Higher Engineering Mathematics, Sixth Edition

Power series methods of solving ordinary differential equations 495

Note that ify=lnx,y′=

1

x

; if in equation (7),

n=1theny′=(− 1 )^0

( 0 )!

x^1

(− 1 )^0 =1andify′=

1

x

then(0)!= 1 (Check that

(− 1 )^0 =1and( 0 )!=1 on a calculator).

Now try the following exercise

Exercise 193 Further problemson higher

order differential coefficients as series

Determine the following derivatives:

1. (a)y(^4 )wheny=e^2 x(b)y(^5 )wheny=8e

t

2

[(a) 16e^2 x(b)

1

4

e

t

(^2) ]

2. (a)y(^4 )wheny=sin3t
   (b)y(^7 )wheny=

1

50

sin5θ

[(a) 81sin3t (b)− 1562 .5cos5θ]

3. (a)y(^8 )wheny=cos2x
   (b)y(^9 )wheny=3cos

2

3

t

[

(a)256cos2x (b)−

29

38

sin

2

3

t

]

4. (a)y(^7 )wheny= 2 x^9 (b)y(^6 )wheny=
   t^7
   8
   [(a)(9!)x^2 (b) 630t]


5. (a)y(^7 )wheny=

1

4

sinh2x

(b)y(^6 )wheny=2sinh3x

[(a) 32 cosh 2x (b) 1458 sinh 3x]

6. (a)y(^7 )wheny=cosh 2x
   (b)y(^8 )wheny=

1

9

cosh 3x

[(a) 128 sinh 2x (b) 729 cosh 3x]

7. (a)y(^4 )wheny=2ln3θ
   (b)y(^7 )wheny=

1

3

ln2t

[

(a)−

6

θ^4

(b)

240

t^7

]

52.3 Leibniz’s theorem

If y=uv (8)

whereuandvare each functions ofx,thenbyusingthe

product rule,

y′=uv′+vu′ (9)

y′′=uv′′+v′u′+vu′′+u′v′

=u′′v+ 2 u′v′+uv′′ (10)

y′′′=u′′v′+vu′′′+ 2 u′v′′+ 2 v′u′′+uv′′′+v′′u′

=u′′′v+ 3 u′′v′+ 3 u′v′′+uv′′′ (11)

y(^4 )=u(^4 )v+ 4 u(^3 )v(^1 )+ 6 u(^2 )v(^2 )

+ 4 u(^1 )v(^3 )+uv(^4 ) (12)

From equations (8) to (12) it is seen that

(a) then’th derivative ofudecreases by 1 moving

from left to right,

(b) then’th derivativeofvincreases by 1 moving from

left to right,

(c) thecoefficients1,4,6,4,1 are thenormal binomial

coefficients (see page 58).

In fact,(uv)(n)may be obtainedby expanding(u+v)(n)

using the binomial theorem (see page 59), where the

‘powers’ are interpreted as derivatives. Thus, expanding

(u+v)(n)gives:

y(n)=(uv)(n)=u(n)v+nu(n−1)v(1)

+

n(n−1)

2!

u(n−2)v(2)

+

n(n−1)(n−2)

3!

u(n−3)v(3)+··· (13)

Equation (13) is a statement of Leibniz’s theorem,

which can be used to differentiate a productntimes.

The theorem is demonstrated in the following worked

problems.

Problem 1. Determiney(n)wheny=x^2 e^3 x.

For a producty=uv, the function taken as

(i) uis the one whose nth derivative can readily be

determined (from equations (1) to (7)),

(ii) vis the one whose derivative reduces to zero after

a few stages of differentiation.