498 Higher Engineering Mathematics
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) 0n=1, (y′′′) 0 =− 3 (y′) 0n=2,(y(^4 )) 0 =− 4 (y′′) 0 =− 4 {− 2 (y) 0 }= 2 × 4 (y) 0n=3,(y(^5 )) 0 =− 5 (y′′′) 0 =− 5 {− 3 (y′) 0 }= 3 × 5 (y′) 0n=4,(y(^6 )) 0 =− 6 (y(^4 )) 0 =− 6 { 2 × 4 (y) 0 }=− 2 × 4 × 6 (y) 0n=5,(y(^7 )) 0 =− 7 (y(^5 )) 0 =− 7 { 3 × 5 (y′) 0 }=− 3 × 5 × 7 (y′) 0n=6,(y(^8 )) 0 =− 8 (y(^6 )) 0 =− 8 {− 2 × 4 × 6 (y) 0 }= 2 × 4 × 6 × 8 (y) 0(iv) Maclaurin’s theorem from page 69 may be
written as:y=(y) 0 +x(y′) 0 +x^2
2!(y′′) 0 +x^3
3!(y′′′) 0+x^4
4!(y(^4 )) 0 + ···Substituting the above values into Maclaurin’s
theorem gives:y=(y) 0 +x(y′) 0 +x^2
2!{− 2 (y) 0 }+x^3
3!{− 3 (y′) 0 }+x^4
4!{ 2 × 4 (y) 0 }+
x^5
5!{ 3 × 5 (y′) 0 }+
x^6
6!{− 2 × 4 × 6 (y) 0 }+x^7
7!{− 3 × 5 × 7 (y′) 0 }+x^8
8!{ 2 × 4 × 6 × 8 (y) 0 }(v) Collecting similar terms together gives:y=(y) 0{
1 −2 x^2
2!+2 × 4 x^4
4!−2 × 4 × 6 x^6
6!+2 × 4 × 6 × 8 x^8
8!− ···}
+(y′) 0{
x−3 x^3
3!+3 × 5 x^5
5!−3 × 5 × 7 x^7
7!+ ···}i.e.y=(y) 0{
1 −x^2
1+x^4
1 × 3−x^6
3 × 5+x^8
3 × 5 × 7− ···}+(y′) 0 ×{
x
1−
x^3
1 × 2+
x^5
2 × 4−x^7
2 × 4 × 6+···}The boundary conditions are that atx= 0 ,y= 1
anddy
dx=2, i.e.(y) 0 =1and(y′) 0 =2.Hence, the power series solution of the differen-tial equation:d^2 y
dx^2+xdy
dx+ 2 y=0is:y={
1 −x^2
1+x^4
1 × 3−x^6
3 × 5+x^8
3 × 5 × 7−···}
+ 2{
x
1−x^3
1 × 2+x^5
2 × 4−x^7
2 × 4 × 6+···}Problem 6. Determine the power series solution
of the differential equation:
d^2 y
dx^2+dy
dx+xy=0 given the boundary conditionsthat atx=0,y=0anddy
dx=1, using
Leibniz–Maclaurin’s method.Following the above procedure:(i) The differential equation is rewritten as:
y′′+y′+xy=0and from the Leibniz theorem of