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I would be incrediabely gratful if someone could go through step by step and explain how to do this question, as i'm rather stuck -and the lecture notes have a lot to be desired!

'Find $z(x,y)$ explicitly if

$$-yz_x+xz_y=\frac{xz}{\sqrt{x^2+y^2}}$$

and $z(x,0)=1$ for $x \geq 1$

So i'm trying to go about solving the characteristics of this, the equations

$$\frac{dx}{ds}=-y$$

$$\frac{dy}{ds}=x$$

$$\frac{dz}{ds}=\frac{xz}{\sqrt{x^2+y^2}}$$

Then use the solution along the curve $l(t)=(t,0,1)$ for $t\geq 1$ to find the coefficients of the solutions to the equation in terms of t. We would then have the solution space in terms of parameters $t,s$ correct? Then we try to eliminate these to get the space in terms of $x,y,z$. I can't seem to solve the equations however.

Any help is greatly appreciated.

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double dollar signs for big fractions, or '\displaystyle' in single dollar signs (this won't be centered) –  David Mitra Nov 14 '11 at 19:21
    
@DavidMitra: Cheers! that's more readable. –  Freeman Nov 14 '11 at 19:22
    
Wow, huge now. I edited this myself, before you did. You should omit the \large tags. –  David Mitra Nov 14 '11 at 19:24
    
@DavidMitra: Whoops! –  Freeman Nov 14 '11 at 19:26

1 Answer 1

up vote 1 down vote accepted

You've got the idea right.

The characterics starting on the positive $x$ axis (at $(x,y)=(t,0)$ to be precise) when $s=0$ are $(x(s),y(s))=(t \cos s,t \sin s)$. Then the equation for $z$ reduces to $dz/ds=z \cos s$ (independent of $t$, as it turns out). Separation of variables gives $$ \int_{1}^{z(s)} \frac{dz}{z} = \int_0^s \cos s \, ds. $$ (I'm being a little sloppy with the notation here, using the same name for the dummy variable as in the upper limit of integration.) Integrating this, we get $\ln z(s)=\sin s$, hence $$ z(x,y) = \exp\frac{y}{\sqrt{x^2+y^2}}. $$

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Thanks so much, this is excellent! –  Freeman Nov 15 '11 at 10:21

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