Given function is-
F(2x+3y+4z,x2+y2−2)=0
Here, u=2x+3y+4z,v=x2+y2−2
Now differentiate f w.r.t x-
dudf.dxdu+dvdf.dxdv=0 −(1)
Also differentiate w.r.t y-
dudf.dydu+dvdf.dydv=0 −(2)
Now eleminating function f between eqn.(1) and (2) and we get the required PDE as-
∣∣dxdudydudxdvdydv∣∣=0
∣∣dxdudydu2x2y∣∣=0
2ydxdu−2xdydu=0
Hence The required PDE is 2ydxdu−2xdydu=0
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