Form a PDE by eliminating the arbitrary function

Question

Form a PDE by eliminating the arbitrary function

(i) $\displaystyle F(xy+z^2,x+y+z)=0 $

(ii) $\displaystyle F(x+y+z,x^2+y^2+z^2)=0 $

How do I proceed for any of these?

Answer 1

We set $u=xy+z^2,v=x+y+z$, then the operation of $d$ on (1) leads to: $$dF(u,v)=\frac{\partial F(u,v)}{\partial u}du+\frac{\partial F(u,v)}{\partial v} dv $$

Thus $$0=dF(u,v)$$ $$\implies 0=du=d(xy+z^2)......(3)$$ $$\text {and } 0=dv=d(x+y+z)......(4)$$

From (3) and (4) we have: $$xdy+ydx+2zdz=0......(5)$$ $$dx+dy+dz=0......(6)$$

Thus getting rid of $dy$ we obtain: $$x(-dx-dz)+ydx+2zdz=0 \implies \frac{y-x}{x-2z}dx=dz......(7)$$

Similarly: $$xdy+y(-dy-dz)+2zdz=0 \implies \frac{x-y}{y-2z}dy=dz......(8)$$

From (7) and (8) we obtain: $$\frac{dx}{\frac{x-2z}{y-x}}=\frac{dy}{\frac{y-2z}{x-y}}=dz \implies \frac{dx}{x-2z}=\frac{dy}{2z-y}=\frac{dz}{y-x}$$

Finally we obtain the desired PDE for $z(x,y)$:

$$(x-2z)\frac{\partial z}{\partial x}-(y-2z)\frac{\partial z}{\partial y}=y-x$$

The other implicit equation can be treated in a similar fashion.

-mike

Answer 2

Let $u = xy + z^2, v = x + y + z$

$F(u, v) = 0$

$F_x = 0\\ F_u u_x + F_v v_x = 0\\ F_u \ (y + 2z z_x) + F_v \ (1 + z_x)= 0$

$F_y = 0\\ F_u u_y + F_v v_y = 0\\ F_u \ (x + 2z z_y) + F_v \ (1 + z_y)= 0$

Equating the value of $\frac{-F_u}{F_v} $ from above, $$ \frac{1 + z_x}{y + 2z z_x} = \frac{1 + z_y}{x + 2z z_y}\\ x - y + z_y(2z - y) + z_x(x - 2z) = 0$$

Answer 3

$F(xy+z^2,x+y+z)=0$ can be written as

$Q(xy+z^2)=x+y+z$

Differentiating wrt x,

$Q' \ (y+2zz_x) = 1+ z_x$

Now wrt y,

$Q' \ (x+2zz_y) = 1 + z_y$

Eliminating $Q'$,

$$\frac{y+2 zz_x}{x + 2zz_y} = \frac{1+z_x}{1+z_y}$$