Given function is-
F ( 2 x + 3 y + 4 z , x 2 + y 2 − 2 ) = 0 F(2x+3y+4z,x^2+y^2-2)=0 F ( 2 x + 3 y + 4 z , x 2 + y 2 − 2 ) = 0
Here, u = 2 x + 3 y + 4 z , v = x 2 + y 2 − 2 u=2x+3y+4z,v=x^2+y^2-2 u = 2 x + 3 y + 4 z , v = x 2 + y 2 − 2
Now differentiate f w.r.t x-
d f d u . d u d x + d f d v . d v d x = 0 − ( 1 ) \dfrac{df}{du}.\dfrac{du}{dx}+\dfrac{df}{dv}.\dfrac{dv}{dx}=0~~~~~~~~~-(1) d u df . d x d u + d v df . d x d v = 0 − ( 1 )
Also differentiate w.r.t y-
d f d u . d u d y + d f d v . d v d y = 0 − ( 2 ) \dfrac{df}{du}.\dfrac{du}{dy}+\dfrac{df}{dv}.\dfrac{dv}{dy}=0~~~~~~~~~-(2) d u df . d y d u + d v df . d y d v = 0 − ( 2 )
Now eleminating function f between eqn.(1) and (2) and we get the required PDE as-
∣ d u d x d v d x d u d y d v d y ∣ = 0 \begin{vmatrix}\dfrac{du}{dx}&\dfrac{dv}{dx}\\\\\dfrac{du}{dy}&\dfrac{dv}{dy}\end{vmatrix}=0 ∣ ∣ d x d u d y d u d x d v d y d v ∣ ∣ = 0
∣ d u d x 2 x d u d y 2 y ∣ = 0 \begin{vmatrix} \dfrac{du}{dx}& 2x\\\\\dfrac{du}{dy}&2y\end{vmatrix}=0 ∣ ∣ d x d u d y d u 2 x 2 y ∣ ∣ = 0
2 y d u d x − 2 x d u d y = 0 2y\dfrac{du}{dx}-2x\dfrac{du}{dy}=0 2 y d x d u − 2 x d y d u = 0
Hence The required PDE is 2 y d u d x − 2 x d u d y = 0 2y\dfrac{du}{dx}-2x\dfrac{du}{dy}=0 2 y d x d u − 2 x d y d u = 0
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