A-56

Differential Equations

Differential equation

Method of solution

Euler or Cauchy equation

x

2

d

2

y

dx +

bx

dy
dx +

cy = S(x)

Putting x = e

t

, the equation becomes

d

2

y

dt

2

+ (b − 1)

dy

dt +

cy = S(e

t

)

and can then be solved as a linear second order
equation.

Bessel’s equation

x

2

d

2

y

dx

2

+ x

dy
dx +

(λ

2

x

2

− n

2

)y = 0

y = c

1

J

n

(λx) + c

2

Y

n

(λx)

Transformed Bessel’s equation

x

2

d

2

y

dx

2

+ (2p + 1)x

dy
dx +

�

α

2

x

2r

+ β

2

�

y = 0

y = x

−p

�

c

1

J

q/r

� α

r

x

r

�

+ c

2

Y

q/r

� α

r

x

r

��

where q =

�

p

2

− β

2

.

Legendre’s equation

(1 − x

2

)

d

2

y

dx

2

− 2x

dy
dx +

n(n + 1)y = 0

y = c

1

P

n

(x) + c

2

Q

n

(x)

Separation of variables

f

1

(x)g

1

(y) dx + f

2

(x)g

2

(y) dy = 0

� f

1

(x)

f

2

(x)

dx +

� g

2

(y)

g

1

(y)

dy = c

Exact equation

M(x, y) dx + N(x, y) dy = 0

where ∂ M/∂y = ∂ N/∂x

�

M∂x +

� �

n −

∂

∂

y

�

M∂x

�

dy = c where ∂x indicates

that the integration is to be performed with respect to x
keeping y constant.

Linear first order equation

dy
dx +

P(x)y = Q(x)

ye

�

P dx

=

�

Qe

�

P dx

dx + c

Bernoulli’s equation

dy
dx +

P(x)y = Q(x)y

n

ve

(1−n)

�

P dx =

�

Qe

(1−n)

�

P dx

dx + c

where v = y

1−n

.

If n = 1, then the solution is ln y =

�

(Q− P) dx + c.

Homogeneous equation

dy
dx =

F

� y

x

�

ln x =

�

dv

F (v)−v

+ c where v = y/x.

If F (v) = v, then the solution is y = cx.

Reducible to homogeneous

(a

1

x + b

1

y + c

1

) dx

+(a

2

x + b

2

y + c

2

) dy = 0

with

a

1

a

2

�=

b

1

b

2

Set u = a

1

x + b

1

y + c

1

and v = a

2

x + b

2

y + c

2

. Then

eliminate x and y and the equation becomes
homogenous.

Reducible to separable

(a

1

x + b

1

y + c

1

) dx

+(a

2

x + b

2

y + c

2

) dy = 0

with

a

1

a

2

=

b

1

b

2

Set u = a

1

x + b

1

y. Then eliminate x or y and the

equation becomes separable.

FOURIER SERIES

1. If f (x) is a bounded periodic function of period 2L (i.e., f (x + 2L) = f (x)), and satisfies the Dirichlet conditions:

(a) In any period f (x) is continuous, except possibly for a finite number of jump discontinuities.

(b) In any period f (x) has only a finite number of maxima and minima.

Then f (x) may be represented by the Fourier series

a

0

2 +

∞

�

n=1

�

a

n

cos

nπ x

L +

b

n

sin

nπ x

L

�

where a

n

and b

n

are as determined below. This series will converge to f (x) at every point where f (x) is continuous, and to

f (x

+

) + f (x

−

)

2

(i.e., the average of the left-hand and right-hand limits) at every point where f (x) has a jump discontinuity.

a

n

=

1
L

�

L

−L

f (x) cos

nπ x

L

dx,

n = 0, 1, 2, 3, . . . ,

b

n

=

1
L

�

L

−L

f (x) sin

nπ x

L

dx,

n = 1, 2, 3, . . .

We may also write

a

n

=

1
L

�

α

+2L

α

f (x) cos

nπ x

L

dx and b

n

=

1
L

�

α

+2L

α

f (x) sin

nπ x

L

dx

where α is any real number. Thus if α = 0,

a

n

=

1
L

�

2L

0

f (x) cos

nπ x

L

dx,

n = 0, 1, 2, 3, . . . ,

b

n

=

1
L

�

2L

0

f (x) sin

nπ x

L

d,

n = 1, 2, 3, . . .

2. If in addition to the restrictions in (1), f (x) is an even function (i.e., f (−x) = f (x)), then the Fourier series reduces to

a

0

2 +

∞

�

n=1

a

n

cos

nπ x

L

That is, b

n

= 0. In this case, a simpler formula for a

n

is

a

n

=

2
L

�

L

0

f (x) cos

nπ x

L

dx,

n = 0, 1, 2, 3, . . .

3. If in addition to the restrictions in (1), f (x) is an odd function (i.e., f (−x) = − f (x)), then the Fourier series reduces to

∞

�

n=1

b

n

sin

nπ x

L

That is, a

n

= 0. In this case, a simpler formula for the b

n

is

b

n

=

2
L

�

L

0

f (x) sin

nπ x

L

dx,

n = 1, 2, 3, . . .

4. If in addition to the restrictions in (2) above, f (x) = − f (L − x), then a

n

will be 0 for all even values of n, including n = 0.

Thus in this case, the expansion reduces to

∞

�

m=1

a

2m−1

cos

(2m − 1)πx

L

A-57

A-58

Fourier Series

5. If in addition to the restrictions in (3) above, f (x) = f (L − x), then b

n

will be 0 for all even values of n. Thus in this case,

the expansion reduces to

∞

�

m=1

b

2m−1

sin

(2m − 1)πx

L

(The series in (4) and (5) are known as odd-harmonic series, since only the odd harmonics appear. Similar rules may be stated
for even-harmonic series, but when a series appears in the even-harmonic form, it means that 2L has not been taken as the
smallest period of f (x). Since any integral multiple of a period is also a period, series obtained in this way will also work,
but in general computation is simplified if 2L is taken to be the smallest period.)

6. If we write the Euler definitions for cos θ and sin θ, we obtain the complex form of the Fourier series known either as the

“Complex Fourier Series” or the “Exponential Fourier Series” of f (x). It is represented as

f (x) =

1
2

n=+∞

�

n=−∞

c

n

e

iω

n

x

where

c

n

=

1
L

�

L

−L

f (x) e

−iω

n

x

dx,

n = 0, ±1, ±2, ±3, . . .

with ω

n

=

nπ

L

for n = 0, ±1, ±2, . . . The set of coefficients c

n

is often referred to as the Fourier spectrum.

7. If both sine and cosine terms are present and if f (x) is of period 2L and expandable by a Fourier series, it can be represented

as

f (x) =

a

0

2 +

∞

�

n=1

c

n

sin

� nπx

L +

φ

n

�

,

where

a

n

= c

n

sin φ

n

,

b

n

= c

n

cos φ

n

,

c

n

=

�

a

2

n

+ b

2

n

,

φ

n

= arctan

� a

n

b

n

�

It can also be represented as

f (x) =

a

0

2 +

∞

�

n=1

c

n

cos

� nπx

L +

φ

n

�

,

where

a

n

= c

n

cos φ

n

,

b

n

= −c

n

sin φ

n

,

c

n

=

�

a

2

n

+ b

2

n

,

φ

n

= arctan

�

−

b

n

a

n

�

where φ

n

is chosen so as to make a

n

, b

n

, and c

n

hold.

8. The following table of trigonometric identities should be helpful for developing Fourier series.

n

n even

nodd

n/2 odd

n/2 even

sin nπ

0

0

0

0

0

cos nπ

(−1)

n

+1

−1

+1

+1

∗ sin

nπ

2

0

(−1)

(n−1)/2

0

0

∗ cos

nπ

2

(−1)

n/2

0

−1

+1

sin

nπ

4

√

2

2

(−1)

(n

2

+4n+11)/8

(−1)

(n−2)/4

0

*A useful formula for sin

nπ

2

and cos

nπ

2

is given by

sin

nπ

2 =

(i)

n+1

2

[(−1)

n

− 1] and cos

nπ

2 =

(i)

n

2

[(−1)

n

+ 1],

where i

2

= −1.

Auxiliary Formulas for Fourier Series

1 =

4

π

�

sin

π

x

k +

1
3

sin

3π x

k +

1
5

sin

5π x

k + · · ·

�

[0 < x < k]

x =

2k

π

�

sin

π

x

k −

1
2

sin

2π x

k +

1
3

sin

3π x

k − · · ·

�

[−k < x < k]

x =

k
2 −

4k

π

2

�

cos

π

x

k +

1

3

2

cos

3π x

k +

1

5

2

cos

5π x

k + · · ·

�

[0 < x < k]

x

2

=

2k

2

π

3

��

π

2

1 −

4
1

�

sin

π

x

k −

π

2

2

sin

2π x

k +

�

π

2

3 −

4

3

3

�

sin

3π x

k

−

π

2

4

sin

4π x

k +

�

π

2

5 −

4

5

3

�

sin

5π x

k + · · ·

�

[0 < x < k]

x

2

=

k

2

3 −

4k

2

π

2

�

cos

π

x

k −

1

2

2

cos

2π x

k +

1

3

2

cos

3π x

k −

1

4

2

cos

4π x

k + · · ·

�

[−k < x < k]

1 −

1
3 +

1
5 −

1
7 + · · · =

π

4

1 −

1

2

2

+

1

3

2

+

1

4

2

+ · · · =

π

2

6

1 −

1

2

2

+

1

3

2

−

1

4

2

+ · · · =

π

2

12

1 +

1

3

2

+

1

5

2

−

1

7

2

+ · · · =

π

2

8

1

2

2

+

1

4

2

+

1

6

2

+

1

8

2

+ · · · =

π

2

24

FOURIER EXPANSIONS FOR BASIC PERIODIC FUNCTIONS

f (x) =

4

π

�

n=1,3,5...

1

n

sin

nπ x

L

f (x) =

2

π

∞

�

n=1

(−1)

n

n

�

cos

nπc

L

− 1

�

sin

nπ x

L

f (x)=

c

L

+

2

π

∞

�

n=1

(−1)n

n

sin

nπc

L

cos

nπ x

L

f (x) =

2

L

∞

�

n=1

sin

nπ

2

sin(

1

2

nπc/L)

1

2

nπc/L

sin

nπ x

L

A-59

A-58

Fourier Series

5. If in addition to the restrictions in (3) above, f (x) = f (L − x), then b

n

will be 0 for all even values of n. Thus in this case,

the expansion reduces to

∞

�

m=1

b

2m−1

sin

(2m − 1)πx

L

(The series in (4) and (5) are known as odd-harmonic series, since only the odd harmonics appear. Similar rules may be stated
for even-harmonic series, but when a series appears in the even-harmonic form, it means that 2L has not been taken as the
smallest period of f (x). Since any integral multiple of a period is also a period, series obtained in this way will also work,
but in general computation is simplified if 2L is taken to be the smallest period.)

6. If we write the Euler definitions for cos θ and sin θ, we obtain the complex form of the Fourier series known either as the

“Complex Fourier Series” or the “Exponential Fourier Series” of f (x). It is represented as

f (x) =

1
2

n=+∞

�

n=−∞

c

n

e

iω

n

x

where

c

n

=

1
L

�

L

−L

f (x) e

−iω

n

x

dx,

n = 0, ±1, ±2, ±3, . . .

with ω

n

=

nπ

L

for n = 0, ±1, ±2, . . . The set of coefficients c

n

is often referred to as the Fourier spectrum.

7. If both sine and cosine terms are present and if f (x) is of period 2L and expandable by a Fourier series, it can be represented

as

f (x) =

a

0

2 +

∞

�

n=1

c

n

sin

� nπx

L +

φ

n

�

,

where

a

n

= c

n

sin φ

n

,

b

n

= c

n

cos φ

n

,

c

n

=

�

a

2

n

+ b

2

n

,

φ

n

= arctan

� a

n

b

n

�

It can also be represented as

f (x) =

a

0

2 +

∞

�

n=1

c

n

cos

� nπx

L +

φ

n

�

,

where

a

n

= c

n

cos φ

n

,

b

n

= −c

n

sin φ

n

,

c

n

=

�

a

2

n

+ b

2

n

,

φ

n

= arctan

�

−

b

n

a

n

�

where φ

n

is chosen so as to make a

n

, b

n

, and c

n

hold.

8. The following table of trigonometric identities should be helpful for developing Fourier series.

n

n even

nodd

n/2 odd

n/2 even

sin nπ

0

0

0

0

0

cos nπ

(−1)

n

+1

−1

+1

+1

∗ sin

nπ

2

0

(−1)

(n−1)/2

0

0

∗ cos

nπ

2

(−1)

n/2

0

−1

+1

sin

nπ

4

√

2

2

(−1)

(n

2

+4n+11)/8

(−1)

(n−2)/4

0

*A useful formula for sin

nπ

2

and cos

nπ

2

is given by

sin

nπ

2 =

(i)

n+1

2

[(−1)

n

− 1] and cos

nπ

2 =

(i)

n

2

[(−1)

n

+ 1],

where i

2

= −1.

Auxiliary Formulas for Fourier Series

1 =

4

π

�

sin

π

x

k +

1
3

sin

3π x

k +

1
5

sin

5π x

k + · · ·

�

[0 < x < k]

x =

2k

π

�

sin

π

x

k −

1
2

sin

2π x

k +

1
3

sin

3π x

k − · · ·

�

[−k < x < k]

x =

k
2 −

4k

π

2

�

cos

π

x

k +

1

3

2

cos

3π x

k +

1

5

2

cos

5π x

k + · · ·

�

[0 < x < k]

x

2

=

2k

2

π

3

��

π

2

1 −

4
1

�

sin

π

x

k −

π

2

2

sin

2π x

k +

�

π

2

3 −

4

3

3

�

sin

3π x

k

−

π

2

4

sin

4π x

k +

�

π

2

5 −

4

5

3

�

sin

5π x

k + · · ·

�

[0 < x < k]

x

2

=

k

2

3 −

4k

2

π

2

�

cos

π

x

k −

1

2

2

cos

2π x

k +

1

3

2

cos

3π x

k −

1

4

2

cos

4π x

k + · · ·

�

[−k < x < k]

1 −

1
3 +

1
5 −

1
7 + · · · =

π

4

1 −

1

2

2

+

1

3

2

+

1

4

2

+ · · · =

π

2

6

1 −

1

2

2

+

1

3

2

−

1

4

2

+ · · · =

π

2

12

1 +

1

3

2

+

1

5

2

−

1

7

2

+ · · · =

π

2

8

1

2

2

+

1

4

2

+

1

6

2

+

1

8

2

+ · · · =

π

2

24

FOURIER EXPANSIONS FOR BASIC PERIODIC FUNCTIONS

f (x) =

4

π

�

n=1,3,5...

1

n

sin

nπ x

L

f (x) =

2

π

∞

�

n=1

(−1)

n

n

�

cos

nπc

L

− 1

�

sin

nπ x

L

f (x)=

c

L

+

2

π

∞

�

n=1

(−1)n

n

sin

nπc

L

cos

nπ x

L

f (x) =

2

L

∞

�

n=1

sin

nπ

2

sin(

1

2

nπc/L)

1

2

nπc/L

sin

nπ x

L

A-59