Higher Engineering Mathematics, Sixth Edition

Second order differential equations of the forma

d^2 y

dx^2 +b

dy

dx+cy=^0479

(d) Whent=0,y=3 hence 3=( 0 +B)e^0 ,i.e.B=3.

Sincey=(At+B)e

4

3 t

then

dy

dt

=(At+B)

(

4

3

e

4

3 t

)

+Ae

4

3 t,bythe

product rule.

When t= 0 ,

dy

dt

= 3

thus 3=( 0 +B)

4

3

e^0 +Ae^0

i.e. 3=

4

3

B+Afrom which,A=−1, since

B=3.

Hence the particular solution is

y=(−t+3)e

4

3 tory=(3−t)e

4

3 t

Problem 3. Solve the differential equation

d^2 y

dx^2

+ 6

dy

dx

+ 13 y=0, given that whenx=0,y= 3

and

dy

dx

=7.

Using the procedure of Section 50.2:

(a)

d^2 y

dx^2

+ 6

dy

dx

+ 13 y=0 in D-operator form is

(D^2 +6D+ 13 )y=0, where D≡

d

dx

(b) Substitutingmfor D gives the auxiliary equation

m^2 + 6 m+ 13 =0.

Using the quadratic formula:

m=

− 6 ±

√

[( 6 )^2 − 4 ( 1 )( 13 )]

2 ( 1 )

=

− 6 ±

√

(− 16 )

2

i.e.m=

− 6 ±j 4

2

=− 3 ±j 2

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

tion is

y=e−^3 x(Acos 2x+Bsin2x)

(d) Whenx=0,y=3, hence

3 =e^0 (Acos 0+Bsin0),i.e.A=3.

Sincey=e−^3 x(Acos 2x+Bsin2x)

then

dy

dx

=e−^3 x(− 2 Asin 2x+ 2 Bcos2x)

−3e−^3 x(Acos 2x+Bsin2x),

by the product rule,

=e−^3 x[( 2 B− 3 A)cos 2x

−( 2 A+ 3 B)sin2x]

Whenx=0,

dy

dx

=7,

hence 7=e^0 [( 2 B− 3 A)cos 0−( 2 A+ 3 B)sin0]

i.e. 7= 2 B− 3 A, from which,B=8, sinceA=3.

Hence the particular solution is

y=e−^3 x(3cos2x+8sin2x)

Since, from Chapter 17, page 165,

acosωt+bsinωt=Rsin(ωt+α),where

R=

√

(a^2 +b^2 )andα=tan−^1

a

b

then

3cos2x+8sin2x

=

√

( 32 + 82 )sin( 2 x+tan−^138 )

=

√

73sin( 2 x+ 20. 56 ◦)

=

√

73sin( 2 x+ 0. 359 )

Thus the particular solution may also be

expressed as

y=

√

73 e−3xsin( 2 x+0.359)

Now try the following exercise

Exercise 187 Further problemson

differential equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy= 0

In Problems 1 to 3, determine the general solution

of the given differential equations.

1. 6

d^2 y

dt^2

−

dy

dt

− 2 y= 0

[

y=Ae

2

3 t+Be−

1

2 t

]

2. 4

d^2 θ

dt^2

+ 4

dθ

dt

+θ= 0

[

θ=(At+B)e−

(^12) t]