Question

Find the partial differential equations by eliminating the arbitrary function from the following: xy + yz + zx = f( z /(x + y) )

Answer 1

x' + $\frac{x}{(x + y)}$ = y' + $\frac{y}{x+y}$

Explanation:

To eliminate the arbitrary function and find the partial differential equation, we need to differentiate the given equation with respect to x and y separately and then manipulate the resulting equations to eliminate the arbitrary function.

Differentiating the equation xy + yz + zx = f($\frac{z}{(x + y}$)) with respect to x, we get:

y + z + zx' = f'($\frac{z}{(x + y)}$) * ($\frac{z}{(x + y)}$)' * (-$\frac{z}{(x + y)}$²) * (1 + y')

Differentiating the equation with respect to y, we get:

x + z + zy' = f'($\frac{z}{(x + y)}$* $\frac{z}{(x + y)}$' * (-$\frac{z}{(x + y)}$²) * (1 + x')

Now, we can eliminate the arbitrary function f'($\frac{z}{(x + y)}$*$\frac{z}{(x + y)}$)' by equating the two resulting equations:

y + z + zx' = x + z + zy'

Simplifying this equation, we get:

x' - y' = $\frac{y}{x+y}$ - $\frac{x}{(x + y)}$

Finally, rearranging the terms, we have:

x' + $\frac{x}{(x + y)}$ = y' + $\frac{y}{x+y}$

This is the partial differential equation obtained by eliminating the arbitrary function from the given equation xy + yz + zx = f($\frac{z}{(x + y)}$).

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