Mathematical Physics Vol 1

7.8 Examples

387

becomes a total differential, and then integrate it.

Solution The condition of integrability (7.46), in this case ( R = 0), is reduced to the linear partial equation v ∂ Q ∂ x − ∂ P ∂ y = P ∂ v ∂ y − Q ∂ v ∂ x ,

to which we assign a system of differential equation d x − Q = d y P = d v v ( Q x − P y ) .

(7.296)

In our case: P = 2 x 3 y − y 2 , Q = − 2 x 4 − xy , from where we obtain P y = 2 x

3 − 2 y and

3 − y , that is

Q x = − 8 x

3 + y .

Q x − P y = − 10 x

(7.297)

Now (7.296) can be written in the form d v v = Q x − P y − Q

Q x − P y P

d x =

d y ,

that is, when we replace (7.297) d v v = −

10 x 3 + y

10 x 3 + y 2 x 3 y − y 2

d x = −

d y .

(7.298)

2 x 4 + xy

Let us first determine the quotients

2 x 4 + xy : − 10 x 3 + y = − 1 5 2 x 3 y − y 2 : − 10 x 3 + y = − 1 5

6 / 5 xy − 10 x 3 + y , 4 / 5 y 2 − 10 x 3 + y .

x +

(7.299)

y −

(7.300)

Let us first analyze the remainders of divisions in relations (7.299) and (7.300). If we multiply the remainder in (7.299) by 4 y , and the remainder in (7.300) by 6 x their absolute value become the same. The, bearing in mind that d x a = d y b = d x + d y a + b , we can eliminate the remainders by appropriate adjustments. If we proceed like that , we obtain d v v = 4 y d x − 4 / 5 xy +( 24 / 5 xy 2 ) / ( − 10 x 3 + y ) = 6 x d y − 6 / 5 xy − ( 24 / 5 xy 2 ) / ( − 10 x 3 + y ) = = 4 y d x + 6 x d y − 10 / 5 xy ⇒ d v v = − 2 d x x − 3 d y y ⇒ ln v = ln x − 2 + ln y − 3 + ln c = ln cx − 2 y − 3 ⇒ v = cx − 2 y − 3 .