Question

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 \( x+y=1 \)

Answer 1

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

Answer 2

the solution looks like this ...

Answer 3

what kind of surface is that ... man i thought it would be a circle.

Answer 4

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

Answer 5

i'm supposed to do it by charpit's method ... an charpit's method is supposed to be easy than lagrange's method ... lol

Answer 6

i exactly did the same thing and got the same result!!
x^2+y^2+z^2=0 !!

Answer 7

but this is wrong because it gives x=y=z=0

Answer 8

The lagrange method works ... but this is too ugly ... I have to eliminate z from 4th order equation

Answer 9

this 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

Answer 10

man i cant get that answer with charpit

Answer 11

@experimentX santosh where are u....lol

Answer 12

still here man ... fishing answer from another site.

Answer 13

what did you get for answer?
i got
z = a x + phi(a) y + c

Answer 14

had been doing
\[ z = k_1x+k_2y+k_3 \]

Answer 15

k2 should be some function of k1 ... i guess there it would reduce my trouble by half

Answer 16

man let me try again..

Answer 17

let me try it again too

Answer 18

The intersection of the sphere and the plane will be a curve. The equation of the curve will solve \[x^2+(1-x)^2+z^2=25\]

Answer 19

I need to find the particular integral of the given DE passing through both of these curves.

Answer 20

no luck ... can't find \( \phi(k_1) \)

Answer 21

\[ \frac{\partial z}{\partial{x}}=\frac{2x-2(1-x)}{-2z}=\frac{2x-1}{-z}\]Similarly solve for y and \[ \frac{\partial y}{\partial{z}}\]

Answer 22

and substitute it there?

Answer 23

Sure

Answer 24

Looks like I messed up with Lagrange sol ... it didn't seem that difficult
\[z/x = \phi(y/x)\]

Answer 25

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.