}+ 2ex(Csin 2x +D cos2x)

= 3ex cos2xEquating coefficients of ex sin 2xgives:

3C 4D 2C+ 4D+ 2C= 0i.e. 3C= 0, from which,C= 0.Equating coefficients of ex cos2xgives:

4C 3D 4C 2D+ 2D = 3i.e. 3D= 3, from which,D=1.

Hence the P.I., = ex(cos2x).

(vi) The general solution,y= u+ v, i.e.y = ex(A cosx+B sinx) ex cos 2x

(vii) Whenx= 0,y= 2 thus2= e0(A cos0 +B sin 0)

e0

cos0i.e. 2= A 1, from which, A = 3

dy

dx= ex(A sinx +B cosx)

+ ex(A cosx +B sinx) [ex(2si n2x) + ex cos2x]

When x= 0, dydx

= 3thus 3= e0(A sin 0 +B cos 0)

+ e0(A cos0 +B sin 0) e0(2sin0) e0 cos0

i.e. 3= B+A 1, from which,B= 1, since A = 3


y = ex(3 cosx + sinx) ex cos 2x


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Exercise 193 Further problems on secondorder differential equations of the form

ad2y

dx2+b

dy

dx

+cy=f(x) where f(x) is a sum

or product


1. 8d2y

dx2 6

dy

dx+y= 2x+ 40 sinxy = Ae x4 +Be x2 + 2x + 12

+8

17(6 cosx 7sinx)

2.d2y

d2 3

dy

d+ 2y = 2si n2 4cos2

y=Ae2+Be+ 12

(si n2 + cos2 )

3.d2y

dx2+

dy

dx 2y = x2 + e2x

y = Aex +Be2x 3

4

12x 1

2x2 + 1

4e2x

4.d2y

dt2 2

dy

dt+ 2y = etsin t

y= et(A cos t+B sin t) t2 etcos tIn Problems 5 to 6 find the particular solutionsof the given differential equations.

5.d2y

dx2 7

dy

dx+ 10y= e2x + 20; when x = 0,

y= 0 anddy

dx=

1

3y=

4

3e5x

10

3e2x

1

3xe2x + 2

6. 2d2y

dx2

dy

dx 6y= 6ex cosx; when x= 0,

y=21

29and d

y

dx=6

20

29y = 2e 32x 2e2x

+3ex

29(3 sinx 7cosx)


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