Or, substituting for u and v their values in a and y,

x

≈ = { (1 + y®) 3 + y } − ¦ { (1+y' )^ − y} log x + ƒ

Equations of the second and higher orders may sometimes be reduced by transformations similar to those employed in Chap. IV. Sect. 2.

By means of the same transformation as in the last example we find

so that

(14)

the integral of which is v = (x + y)ap(x),

≈ = fdx(x + y)ap(x) + ¥(y).

Integrate the equation

Assume dx xdu, dy = ydv; then by Ex. (6) of Chap.

=

III. Sect. 1, of the Diff. Calc. we have generally

...

d

But by a known theorem of Vandermonde if

[x]' = x(x − 1)... (x − r + 1),

[x]”+n[x]”-1 [y] +

n (n - 1)

1.2

· [x]"−2 [y]2 + &c. + [y]” = [x + y]”.

Therefore, as the symbols of differentiation are subject to the same laws of combination as the algebraical symbols, the differential equation may be written

≈ = 4% (v P(v − u) + e" P¡ (v − u) + &c. + €(n−1)" Pn-1 (v − u) ;

0;

the integral of which (see Ex. (11) of the preceding section) is

≈ = e−(av+bu) [dv €a” fdu e3“V + €

av

bu

1

-1

or ≈ = ̧3 ̧ï fdy ya-1 fdæ a3¬1V + — a ƒ (x) + —, F' (y).

ya

1

1

f

y

≈ = $(2) + ↓ (xy).

+

=

d2z a2

d2x 2 dz
dx2 x dx dy

process as in Ex. (9) of Chap. IV. Sect. 2, under the form

and thence by the same process as in Ex. (10) of Chap. IV.

and therefore

1

x=

1

= { f′ ( x + ay) + y' (x − ay) } − } { p(x+ay) + \ (x−ay)}.

x

This equation occurs in the Theory of Sound. Airy's Tracts, p. 271.

d2z a2 d2 z

See

(19) Let

= 0.

This equation is of the same form as that in Ex. (6) of Chap. v., and its integral will be found from that given

d

dy

there by putting a for c, and changing the arbitrary constants into arbitrary functions of y.

Hence we find

may in the same way be deduced from that of Ex. (8) of the same Chapter: the result is

≈ = x {F' (y + 3 ax13) + ƒ' (y − sax})}

The integral of this equation may be deduced from that

in Ex. (10) of Chap. v. by putting

gives us

x=

1

ах

d

a2

for q. This

dy2

{F (y − a x) − ƒ (y + ax) } + F' (y − a x) + ƒ’(y + ax).