<s>[INST] How to know multipliers in solving PDE Equations? Solve $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}.$$ Solve the similtaneous equation by using method of multipliers. How can we choose these multipliers? Is there any specific method to know these multipliers?
 [/INST] Try to find triplets (multipliers) $l,m,n$ such that in,
$$\frac{dx}{y+z}=\frac{dy}{z+x}=\dfrac{dz}{x+y}=\dfrac{l dx+m dy+n dz}{l(y+z)+m(z+x)+n (x+y)}$$
make the denominator vanish
Set $l=y-z,m=z-x,n=x-y$
Checking for the denominator to vanish: $(y-z)(y+z)+(z-x)(z+x)+(x-y)(x+y)=y^2-z^2+z^2-x^2+x^2-y^2=0$
$$\frac{dx}{y+z}=\frac{dy}{z+x}=\dfrac{dz}{x+y}=\dfrac{(y-z)dx+(z-x)dy+(x-y)dz}{0}$$
We have to solve $(y-z)dx+(z-x)dy+(x-y)dz=0$ Rearranging,
$(y-z)dx+(z-y+y-x)dy+(x-y)dz=0$
$(y-z)dx-(y-z)dy-(x-y)dy+(x-y)dz=0$
$(y-z)(dx-dy)-(x-y)(dy-dz)=0$
$\dfrac{(dx-dy)(y-z)-(x-y)(dy-dz)}{(y-z)^2}=0$
Finally
$\dfrac{x-y}{y-z}=c_1$
Choosing another triplet in a similar fashion, you can get the second equation.
The specifity of the method is the posing of the quotient. With respecto to something more systematic, I don't think there is an algorithm to find them.
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