Higher Engineering Mathematics

POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 493

I

3. (a)y(8)wheny=cos 2x

(b)y(9)wheny=3 cos

2

3

t

[

(a) 256 cos 2x (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

sinh 2x

(b)y(6)wheny=2 sinh 3x

[(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=2ln 3θ

(b)y(7)wheny=

1

3

ln 2t

[

(a)−

6

θ^4

(b)

240

t^7

]

52.3 Leibniz’s theorem

If y=uv (8)

whereuandvare each functions ofx, then by using
the 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 derivative ofvincreases by 1 moving
from left to right
(c) the coefficients 1, 4, 6, 4, 1 are the normal
binomial coefficients (see page 58)

In fact, (uv)(n) may be obtained by 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 theo-

rem, which can be used to differentiate a productn

times. 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.

Thus, wheny=x^2 e^3 x,v=x^2 , since its third deriva-

tive is zero, andu=e^3 xsince thenth derivative is

known from equation (1), i.e. 3neax

Using Leinbiz’s theorem (equation (13),

y(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)+ ···

where in this casev=x^2 ,v(1)= 2 x,v(2)=2 and

v(3)= 0

Hence, y(n)=(3ne^3 x)(x^2 )+n(3n−^1 e^3 x)(2x)

+

n(n−1)

2!

(3n−^2 e^3 x)(2)

+

n(n−1)(n−2)

3!

(3n−^3 e^3 x)(0)

= 3 n−^2 e^3 x(3^2 x^2 +n(3)(2x)

+n(n−1)+0)

i.e. y(n)=e^3 x 3 n−^2 (9x^2 + 6 nx+n(n−1))

Problem 2. Ifx^2 y′′+ 2 xy′+y=0 show that:

xy(n+2)+2(n+1)xy(n+1)+(n^2 +n+1)y(n)= 0