Page images
PDF
EPUB

and from () it appears that

2 1+ qx

qx This appears to be the solution of the equation, but it does not satisfy it unless C, = 0, when it becomes

[ocr errors]
[ocr errors]

dx

y = C2 (1+

qa which is only a particular integral, and therefore incomplete.

This arises from our implying in the use of equation (4) that nbn + qbn-1 = 0 is generally true, whereas the equation

(n − 1) (nbn + qb,-1) = 0, derived from the auxiliary equation

ds
+9 0,

dx2 shews that b, is not necessarily connected with bu, since it may be satisfied by n = - 1.

To complete the solution, we have from (2) which is always true

boy and from (4) which is true for n = -1, we have

2 = b.it

9

2 b., = 0,

a-1 =

bo 9

9 These quantities are independent of a, a,, &c., therefore writing C, for a, as it is an arbitrary constant,

2 y = C, 1

9x is a particular integral of the proposed equation, and the complete solution is

2

2 y = C, 11

+ C, (1 +

[ocr errors]
[ocr errors]

boy

2

or as

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

=

[ocr errors]

dom

1

1

2n-)

[ocr errors]
[ocr errors]

(14) To integrate

pm dm-ly + k" y

dun where p is an integer. Assume y= (a,x"), then if

d (0,2") = 2*(p+1)-1

E (6,2"), a, = (n - pm + 1)... (n - m + 1) b. But from the given equation n (n − 1) ... (n m + 2) {n m (P + 1) +1} an + k" On-- = 0, from which

n(n-1)... (11 m + 2) ... (n - m + 1) b, + k" bn-m = 0.

But this is the equation which would result from substituting (bną") in

dy

+ k" y = 0;

dz. therefore E (b,«) is the solution of this last equation, and is therefore known. Calling it X, we have

d X
Y = (an2") = xmp+1)–1

X"-1 dc
Let m = 2, p = 2, then the integral of

4 dy

+ ky = 0
d x2
d

(k x + a)
is

C cos

dx y = C{(3 – kx®) cos (k x + a) + 3 k x sin (k x + a)}.

Ellis, Cum. Math. Jour. Vol. 11. p. 202.

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

2

Y = x5

[ocr errors]

(

ܪ

or

CHAPTER VI.

PARTIAL DIFFERENTIAL EQUATIONS.

Sect. 1. Linear Equations with Constant Coefficients.

By the method of the separation of symbols the integration of Linear Partial Differential Equations is reduced to the same processes as those for the integration of ordinary differential equations of the same class. Hence the theory which is given in the beginning of Chap. iv. is equally applicable to the present subject, and it is unnecessary to repeat it here; I shall therefore content myself with referring to what has been previously said in the Chapter alluded to, adding that every differential equation of this class between two variables has an exact analogue among partial differential equations of the same class, and that the form of the solution of the latter is the same as that of the former. On this point one remark may be made which is of considerable importance in the interpretation of our results. As in the solution of ordinary differential equations we continually meet with expressions of the form

Ceas, so in partial differential equations we shall find expressions of the form

d

[ocr errors]

Φ(y), in which the arbitrary function takes the place of the arbitrary constant. Now as the preceding formula is the symbolical expression for Taylor's Theorem, we know that

d

dy E

φω) = φg + αα). Hence, in the solution of partial differential equations, arbitrary functions of binomials play the same parts as arbitrary constants multiplied by exponentials do in equations between two variables.

[blocks in formation]

dy

d Now supposing x to be the independent variable, and a constant, with respect to it, by the Theorem given in Ex. (11), Chap. xv. of the Diff. Calc. this is equivalent to

[ocr errors][subsumed][ocr errors][ocr errors][merged small][ocr errors]

or, effecting the integration, and adding an arbitrary function of y, instead of an arbitrary constant,

[merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small][merged small][merged small][ocr errors][ocr errors]

or, as the form of ø is arbitrary, we may for

[ocr errors][merged small][ocr errors][merged small][merged small]

It is obvious that if we had taken y for our independent

d variable, and considered as a constant with respect to it,

da we should have had

cy

+ (bx - ay).

[ocr errors][merged small][merged small][ocr errors]

-1

ms €

a

cos ry

dy

d
location
dx

po ceny sdxe-** cos ry.
But by Taylor's Theorem

e-4* à cos ry = cos r(y a x); therefore

* = *** Sdxe": cos r(y ax); and, integrating with respect to X, % = e des {m cosr(y ax) - arsinr(y ax)} **dy (y);

m2 + a2 pa
(m cos ry
- ar sin ry)

+ P(y + ax).
mo + a pa

d

[ocr errors]
[ocr errors]

or,

The same method is applicable to any number of independent variables.

[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

If we expand the operating factor in ascending powers of

d d b

we shall have
dy
d

d
d d

d -2 d d
b

; dy dx

d x) the other terms being neglected, because when the operations

« PreviousContinue »