Question

Ive (y-Z) P+(x-y) q=z-x​

Answer 1

Answer:

Step-by-step explanation:

I have given it a go and found one part of the answer, which is obtained by simply using multipliers 1,1,1 and hence u=x+y+z.

In the second part, I use multipliers x,z,y in order, and hence the denominator reduces to 0, therefore the numerator is equal to 0, thus

xdx+zdy+ydz=0

xdx+d(zy+yz)=0

xdx+d(2yz)=0

xdx+2d(yz)=0.

On integration we get x22+2yz=v, but in my book the answer is x22+yz=v

Note that zdy+ydz=d(yz), not d(zy+yz). This solves your problem.