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 2 years ago
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 \( x+y=1 \)
 2 years ago
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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experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0the 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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0the solution looks like this ...

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0what kind of surface is that ... man i thought it would be a circle.

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2man 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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0i'm supposed to do it by charpit's method ... an charpit's method is supposed to be easy than lagrange's method ... lol

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0i exactly did the same thing and got the same result!! x^2+y^2+z^2=0 !!

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2but this is wrong because it gives x=y=z=0

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0The lagrange method works ... but this is too ugly ... I have to eliminate z from 4th order equation

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0this is my last problem from first order DE ... I'm moving on to second order DE after this Q http://www.mathresources.com/products/mathresource/maa/charpits_method.html

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2man i cant get that answer with charpit

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2@experimentX santosh where are u....lol

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0still here man ... fishing answer from another site.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0what did you get for answer? i got z = a x + phi(a) y + c

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0had been doing \[ z = k_1x+k_2y+k_3 \]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0k2 should be some function of k1 ... i guess there it would reduce my trouble by half

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2man let me try again..

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0let me try it again too

Valpey
 2 years ago
Best ResponseYou've already chosen the best response.0The intersection of the sphere and the plane will be a curve. The equation of the curve will solve \[x^2+(1x)^2+z^2=25\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0I need to find the particular integral of the given DE passing through both of these curves.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0no luck ... can't find \( \phi(k_1) \)

Valpey
 2 years ago
Best ResponseYou've already chosen the best response.0\[ \frac{\partial z}{\partial{x}}=\frac{2x2(1x)}{2z}=\frac{2x1}{z}\]Similarly solve for y and \[ \frac{\partial y}{\partial{z}}\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0and substitute it there?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0Looks like I messed up with Lagrange sol ... it didn't seem that difficult \[z/x = \phi(y/x)\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0the 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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