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solve: \[ x\; {\partial z \over \partial x} + y\; {\partial z \over \partial y} = z \] that passes though \( x^2+y^2+z^2=25 \) and \( x+y=1 \)

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the answer according to book is \[ 25(x+y) = x^2+y^2+z^2\] looks like intersection of two surfaces more than solution of DE
the solution looks like this ...
1 Attachment
what kind of surface is that ... man i thought it would be a circle.

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Other answers:

man how can we solve something like this? from the \(x^2+y^2+z^2=25\) we have \[{\partial z \over \partial x}=-\frac{x}{z}\]\[{\partial z \over \partial y}=-\frac{y}{z}\]put back in the original equation\[x^2+y^2=-z^2\]lol
i'm supposed to do it by charpit's method ... an charpit's method is supposed to be easy than lagrange's method ... lol
i exactly did the same thing and got the same result!! x^2+y^2+z^2=0 !!
but this is wrong because it gives x=y=z=0
The lagrange method works ... but this is too ugly ... I have to eliminate z from 4th order equation
this is my last problem from first order DE ... I'm moving on to second order DE after this Q
man i cant get that answer with charpit
@experimentX santosh where are
still here man ... fishing answer from another site.
what did you get for answer? i got z = a x + phi(a) y + c
had been doing \[ z = k_1x+k_2y+k_3 \]
k2 should be some function of k1 ... i guess there it would reduce my trouble by half
man let me try again..
let me try it again too
The intersection of the sphere and the plane will be a curve. The equation of the curve will solve \[x^2+(1-x)^2+z^2=25\]
I need to find the particular integral of the given DE passing through both of these curves.
no luck ... can't find \( \phi(k_1) \)
\[ \frac{\partial z}{\partial{x}}=\frac{2x-2(1-x)}{-2z}=\frac{2x-1}{-z}\]Similarly solve for y and \[ \frac{\partial y}{\partial{z}}\]
and substitute it there?
Looks like I messed up with Lagrange sol ... it didn't seem that difficult \[z/x = \phi(y/x)\]
the problem reduced to \[ z = ax +by \\ x^2+y^2+z^2=25\\ x+y=1\] Need to eliminate 'a' and 'b' from these equations.

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