Higher Engineering Mathematics

486 DIFFERENTIAL EQUATIONS

3.

d^2 y

dx^2

+ 9 y=26e^2 x

[y=Acos 3x+Bsin 3x+2e^2 x]

4. 9

d^2 y

dt^2

− 6

dy

dt

+y=12e

t

3

[

y=(At+B)e

1

3 t+^23 t^2 e

1

3 t

]

In problems 5 and 6 find the particular solutions

of the given differential equations.

5. 5

d^2 y

dx^2

+ 9

dy

dx

− 2 y=3ex; whenx=0,y=

1

4

and

dy

dx

=0.

[

y=

5

44

(

e−^2 x−e

1

5 x

)

+

1

4

ex

]

6.

d^2 y

dt^2

− 6

dy

dt

+ 9 y=4e^3 t; whent=0,y= 2

and

dy

dt

=0[y=2e^3 t(1− 3 t+t^2 )]

51.5 Worked problems on differential

equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy=f(x)wheref(x)

is a sine or cosine function

Problem 7. Solve the differential equation

2

d^2 y

dx^2

+ 3

dy

dx

− 5 y=6 sin 2x.

Using the procedure of Section 51.2:

(i) 2

d^2 y

dx^2

+ 3

dy

dx

− 5 y=6 sin 2xin D-operator form

is (2D^2 +3D−5)y=6 sin 2x

(ii) The auxiliary equation is 2m^2 + 3 m− 5 =0,

from which,

(m−1)(2m+5)=0,

i.e. m=1orm=−^52

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

u=Aex+Be−

5

2 x.

(iv) Let the P.I., v=Asin 2x+Bcos 2x (see

Table 51.1(d)).

(v) Substitutingv=Asin 2x+Bcos 2xinto

(2D^2 +3D−5)v=6 sin 2xgives:

(2D^2 +3D−5)(Asin 2x+Bcos 2x)=6 sin 2x.

D(Asin 2x+Bcos 2x)

= 2 Acos 2x− 2 Bsin 2x

D^2 (Asin 2x+Bcos 2x)

=D(2Acos 2x− 2 Bsin 2x)

=− 4 Asin 2x− 4 Bcos 2x

Hence (2D^2 +3D−5)(Asin 2x+Bcos 2x)

=− 8 Asin 2x− 8 Bcos 2x+ 6 Acos 2x

− 6 Bsin 2x− 5 Asin 2x− 5 Bcos 2x

=6 sin 2x

Equating coefficient of sin 2xgives:

− 13 A− 6 B= 6 (1)

Equating coefficients of cos 2xgives:

6 A− 13 B=0(2)

6 ×(1)gives : − 78 A− 36 B= 36 (3)

13 ×(2)gives : 78 A− 169 B=0(4)

(3)+(4)gives : − 205 B= 36

from which, B=

− 36

205

SubstitutingB=

− 36

205

into equation (1) or (2)

givesA=

− 78

205

Hence the P.I.,v=

− 78

205

sin 2x−

36

205

cos 2x.

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

y=Aex+Be−

5

2 x

−

2

205

(39 sin 2x+18 cos 2x)

Problem 8. Solve

d^2 y

dx^2

+ 16 y=10 cos 4x

giveny=3 and

dy

dx

=4 whenx=0.