Question

x²(y-z)p+y²(z-x)q=z²(x-y)​

Answer 1

Answer:

Answer on Question #43673 - Math - Differential Calculus|Equation

Problem.

Solve. (X²-yz)p +(y²-zx)q =z²-

xy

Remark.

We suppose that

푝=

휕푧

휕푥

and

푞 =

휕푧

휕푦

.

We need to solve

(

푥

2

−푦푧

)

휕푧

휕푥

+

(

푦

2

−

푧푥

)

휕푧

휕푦

=푧

2

−

푥푦

.

Solution.

The

auxiliary equations

are

푑푥

푥

2

−

푦푧

=

푑푦

푦

2

−

푧푥

=

푑푧

푧

2

−

푥푦

.

Hence

푑푥−푑푦

(

푥

2

−푦푧)−(

푦

2

−

푧푥

)

=

푑푦−푑푧

(

푦

2

−

푧푥

)−(

푧

2

−

푥푦

)

=

푑푧−푑푥

(

푧

2

−

푥푦

)−(푥

2

−푦푧)

or

푑(

푥−푦)

(

푥−푦)

(

푥+푦+푧)

=

푑(

푦−푧)

(

푦−푧)

(

푥+푦+푧)

=

푑(

푧−푥)

(

푧−푥)

(

푥+푦+푧)

.

Therefore

푑(푥−푦)

푥−푦

=

푑(

푦−푧)

푦−푧

=

푑(

푧−푥)

푧−푥

The solutions of the equations are

ln

|

푥−푦

|

=

ln

|

푦−푧

|

+

ln

퐶

1

,

ln

|

푦−푧

|

=

ln

|

푧−푥

|

+

ln

퐶

2

or

푥−푦

푦−푧

=퐶

1

,

푦−푧

푧−푥

=퐶

2

.

So

the general solution of

the equation is

휙

(

푥−푦

푦−푧

,

푦−푧

푧−푥

)

=0,

where

휙

is a differentiable arbitrary function.

Answer:

The general solution of

equation is

휙

(

푥−푦

푦−푧

,

푦−푧

푧−푥

)

=0,

where

휙

is a differentiable arbitrary function.