A community for students.
Here's the question you clicked on:
 0 viewing
experimentX
 4 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 \( x+y=1 \)
experimentX
 4 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 \( x+y=1 \)

This Question is Closed

experimentX
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0the solution looks like this ...

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0man 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
 4 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
 4 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 !!

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

experimentX
 4 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
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0man i cant get that answer with charpit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX santosh where are u....lol

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

experimentX
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0had been doing \[ z = k_1x+k_2y+k_3 \]

experimentX
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0man let me try again..

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

Valpey
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0no luck ... can't find \( \phi(k_1) \)

Valpey
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0and substitute it there?

experimentX
 4 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
 4 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.