Power series methods of solving ordinary differential equations 495
Note that if y = ln x, y =
1
x
; if in equation (7),
n = 1 then y
= (−1)
0 (0)!
x 1
(−1) 0 = 1 and if y =
1
x
then (0)!= 1 (Check that
(−1) 0 = 1 and (0)! = 1 on a calculator).
Now try the following exercise
Exercise 193 Further problems on higher
order differential coefficients as series
Determine the following derivatives:
1. (a) y (4) when y = e 2x (b) y (5) when y = 8 e
t
2
[(a) 16 e 2x (b)
1
4
e
t
2 ]
2. (a) y (4) when y = sin 3t
(b) y (7) when y =
1
50
sin 5θ
[(a) 81 sin3t (b) −1562.5 cos5θ]
3. (a) y (8) when y = cos 2x
(b) y (9) when y = 3 cos
2
3
t
(a) 256 cos2x (b) −
2 9
3 8 sin
2
3
t
4. (a) y (7) when y = 2x 9 (b) y (6) when y =
t 7
8
[(a) (9! )x 2 (b) 630 t ]
5. (a) y (7) when y =
1
4
sinh 2x
(b) y (6) when y = 2 sinh3x
[(a) 32 cosh 2x (b) 1458 sinh 3x]
6. (a) y (7) when y = cosh 2x
(b) y (8) when y =
1
9
cosh 3x
[(a) 128 sinh 2x (b) 729 cosh 3x]
7. (a) y (4) when y = 2ln 3θ
(b) y (7) when y =
1
3
ln 2t
(a) −
6
θ 4 (b)
240
t 7
52.3 Leibniz’s theorem
If
y = uv
(8)
where u and v are each functions of x, then by using the
product rule,
y
= uv
+ vu
(9)
y
= uv
+ v
u
+ vu
+ u
v
= u
v + 2u
v
+ uv
(10)
y
= u
v
+ vu
+ 2u
v
+ 2v
u
+ uv
+ v
u
= u
v + 3u
v
+ 3u
v
+ uv
(11)
y
(4)
= u
(4)
v + 4u
(3)
v
(1)
+ 6u
(2)
v
(2)
+ 4u
(1)
v
(3)
+ uv
(4)
(12)
From equations (8) to (12) it is seen that
(a) the n’th derivative of u decreases by 1 moving
from left to right,
(b) the n’th derivative of v increases by 1 moving from
left to right,
(c) the coefficients 1, 4, 6, 4, 1 are the normal binomial
coefficients (see page 58).
In fact, (uv) (n) may be obtained by expanding (u + v) (n)
using the binomial theorem (see page 59), where the
‘powers’ are interpreted as derivatives. Thus, expanding
(u + v) (n) gives:
y
(n)
= (uv)
(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)
+· · ·
(13)
Equation (13) is a statement of Leibniz’s theorem,
which can be used to differentiate a product n times.
The theorem is demonstrated in the following worked
problems.
Problem 1. Determine y (n) when y = x 2 e 3x .
For a product y = uv, the function taken as
(i) u is the one whose nth derivative can readily be
determined (from equations (1) to (7)),
(ii) v is the one whose derivative reduces to zero after
a few stages of differentiation.
Note that if y = ln x, y =
1
x
; if in equation (7),
n = 1 then y
= (−1)
0 (0)!
x 1
(−1) 0 = 1 and if y =
1
x
then (0)!= 1 (Check that
(−1) 0 = 1 and (0)! = 1 on a calculator).
Now try the following exercise
Exercise 193 Further problems on higher
order differential coefficients as series
Determine the following derivatives:
1. (a) y (4) when y = e 2x (b) y (5) when y = 8 e
t
2
[(a) 16 e 2x (b)
1
4
e
t
2 ]
2. (a) y (4) when y = sin 3t
(b) y (7) when y =
1
50
sin 5θ
[(a) 81 sin3t (b) −1562.5 cos5θ]
3. (a) y (8) when y = cos 2x
(b) y (9) when y = 3 cos
2
3
t
(a) 256 cos2x (b) −
2 9
3 8 sin
2
3
t
4. (a) y (7) when y = 2x 9 (b) y (6) when y =
t 7
8
[(a) (9! )x 2 (b) 630 t ]
5. (a) y (7) when y =
1
4
sinh 2x
(b) y (6) when y = 2 sinh3x
[(a) 32 cosh 2x (b) 1458 sinh 3x]
6. (a) y (7) when y = cosh 2x
(b) y (8) when y =
1
9
cosh 3x
[(a) 128 sinh 2x (b) 729 cosh 3x]
7. (a) y (4) when y = 2ln 3θ
(b) y (7) when y =
1
3
ln 2t
(a) −
6
θ 4 (b)
240
t 7
52.3 Leibniz’s theorem
If
y = uv
(8)
where u and v are each functions of x, then by using the
product rule,
y
= uv
+ vu
(9)
y
= uv
+ v
u
+ vu
+ u
v
= u
v + 2u
v
+ uv
(10)
y
= u
v
+ vu
+ 2u
v
+ 2v
u
+ uv
+ v
u
= u
v + 3u
v
+ 3u
v
+ uv
(11)
y
(4)
= u
(4)
v + 4u
(3)
v
(1)
+ 6u
(2)
v
(2)
+ 4u
(1)
v
(3)
+ uv
(4)
(12)
From equations (8) to (12) it is seen that
(a) the n’th derivative of u decreases by 1 moving
from left to right,
(b) the n’th derivative of v increases by 1 moving from
left to right,
(c) the coefficients 1, 4, 6, 4, 1 are the normal binomial
coefficients (see page 58).
In fact, (uv) (n) may be obtained by expanding (u + v) (n)
using the binomial theorem (see page 59), where the
‘powers’ are interpreted as derivatives. Thus, expanding
(u + v) (n) gives:
y
(n)
= (uv)
(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)
+· · ·
(13)
Equation (13) is a statement of Leibniz’s theorem,
which can be used to differentiate a product n times.
The theorem is demonstrated in the following worked
problems.
Problem 1. Determine y (n) when y = x 2 e 3x .
For a product y = uv, the function taken as
(i) u is the one whose nth derivative can readily be
determined (from equations (1) to (7)),
(ii) v is the one whose derivative reduces to zero after
a few stages of differentiation.