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The first and second conditions agree in giving p = -1, and from the second and third we find

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and the equation is then reduced to

dx
do + box? a’ +2 d.x.

?

(15) If m =
If m= - 4, the equation becomes

dx a'd.x
dx + 6'72

r2

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in which the variables are separated. The integral is ab +

2 ab

Ce *
ab

x + b*x* y
If in the equation
do

1
de + b2 g = a co +2 dx we assume x = -

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and in this equation making (m + 3) **+ d.r = dv, and for
shortness putting
Q?

6
m + 4

n,

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m + 3

m + 3

m + 3

it is reduced to

du + Biu' dv = a’u" dv, which is similar to the proposed equation, and is therefore integrable if n= - 4, or m = - - g. If n be not equal to - 4, we may transform this equation by the same assumptions as before, when we shall obtain an equation of the form

du' + B'? udv' = a?o'"'do',

which is integrable if n' = – 4, furnishing a corresponding value for m. In this way we may proceed, continually transforming the equation and finding values of m which render Riccati's equation integrable. It will be found that these values are included in the formula

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m =

9 1

1

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2 r. p being an integer. Another series of values for m may be found by making in the original equation, when it becomes du + a' uox" dx =

6dx ;
and this being transformed by the assumptions

a
BP,
= a*,

n,
m + 1

m + 1 du + Biu'dv = a'u" dv, which is similar to the proposed equation and integrable if

4r n be of the form

that is, if 2r

m

20° +1

= 0,

m + 1

we find

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or

m + 1 2r 1'

2r + 1 Hence all the values of m are included in the formula

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(18) The equation

dy + ay" r" dx + br." y dr = 0

can be made homogeneous if

(p + 1) (1 - 9) = (m + 1) (1 – n), by the assumption

P+1

I-n

m +1
19

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or

There is an exception to this if n = 1 and q = 1; but in this case the equation becomes

dy + y (a XP + 6.2") dx = 0, in which the variables are already separated.

dy

dx

a

+6

+ e

= 0.

(19) Let aydx + bædy + 2" yu (cydx + exdy) = 0);
dividing by xy we have
dx

dy
+ x1
y

y
From this it appears that the assumptions

x"y' = U, Xy will simplify the equation. It becomes after these substitutions

du

0,

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Assume

x + y = U,

a’du whence

dy

a? + u? in which the variables are separated. The integral is

y+c Y + = a tan

a

(21) Let (y – w) (1 + x^) dy = n (1 + yo) dx.
To separate the variables assume

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y

1 + 0 0 when the equation becomes dx

udu
1 + x2

(1 + d) {+ 1 + 4)}}"
To integrate this put 1+ u^ = t', which gives

dt
1 + x2 t{nt + (t? – 1)}}}

I + 8? and again putting t=

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we find

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28

dx 2ds

2nds
1 + v?
1 + ?

(n + 1) + (n − 1) 82' which is easily integrable.

Euler, Calc. Integ. Vol. 1. p. 270. Sect. 4. Equations which involve y and its differentials in powers and products.

I. Equations of the form

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are to be resolved (when possible) into the simple factors

dy
U2

U,

0; \dx and each of these is to be integrated separately. Any one of these integrals, or the product of any number of them, will be an integral of the proposed equation.

dy

2

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y =

(1) Let

a” = 0.
dx)
dy

dy
Here

a = 0,

+ Q = 0; dx

dx therefore y = ax + C,

a x + C12 are both integrals : also

(y a X c) (y + ax – C) = 0. If we suppose c and c to be the same, this may be put under the form

(y c)= așa?.

dy (2) Let y

4a2 = 0.

2

or

2

Let Y

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or

The integrals are
y = 4 ax + C,

y' = – 4 ax + Co, and

(y4 a x c) (yo + 4a x – c,) = 0;
(yo c)= 16 ax,

when c = c.
(dy dy
(3)

+ 2x

y. dx

dx The integrals are

(v? + y')} = x + C, (x? + y?)! = – X + C1, and {(x2 + y2)! – x - c} {(x® + yo)! + ix-c} = 0;

yo = 2cx + c', when c; = c. (4) Let dy) +

x y = 0. d x) .dx.

dx The factors in this case are

dy

= 0; dx

dx
and the integrals are
a3

1
y + C,
YE E

y

+ C2, 3

dy

20

y
(N - ) (-- -) ( + -c.) = 0

and

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