Question

in general solution of p- q = ln(x + y)is
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Answer 1

Step-by-step explanation:

p-q = log(x+y)

This is a Lagrange's equation whoes general formula is pP+qQ=R

This can be solved by the formula

(dx/P)=(dy/Q)=(dz/R)

Here, P=1 ,Q= -1 ,R=log(x+y)

Now, {dx/1}={dy/-1}={dz/log(x+y)}

From the 1st two ratio,

dx= -dy

=> dx+dy=0

Integrating above equation we get,

x+y=C ―――(1)

Now from 1st and 3rd ration,

dx=dz/log(x+y)

=>dx=dz/logC by using equation (1)

=>(logC)dx =dz ――――(2)

=>( logC)x=z+K. {Integrating above equation (2)}

=>xlog(x+y)=z +K――――(3)

Equation (1)&(3) together yeilds the solution.

Answer 2

Answer:

Step-by-step explanation:

Concept:

The linear equation of Lagrange. Lagrange's Linear Equation is a partial differential equation of the form$Pp+Qq = R$, where$P, Q, and R$ are functions of$x, y, and z$. This equation is of first order and linear in $p and q$.

The idea behind stationary activity. a variational integral principle operating under the influence of potential forces that do not explicitly depend on time and arising in the dynamics of holonomic systems constrained by ideal stationary constraints.

Given:

The equation $p- q = ln(x + y)$

Find:

To find the general solution of $p- q = ln(x + y)$

Solution:

$p-q = log(x+y)$

The general formula for this Lagrange equation is$pP+qQ=R.$

The formula can be used to resolve this.

$(dx/P)=(dy/Q)=(dz/R)$

Here,$P=1 ,Q= -1 ,R=log(x+y)$

$Now, {dx/1}={dy/-1}={dz/log(x+y)}$

Using the first two ratio,

$dx= -dy$

⇒ $dx+dy=0$

Integrating above equation we get,

$x+y = C .....(1)$

From the first and third ratios,

$dx=dz/log(x+y)$

⇒ $dx=dz/logC by using equation (1)$

$(logC)dx =dz .....(2)$

$( logC)x=z+K. {Integrating above equation (2)}$

$xlog(x+y)=z +K .....(3)$

Together, equations (1) and (3) produce the answer.

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