Power series methods of solving ordinary differential equations 497
sin(x + 2π) ≡ sin x, sin
x +
3π
2
≡ −cos x,
and sin (x + π) ≡ −sin x,
then y (5) = x 4 cos x + 20x 3 sin x + 120x 2 (−cos x)
+ 240x(−sin x) + 120 cos x
i.e. y (5) = (x 4 − 120x 2 + 120)cos x
+ (20x 3 − 240x)sin x
Now try the following exercise
Exercise 194 Further problems on
Leibniz’s theorem
Use the theorem of Leibniz in the following
problems:
1. Obtain the n’th derivative of: x 2 y.
x 2 y (n) + 2n x y (n−1) + n(n − 1)y (n−2)
2. If y = x 3 e 2x find y (n) and hence y (3) .
⎡
⎢
⎣
y (n) = e 2x 2 n−3 {8x 3 + 12nx 2
+ n(n − 1)(6x) + n(n − 1)(n − 2)}
y (3) = e 2x (8x 3 + 36x 2 + 36x + 6)
⎤
⎥
⎦
3. Determine the 4th derivative of: y = 2x 3 e −x .
[ y (4) = 2e −x (x 3 − 12x 2 + 36x − 24)]
4. If y = x 3 cos x determine the 5th derivative.
[ y (5) = (60x − x 3 ) sin x +
(15x 2 − 60) cos x]
5. Find an expression for y (4) if y = e −t sin t .
[ y (4) = −4 e −t sin t ]
6. If y = x 5 ln 2x find y (3) .
[ y (3) = x 2 (47 + 60 ln 2x)]
7. Given 2x 2 y + x y + 3y = 0 show that
2x 2 y (n+2) + (4n + 1)x y (n+1) + (2n 2 − n +
3)y (n) = 0.
8. If y = (x 3 + 2x 2 )e 2x determine an expansion
for y (5) .
[ y (5) = e 2x 2 4 (2x 3 + 19x 2 + 50x + 35)]
52.4 Power series solution by the
Leibniz–Maclaurin method
For second order differential equations that cannot be
solved by algebraic methods, the Leibniz–Maclaurin
method produces a solution in the form of infinite
series of powers of the unknown variable. The following simple 5-step procedure may 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 at x = 0,
(iii) determine the values of the derivatives at x = 0,
i.e. find ( y) 0 and ( y ) 0 ,
(iv) substitute in the Maclaurin expansion for
y = f (x) (see page 69, 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 solution
of the differential equation:
d 2 y
dx 2 + x
d y
dx
+ 2y = 0 using Leibniz–Maclaurin’s
method, given the boundary conditions that at
x = 0, y = 1 and
d y
dx
= 2.
Following the above procedure:
(i) The differential equation is rewritten as:
y + x y + 2y = 0 and from the Leibniz theorem
of equation (13), each term is differentiated n
times, which gives:
y
(n+2)
+{y
(n+1)
(x)+n y
(n)
(1)+0}+2 y
(n)
= 0
i.e.
y
(n+2)
+ x y
(n+1)
+ (n + 2) y
(n)
= 0
(14)
(ii) At x = 0, equation (14) becomes:
y
(n+2)
+ (n + 2) y
(n)
= 0
from which, y (n+2) =−(n +2) y (n)
sin(x + 2π) ≡ sin x, sin
x +
3π
2
≡ −cos x,
and sin (x + π) ≡ −sin x,
then y (5) = x 4 cos x + 20x 3 sin x + 120x 2 (−cos x)
+ 240x(−sin x) + 120 cos x
i.e. y (5) = (x 4 − 120x 2 + 120)cos x
+ (20x 3 − 240x)sin x
Now try the following exercise
Exercise 194 Further problems on
Leibniz’s theorem
Use the theorem of Leibniz in the following
problems:
1. Obtain the n’th derivative of: x 2 y.
x 2 y (n) + 2n x y (n−1) + n(n − 1)y (n−2)
2. If y = x 3 e 2x find y (n) and hence y (3) .
⎡
⎢
⎣
y (n) = e 2x 2 n−3 {8x 3 + 12nx 2
+ n(n − 1)(6x) + n(n − 1)(n − 2)}
y (3) = e 2x (8x 3 + 36x 2 + 36x + 6)
⎤
⎥
⎦
3. Determine the 4th derivative of: y = 2x 3 e −x .
[ y (4) = 2e −x (x 3 − 12x 2 + 36x − 24)]
4. If y = x 3 cos x determine the 5th derivative.
[ y (5) = (60x − x 3 ) sin x +
(15x 2 − 60) cos x]
5. Find an expression for y (4) if y = e −t sin t .
[ y (4) = −4 e −t sin t ]
6. If y = x 5 ln 2x find y (3) .
[ y (3) = x 2 (47 + 60 ln 2x)]
7. Given 2x 2 y + x y + 3y = 0 show that
2x 2 y (n+2) + (4n + 1)x y (n+1) + (2n 2 − n +
3)y (n) = 0.
8. If y = (x 3 + 2x 2 )e 2x determine an expansion
for y (5) .
[ y (5) = e 2x 2 4 (2x 3 + 19x 2 + 50x + 35)]
52.4 Power series solution by the
Leibniz–Maclaurin method
For second order differential equations that cannot be
solved by algebraic methods, the Leibniz–Maclaurin
method produces a solution in the form of infinite
series of powers of the unknown variable. The following simple 5-step procedure may 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 at x = 0,
(iii) determine the values of the derivatives at x = 0,
i.e. find ( y) 0 and ( y ) 0 ,
(iv) substitute in the Maclaurin expansion for
y = f (x) (see page 69, 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 solution
of the differential equation:
d 2 y
dx 2 + x
d y
dx
+ 2y = 0 using Leibniz–Maclaurin’s
method, given the boundary conditions that at
x = 0, y = 1 and
d y
dx
= 2.
Following the above procedure:
(i) The differential equation is rewritten as:
y + x y + 2y = 0 and from the Leibniz theorem
of equation (13), each term is differentiated n
times, which gives:
y
(n+2)
+{y
(n+1)
(x)+n y
(n)
(1)+0}+2 y
(n)
= 0
i.e.
y
(n+2)
+ x y
(n+1)
+ (n + 2) y
(n)
= 0
(14)
(ii) At x = 0, equation (14) becomes:
y
(n+2)
+ (n + 2) y
(n)
= 0
from which, y (n+2) =−(n +2) y (n)