y2zp +x2zq=y2x
The subsidiary equations are
y2zdx=x2zdy=y2xdz
Considering the first two terms of the equation we get
y2dx=x2dy
x2dx−y2dy=0
x3−y3=c1
Considering the 1st and last terms of the equation
y2zdx=y2xdz
zdx=xdz
xdx−zdz=0
x2−z2=c2
Solution
f(x3−y3,x2−z2)=0
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