Question

Solve the partial differential equation yq-xp = Z​

Answer 1

Answer:

The solution of xp+yq=z is

A) f(x,y)=0

B) f(x2,y2)=0

C) f(xy,yz)=0

D) f(xy,yz)=0

Correct Answer:

D) f(xy,yz)=0

Description for Correct answer:

xp+yq=z

This is Lagrange's linear equation of the type

Pp+Qq=R

Solution is dxP=dyQ=dzR

Therefore solution of xp+yq=z is given by,

dxx=dyy=dzz

From first two equations,

dxx=dyy⇒∫dxx=∫dyy

⇒logx=logy+logc1

⇒x=yc1⇒xy=c1...(1)

Also from last two equations,

dyy=dzz⇒∫dyy=∫dzz

⇒logy=logz+logc2

⇒y=zc2⇒yz=c2...(2)

From (1) and (2) the solution is f(xy,yz)=0

Step-by-step explanation:

HOPE MY ANSWER Helps

Answer 2

Given:

yq - xp = z

To find:

Solve the equation yq-xp = z

Solution:

yq - xp =z

The alternative equation for the given equation will be:

$\frac{dx}{-x} = \frac{dy}{y} = \frac{dz}{z}$

We know,

$\frac{dx}{-x} = \frac{dy}{y}$

Now, we obtain:

-㏒ x = ㏒ y - ㏒ a

㏒ a = ㏒ y + ㏒ x

㏒ a = ㏒ xy ( ㏒ y + ㏒ x = ㏒ xy)

Removing ㏒ from both sides, we obtain :

a = xy

We know,

$\frac{dy}{y} =\frac{dz}{z}$

Now, we obtain:

㏒ y = ㏒ z + ㏒ b

㏒ b =㏒ y - ㏒ z

㏒b = ㏒ (y/z)  (㏒ a - ㏒b = ㏒ (a/b))

Removing ㏒ from both sides, we obtain:

b = y/z  

Now, from equation a = xy and equation b = y/z   we obtain,  function(  $\frac{y}{z}$ , xy ) = 0