Question

Form pde by eliminating arbitrary function xyz=∅(x^2+y^2+z^2)​

Answer 1

Answer:

We set u=xy+z2,v=x+y+z, then the operation of d on (1) leads to:

dF(u,v)=∂F(u,v)∂udu+∂F(u,v)∂vdv

Thus

0=dF(u,v)

⟹0=du=d(xy+z2)......(3)

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⟹y−xx−2zdx=dz......(7)

Similarly:

xdy+y(−dy−dz)+2zdz=0⟹x−yy−2zdy=dz......(8)

From (7) and (8) we obtain:

dxx−2zy−x=dyy−2zx−y=dz⟹dxx−2z=dy2z−y=dzy−x

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

(x−2z)∂z∂x−(y−2z)∂z∂y=y−x

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

-mike

Answer 2

PDE form by eliminating the arbitary function is $f_x+f_y=0.$

Step 1:Form PDE.

Given- xyz = ∅$(x^{2} +y^{2} +z^{2} )$

$f(x,y,z)$ = ∅$(u,v)=0,$

$f_x=$∅$_uu_x$ +∅$_vv_x$

$f_x$ =$2x$∅$_u$ + $2y$∅$_v$ = 0,

$f_y$ =$2y$∅$_u$ + $2x$∅$_v$ = 0,

$f_z$ = $-2z$∅$_u$ + $2z$∅$_v$ = 0,

Now eliminate ∅$_u$, ∅$_v$ ,

It seems that,

∅$_u$  =   ∅$_v$

$f_x$ =$2x$∅$_u$ - $2y$∅$_u$

$f_y$ =$2y$∅$_u$ -  $2x$∅$_u$

$f_x+f_y=0.$