Higher Engineering Mathematics

THE SOLUTION OF DIFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS 647

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Equating s^2 terms gives: 0=A+C, from

which,A=−1, sinceC=1. Hence,

L−^1

{

9

s^2 (s−3)

}

=L−^1

{

−

1

s

−

3

s^2

+

1

s− 3

}

=− 1 − 3 x+e^3 x, from (i),

(vi) and (iii) of Table 66.1.

i.e.y=e^3 x− 3 x− 1

Problem 4. Use Laplace transforms to solve

the differential equation:

d^2 y

dx^2

− 7

dy

dx

+ 10 y=e^2 x+20, given that when

x=0,y=0 and

dy

dx

=−

1

3

Using the procedure:

(i)L

{

d^2 y

dx^2

}

− 7 L

{

dy

dx

}

+ 10 L{y}=L{e^2 x+ 20 }

Hence [s^2 L{y}−sy(0)−y′(0)]−7[sL{y}

−y(0)]+ 10 L{y}=

1

s− 2

+

20

s

(ii)y(0)=0 andy′(0)=−

1

3

Hence s^2 L{y}− 0 −

(

−

1

3

)

− 7 sL{y}+ 0

+ 10 L{y}=

21 s− 40

s(s−2)

(iii) (s^2 − 7 s+10)L{y}=

21 s− 40

s(s−2)

−

1

3

=

3(21s−40)−s(s−2)

3 s(s−2)

=

−s^2 + 65 s− 120

3 s(s−2)

Hence L{y}=

−s^2 + 65 s− 120

3 s(s−2)(s^2 − 7 s+10)

=

1

3

[

−s^2 + 65 s− 120

s(s−2)(s−2)(s−5)

]

=

1

3

[

−s^2 + 65 s− 120

s(s−5)(s−2)^2

]

(iv)y=

1

3

L−^1

{

−s^2 + 65 s− 120

s(s−5)(s−2)^2

}

−s^2 + 65 s− 120

s(s−5)(s−2)^2

≡

A

s

+

B

s− 5

+

C

s− 2

+

D

(s−2)^2

≡

(

A(s−5)(s−2)^2 +B(s)(s−2)^2

+C(s)(s−5)(s−2)+D(s)(s−5)

)

s(s−5)(s−2)^2

Hence

−s^2 + 65 s− 120

≡A(s−5)(s−2)^2 +B(s)(s−2)^2

+C(s)(s−5)(s−2)+D(s)(s−5)

Whens=0,− 120 =− 20 A, from which,A=6.

Whens=5, 180= 45 B, from which,B=4.

Whens=2, 6=− 6 D, from which,D=−1.

Equatings^3 terms gives: 0=A+B+C, from

which,C=−10.

Hence

1

3

L−^1

{

−s^2 + 65 s− 120

s(s−5)(s−2)^2

}

=

1

3

L−^1

{

6

s

+

4

s− 5

−

10

s− 2

−

1

(s−2)^2

}

=

1

3

[6+4e^5 x−10 e^2 x−xe^2 x]

Thusy= 2 +

4

3

e^5 x−

10

3

e^2 x−

x

3

e^2 x

Problem 5. The current flowing in an electri-

cal circuit is given by the differential equation

Ri+L(di/dt)=E, whereE,LandRare con-

stants. Use Laplace transforms to solve the

equation for current igiven that whent=0,

i=0.

Using the procedure:

(i)L{Ri}+L

{

L

di

dt

}

=L{E}

i.e. RL{i}+L[sL{i}−i(0)]=

E

s