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The integral of this is easily seen to be

x = C, eat + C, eße + Czeyt, a, ß, y being the roots of

23 – (a'b + a'c + 6"c') x + a'b'c + a'bc' = 0. The values of y and s are easily derived from that of x,

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Eliminating y by operating on the first with multiplying the second by a and adding, there results

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where h', k* are the roots of the equation

gi + (a' + b) x + a'b ab'= 0.
Hence we find
ac' -ca

+ C, cos (hx + a) + C, cos (k x + 3);
a'b - ab

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h + b

ki + b Cicos (h x + a)- C, cos(kx +ß).

a

a

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d'y dy dx
+ 2

+

+ 6y + 5x = sin t. dt?

dt dt

Eliminating y we obtain the equation

X = 2 cos 2 t – 4 sin 2t – 2 cost;

{G+-+ o} {G+} {G+"}

or

X = 2 cos 2t - 4 sin 2t - 2 cost;

the integral of which is x = cos 2t – 2 sin 2 t - cos + + C, cos (24+ + a) + C, cos (33t+ß), and from this the value of y is easily found.

Take the system of equations.

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These may be written under the form (a, + 0,0° + a,d' + &c.) x + (0,0 + d + &c.) y = m sin nt, (a, + a,d" + a,d' + &c.) y - (a d + a+ &c.) r = m cos nt, where for convenience the differentials of t are omitted.

Eliminating y we have

{(1, +ad+ a,d' + &c.) + (a, d + ad + &c.)"} x =
m (1, + a,n - a.n- 030° + a, n' + a, n' - &c.) sin nt.

It is obvious from the form of this that the complementary function must be of the form

(4 cost + B sint), where all the values are to be assigned to , which satisfy the equation

(a, -0,1 +0,1 + &c.)-X (a, - 031? + &c.) = 0.

Ilence we have

X = Y(A cost + B sin t) +
m (a, + a, n - 0.0 - 4,2" + 0,0 + (1-12 - &c.)

sin nt; (a, - 0,1" + a, n' + &c.) - n" (a, - 4n' + &c.)"

m sin nt or X=(Acost + B sin t) +

do-a, n-a, n° +ayn+a,n'- &c. Whence also

m cos nt y=-(A sin \t - B cost) +

ao-an-a,n' + azn + a,n'-&c.

The same method is applicable to linear partial differential equations in which the coefficients are constants. The two symbols of differentiation are to be treated as two independent constants, since they do not affect each other, and are both subject to the laws which regulate the combinations of ordinary algebraical symbols.

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d d

d
ta +6) x + c U = 0,
dr
dy

d x

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d d

d
+ a

+ b) ut ' % = 0.
Ida
dy

dr To eliminate u, operate on the first equation with d d

d + a + b, and on the second with c and subtract: d.x

dr we have then

d

d
(a
+ b) (a + b)?
d
dy

I = 0.
dx
1 - cc' do

1-cc

dy

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dy

+2

+

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=

n

1 If we call 1 - (cc')

and it

ፃዜ be divided into the two factors d d d

d
+ m (a

+ na
dc dy do dy

Z = 0;

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the integral of this equation is

z = 0 (y) + xy (y) + far" x (x); whence also u may be determined.

(11) The equations for determining the small disturbances of an elastic medium in three dimensions are

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We might in this case eliminate v and w by a process similar to that used in Ex. (5) of this section ; but the following method is more convenient.

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2

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respectively, and add: then d2 d

d +

+ d x

d

Operate on each of these by Coa) , {G

d du
d dv

d dew +

+ d x dt?

dy dt? dx dt d? du dv

dw
+ +
dt? dr dy dy

1 dag a? dt?

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dt

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can

- 2

2 dr

- 2

dr

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W =

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dx'

d'r

d Hence - a

+

p = 0.

dy
The integral of this is
ar{6)+(+6)}"

-ar{4'+6)+6)}"
p(x,y,x) +

¥ (x, y, z) From this p is determined, and hence we find U, V, W, as d) - 2 dr d

d dt dy'

do When the equations are not linear there is no general method for integrating them; and therefore the means of doing so must be adapted to the particular case under consideration. Two of the more important examples of such equations, which occur in dynamics, are subjoined.

(12) The equations for determining the motion of a particle attracted to a fixed centre of force varying inversely as the square of the distance are HUN

d'y, HY
+
0 (1),

(2),
dt?
203

dt? where you? = x2 + y'.

Multiply the first equation by y and the second by r, and subtract: then

dr

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V

dạy

- Y dt?

= 0.

dt?

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