Higher Engineering Mathematics, Sixth Edition

488 Higher Engineering Mathematics

from which, 2ke^2 x=3e^2 xandk=^32

Hence the P.I.,v=kx^2 e^2 x=^32 x^2 e^2 x.

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

y=(Ax+B)e^2 x+ 23 x^2 e^2 x

Now try the following exercise

Exercise 190 Further problems on

differential equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy=f(x)wheref(x)is an

exponential function

In Problems 1 to 4, find the general solutionsof the

given differential equations.

1.

d^2 y

dx^2

−

dy

dx

− 6 y=2ex

[

y=Ae^3 x+Be−^2 x−^13 ex

]

2.

d^2 y

dx^2

− 3

dy

dx

− 4 y=3e−x

[

y=Ae^4 x+Be−x−^35 xe−x

]

3.

d^2 y

dx^2

+ 9 y=26e^2 x

[y=Acos 3x+Bsin3x+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 problemson

differential equations of the

forma

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=6sin2x.

Using the procedure of Section 51.2:

(i) 2

d^2 y

dx^2

+ 3

dy

dx

− 5 y=6sin2xin D-operator form

is(2D^2 +3D− 5 )y=6sin2x

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

which,

(m− 1 )( 2 m+ 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+Bcos2x (see

Table 51.1(d)).

(v) Substitutingv=Asin2x+Bcos2xinto

(2D^2 +3D− 5 )v=6sin2xgives:

(2D^2 +3D− 5 )(Asin 2x+Bcos2x)=6sin2x.

D(Asin 2x+Bcos2x)

= 2 Acos 2x− 2 Bsin2x

D^2 (Asin 2x+Bcos2x)

=D( 2 Acos 2x− 2 Bsin2x)

=− 4 Asin 2x− 4 Bcos2x

Hence (2D^2 +3D− 5 )(Asin 2x+Bcos2x)

=− 8 Asin 2x− 8 Bcos2x+ 6 Acos 2x

− 6 Bsin2x− 5 Asin2x− 5 Bcos2x

=6sin2x

Equating coefficient of sin2xgives:

− 13 A− 6 B=6(1)