496 DIFFERENTIAL EQUATIONS
(iv) Maclaurin’s theorem from page 67 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= 1anddy
dx=2, i.e. (y) 0 =1 and (y′) 0 =2.Hence, the power series solution of the differ-ential equation:d^2 y
dx^2+xdy
dx+ 2 y=0 is: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 solu-
tion of the differential equation:
d^2 y
dx^2+dy
dx+xy=0 given the boundary con-ditions that atx=0,y=0 anddy
dx=1, using
Leibniz–Maclaurin’s method.Following the above procedure:(i) The differential equation is rewritten as:
y′′+y′+xy=0 and from the Leibniz theorem
of equation (13), each term is differentiated n
times, which gives:y(n+2)+y(n+1)+y(n)(x)+ny(n−1)(1)+ 0 = 0
i.e. y(n+2)+y(n+1)+xy(n)+ny(n−1)= 0
(15)
(ii) Atx=0, equation (15) becomes:y(n+2)+y(n+1)+ny(n−1)= 0from which, y(n+2)=−{y(n+1)+ny(n−1)}
This is therecurrence relationand applies for
n≥ 1
(iii) Substitutingn=1, 2, 3,...will produce a set of
relationships between the various coefficients.
For n=1, (y′′′) 0 =−{(y′′) 0 +(y) 0 }
n=2, (y(4)) 0 =−{(y′′′) 0 +2(y′) 0 }
n=3, (y(5)) 0 =−{(y(4)) 0 +3(y′′) 0 }
n=4, (y(6)) 0 =−{(y(5)) 0 +4(y′′′) 0 }
n=5, (y(7)) 0 =−{(y(6)) 0 +5(y(4)) 0 }
n=6, (y(8)) 0 =−{(y(7)) 0 +6(y(5)) 0 }
From the given boundary conditions, atx=0,y=0, thus (y) 0 =0, and atx=0,dy
dx=1, thus(y′) 0 = 1