By Lagrange’s method the auxiliary equations of the given differential equation are as following:
x−ydx=y−x−zdy=zdz
⟹x−ydx=y−x−zdy=zdz=0dx+dy+dz⟹dx+dy+dz=0⟹x+y+z=c1
Hence from the first fraction, we have x−ydx=y−x−zdy
⟹y−c1+ydy=zdz
Now, by integrating on both sides we get (1/2)log(2y−c1)=log(z)+log(c2)⟹c2=z2y−x−z .
Since x+y+z=c1 and z2y−x−z=c2 are independent solutions of the given differential equation, so a solution of the given differential equation is
c2=f(c1)⟹z2y−x−z=f(x+y+z) ________(1)
On x2+y2=1,z=1 , so put x=cos(t),y=sin(t) and put in solution, we get
sin(t)−cos(t)=f(cos(t)+sin(t)) ________(2)
So, the solution of the given differential equation is given by (1), where the function is given by (2).
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