Question

solve the following partial differential equation p+q=x+y+z​

Answer 1

The solution of given differential equation p+q=x+y+z is Ф(x-y ,$e^{-x}$(2+x+y+z))

Given that;

p+q=x+y+z

To find;

solution of p+q=x+y+z

Solution;

We have lagrange's auxillary equations,

$\frac{dx}{1}$ = $\frac{dy}{1}$ = $\frac{dz}{x+y+z}$

taking the first two terms,

$\frac{dx}{1}$ = $\frac{dy}{1}$

on integrating we get,

∫dx = ∫dy

x = y + c1

x - y = c1 ..... (1)

choosing 1,1,1 as multipliers we get,

$\frac{dx}{1}$ = $\frac{dy}{1}$ = $\frac{dz}{x+y+z}$

Each fractions  = $\frac{dx+dy+dz}{1+1+x+y+z}$

∫1.dx = ∫ $\frac{d(x+y+z)}{2+(x+y+z)}$

let x+y+z = u

then, du= d( x+y+z)

∫1.dx = ∫ $\frac{du}{2+u}$

x = log(2+u) + logc2

x = $\frac{log(2+u)}{c2}$

$e^{x}$ = $\frac{2+u}{c2}$

c2 = $e^{-x}$(2+x+y+z) .....(2)

From equations 1 and 2 we get,

Ф(x-y, $e^{-x}$(2+x+y+z)) is the equation's required solution.

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