4. Deduce in like manner the functional expression for all the integrating factors of the equation

5. Obtain integrating factors for the homogeneous equations:

(2) (8y+10x) dx + (5y + 7x) dy = 0.

(3) (x2+2xy - y3) dx + (y2 + 2xy — x2) dy = 0.

y

dox

(5) (cos + y sin 2) ydz + (z cosy sin) ady = 0.

Ꮳ

Exhibit the corresponding integrals of the above equations.

=

0. What

formula ought here to be employed and to what result does it lead ?

7. Determine an integrating factor of each of the equations (1) (x2y2+xy) ydx + (x2y2 − 1) xdy = 0. ✔

(2) (x3y3+x2y3+ xy +1) ¿dx+ (x3y3 — a2y2— xy + 1) xdy=0. V

CHAPTER V.

ON THE GENERAL DETERMINATION OF THE INTEGRATING FACTORS OF THE EQUATION Mdx + Ndy = 0.

1. PROP. It is required to form a differential equation for determining in the most general manner the integrating factors of the equation Mdx+ Ndy = 0.

Let μ be any integrating factor of the above equation, then u since Mdx+uNdy is by hypothesis an exact differential, we have by Prop. I. Chap. III.

Now this equation involves the partial differential coefficients of u taken with respect to x and y. It is therefore a partial differential equation. We have not the means of solving it generally, and it will hereafter appear that its general solution would demand a previous general solution of the differential equation Mdx + Ndy=0, of which μ is the integrating factor. But there are many cases in which we can solve the equation under some restrictive condition or hypothesis, and the form of the solution obtained will always indicate when the supposed condition or hypothesis is legitimate.

The following are examples of such solutions.

2. Let μ be a function of one of the variables only, e.g.

αμ

=

suppose μ = Þ (~), then since 0, we have from (1)

dy

Now if the second member of this equation is a function of x the equation is integrable, and we have

We have seen that the hypothesis assumed as the basis of the above solution, viz. that the integrating factor μ is a function of x only, is legitimate when the constitution of the functions M and N is such that the expression

is a function of x only. In this case (2) enables us to determine the value of μ.

In like manner the condition under which is a function

of y only, is

dN dM

dx

M

μ

.........

(3x2 + 6xy + 3y3) dx + (2x2 + 3xy) dy= 0 .........................(5) admits of an integrating factor which is a function of x only. Making M= 3x2 + 6xy + 3y2, N = 2x2 + 3xy, we find

and this result being a function of x alone, the determination. of μ as a function of x alone is seen to be possible. From (2) we now find

C being an arbitrary constant.

Now multiplying (5) by Cx, we have

C{(3x3 + 6x3y + 3xy3) dx + (2x3 + 3x3y) dy}

= 0.

The first member of this equation remains a complete differential whatever value we assign to C. If we make C=1, and integrate, we find

The student may obtain also the same result by solving (5) as a homogeneous equation.

The linear differential equation of the first order

P and Q being functions of x, may be solved by the above method.

For, reducing it to the form

(Py− Q) dx + dy = 0........

..(7),

which being a function of x we find from (2)

Multiplying (7) by the factor thus determined, we have

the first member of which is now the exact differential of the function

Equating this expression to an arbitrary constant c, we

find

which agrees with the result of Art. 10, Chap. II.

.(8),

3. Let it be required to determine the conditions under which the equation Mdx+ Ndy = 0, can be made integrable by a factor μ which is a function of the product xy.

Representing xy by v and making μ=4(v), the partial differential equation (1) becomes

Thus the condition sought is that the second member of the