Higher Engineering Mathematics, Sixth Edition

496 Higher Engineering Mathematics

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

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 )=2and

v(^3 )= 0

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

+

n(n− 1 )

2!

( 3 n−^2 e^3 x)( 2 )

+

n(n− 1 )(n− 2 )

3!

( 3 n−^3 e^3 x)( 0 )

= 3 n−^2 e^3 x( 32 x^2 +n( 3 )( 2 x)

+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

Differentiating each term of x^2 y′′+ 2 xy′+y= 0

ntimes, using Leibniz’s theorem of equation (13),

gives:

{

y(n+^2 )x^2 +ny(n+^1 )( 2 x)+

n(n− 1 )

2!

y(n)( 2 )+ 0

}

+{y(n+^1 )( 2 x)+ny(n)( 2 )+ 0 }+{y(n)}= 0

i.e. x^2 y(n+^2 )+ 2 nxy(n+^1 )+n(n− 1 )y(n)

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

i.e. x^2 y(n+^2 )+ 2 (n+ 1 )xy(n+^1 )

+(n^2 −n+ 2 n+ 1 )y(n)= 0

or x^2 y(n+2)+ 2 (n+ 1 )xy(n+1)

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

Problem 3. Differentiate the following

differential equationntimes:

( 1 +x^2 )y′′+ 2 xy′− 3 y=0.

By Leibniz’s equation, equation (13),

{

y(n+^2 )( 1 +x^2 )+ny(n+^1 )( 2 x)+

n(n− 1 )

2!

y(n)( 2 )+ 0

}

+ 2 {y(n+^1 )(x)+ny(n)( 1 )+ 0 }− 3 {y(n)}= 0

i.e.( 1 +x^2 )y(n+^2 )+ 2 nxy(n+^1 )+n(n− 1 )y(n)

+ 2 xy(n+^1 )+ 2 ny(n)− 3 y(n)= 0

or ( 1 +x^2 )y(n+^2 )+ 2 (n+ 1 )xy(n+^1 )

+(n^2 −n+ 2 n− 3 )y(n)= 0

i.e.( 1 +x^2 )y(n+2)+2(n+1)xy(n+1)

+(n^2 +n−3)y(n)= 0

Problem 4. Find the 5th derivative ofy=x^4 sinx.

If y=x^4 sinx, then using Leibniz’s equation with

u=sinxandv=x^4 gives:

y(n)=

[

sin

(

x+

nπ

2

)

x^4

]

+n

[

sin

(

x+

(n− 1 )π

2

)

4 x^3

]

+

n(n− 1 )

2!

[

sin

(

x+

(n− 2 )π

2

)

12 x^2

]

+

n(n− 1 )(n− 2 )

3!

[

sin

(

x+

(n− 3 )π

2

)

24 x

]

+

n(n− 1 )(n− 2 )(n− 3 )

4!

[

sin

(

x

+

(n− 4 )π

2

)

24

]

andy(^5 )=x^4 sin

(

x+

5 π

2

)

+ 20 x^3 sin(x+ 2 π)

+

( 5 )( 4 )

2

( 12 x^2 )sin

(

x+

3 π

2

)

+

( 5 )( 4 )( 3 )

( 3 )( 2 )

( 24 x)sin(x+π)

+

( 5 )( 4 )( 3 )( 2 )

( 4 )( 3 )( 2 )

( 24 )sin

(

x+

π

2

)

Since sin

(

x+

5 π

2

)

≡sin

(

x+

π

2

)

≡cosx,