Higher Engineering Mathematics

POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 495

I

3. Determine the 4th derivative of:y= 2 x^3 e−x

[y(4)=2e−x(x^3 − 12 x^2 + 36 x−24)]

4. Ify=x^3 cosxdetermine the 5th derivative.

[y(5)=(60x−x^3 ) sinx+

(15x^2 −60) cosx]

5. Find an expression fory(4)ify=e−tsint.

[y(4)=−4e−tsint]

6. Ify=x^5 ln 2xfindy(3).

[y(3)=x^2 (47+60 ln 2x)]

7. Given 2x^2 y′′+xy′+ 3 y=0 show that
   2 x^2 y(n+2)+(4n+1)xy(n+1)+(2n^2 −n+
   3)y(n)= 0


8. Ify=(x^3 + 2 x^2 )e^2 xdetermine an expansion
   fory(5).

[y(5)=e^2 x 24 (2x^3 + 19 x^2 + 50 x+35)]

52.4 Power series solution by the

Leibniz–Maclaurin method

For second order differential equations that can-
not be solved by algebraic methods, theLeibniz–
Maclaurin methodproduces a solution in the form
of infinite series of powers of the unknown variable.
The following simple5-step proceduremay be used
in the Leibniz–Maclaurin method:

(i) Differentiate the given equation n times, using

the Leibniz theorem of equation (13),

(ii) rearrange the result to obtain the recurrence

relation atx=0,

(iii) determine the values of the derivatives atx=0,

i.e. find (y) 0 and (y′) 0 ,

(iv) substitute in the Maclaurin expansion for

y=f(x) (see page 67, equation (5)),

(v) simplify the result where possible and apply

boundary condition (if given).

The Leibniz–Maclaurin method is demonstrated,
using the above procedure, in the following worked
problems.

Problem 5. Determine the power series solu-

tion of the differential equation:

d^2 y

dx^2

+x

dy

dx

+ 2 y=0 using Leibniz–Maclaurin’s

method, given the boundary conditions that at

x=0,y=1 and

dy

dx

=2.

Following the above procedure:

(i) The differential equation is rewritten as:

y′′+xy′+ 2 y=0 and from the Leibniz theorem

of equation (13), each term is differentiatedn

times, which gives:

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

i.e. y(n+2)+xy(n+1)+(n+2)y(n)= 0

(14)

(ii) Atx=0, equation (14) becomes:

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

from which, y(n+2)=−(n+2)y(n)

This equation is called arecurrence relation

orrecurrence formula, because each recurring

term depends on a previous term.

(iii) Substituting n=0, 1, 2, 3, ... will produce

a set of relationships between the various

coefficients.

Forn=0, (y′′) 0 =−2(y) 0

n=1, (y′′′) 0 =−3(y′) 0

n=2, (y(4)) 0 =−4(y′′) 0 =− 4 {−2(y) 0 }

= 2 ×4(y) 0

n=3, (y(5)) 0 =−5(y′′′) 0 =− 5 {−3(y′) 0 }

= 3 ×5(y′) 0

n=4, (y(6)) 0 =−6(y(4)) 0 =− 6 { 2 ×4(y) 0 }

=− 2 × 4 ×6(y) 0

n=5, (y(7)) 0 =−7(y(5)) 0 =− 7 { 3 ×5(y′) 0 }

=− 3 × 5 ×7(y′) 0

n=6, (y(8)) 0 =−8(y(6)) 0 =

− 8 {− 2 × 4 ×6(y) 0 }= 2 × 4 × 6 ×8(y) 0