Form Partial Differential Equation from x+y+z= f(x²+y²+z²) | Elimination of arbitrary functions

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RK Eduworld (Classic)

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Your formation of PDE failed because that forms
[(x+zp)/(y+zq)]=[(1+p)/(1+q)]
Which gives (z-y)p+(x-z)q=y-x is required the linear first order PDE.

gudissakusse

Sorry but your solution is incorrect because at 3:06 del(theta)/del(x) is not equal to 2x as you're ignoring z which is dependent on x. del(theta)/del(x) is actually 2x + 2zp and similarly for y also

Somebody-jcdw

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