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Consider: $$x^2p+y^2q=(x+y)z$$ where $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$. Thus by Lagrange's Method

$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{(x+y)z}$$

$$\Rightarrow \frac{dx}{x^2}=\frac{dy}{y^2} \Rightarrow \frac{1}{x}-\frac{1}{y}=k$$

$$\Rightarrow \frac{dx}{x^2}=\frac{dz}{(x+y)z} \Rightarrow -\frac{1}{x}-\frac{\ln(z)}{(x+y)}=c$$

$$\Rightarrow \phi (\frac{1}{x}-\frac{1}{y},x+y+x\ln(z))=c$$

Or if we explore further $$\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-xyz(x+y)}\Rightarrow \frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-x^2yz-xy^2z}\Rightarrow yzdx+xzdy-xydz=0 \Rightarrow xyz=k$$

So can we even write $f(\frac{1}{x}-\frac{1}{y},xyz)=k$. So which one is right?

Soham

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There are no derivatives with respect to $z$ in the equation you wrote, so you are no quite using Lagrange's method correctly. –  Mariano Suárez-Alvarez Aug 27 '12 at 15:43
    
On the other hand, I cannot see how you got from the equation you wrote at the beginning by «exploring further», the equation you wrote in the last paragraph... –  Mariano Suárez-Alvarez Aug 27 '12 at 15:49
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Whatever textbook, I am sure that it never writes things like $\frac{\text{something}}{0}$.... :-) –  Mariano Suárez-Alvarez Aug 27 '12 at 15:57
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My last intervention here: have you remarked that in the equation you have both $z$ and $u$, and, as I observed a couple of times by now, that is probably a typo? –  Mariano Suárez-Alvarez Aug 27 '12 at 16:16
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It seems that the method you're trying to apply is more usually called the method of characteristics? –  joriki Aug 27 '12 at 17:07
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3 Answers 3

up vote 0 down vote accepted

I think that your work using the characteristics method was right at the start including :

$$\frac{dx}{x^2}=\frac{dy}{y^2} \Rightarrow \frac{1}{x}-\frac{1}{y}=k$$

But the implication here was wrong :

$$\frac{dx}{x^2}=\frac{dz}{(x+y)z} \Rightarrow -\frac{1}{x}-\frac{\ln(z)}{(x+y)}=c$$

because the $x$ appears at the denominator of $dz$ as well as the numerator of $dx$ so that I think this should be : $$\frac{dx}{x^2}=\frac{dz}{(x+y)z} \implies \frac{(x+y)\,dx}{x^2}=\frac{dz}z$$ with Joriki's correction and using $\ y=\dfrac 1{\frac 1x-k}=\dfrac x{1-kx}$ we get : $$\left(\frac 1x+\frac 1x+\frac k{1-kx}\right)dx=\frac{dz}z\implies 2\ln(x)-\ln(1-kx)-\ln(z)=C_0$$ $$\implies \ln\left(\frac {x^2}{1-kx}\right)-\ln(z)=\ln(xy)-\ln(z)=C_0$$ (we replaced by $y$ again to remove the $k$ constant) $$\implies \phi\left(\frac{1}{x}-\frac{1}{y},\ \frac {xy}z\right)=C$$ (you had another error here : forgetting the denominator of the second parameter)

After that you wrote :

EDITED:
Or if we explore further $$\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-xyz(x+y)}\Rightarrow \frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-x^2yz-xy^2z}\Rightarrow yzdx+xzdy-xydz=0 \Rightarrow xyz=k$$

If you divide $\ yzdx+xzdy-xydz=0$ by $xyz\ $ you get : $\dfrac {dx}x +\dfrac {dy}y -\dfrac {dz}z = 0\implies\dfrac {xy}z=C_1$

So that the other way to write the solution doesn't differ from the first one : $$\phi\left(\frac 1x -\frac 1y,\ \frac {xy}z\right)=C$$

Because of the arbitrary character of $\phi$ other equivalent parameters could have been obtained/chosen. Like replacing one of the parameters by the product $\dfrac{y-x}z$ or something more elaborate.

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Edited... and Ray, thanks for showing interest in the question. Much appreciated. Thanks –  Soham Aug 27 '12 at 17:06
    
It seems you ended up making a similar mistake as the OP -- you treat $y$ as constant in integrating with respect to $x$ -- see my answer. –  joriki Aug 27 '12 at 17:29
    
@Joriki: Oops you are right. Thanks for the correction ! –  Raymond Manzoni Aug 27 '12 at 17:43
    
@Joriki: Sorry to bother you again with this but do you see something wrong in my (updated) answer (the downvote is of course unexplained !) ? –  Raymond Manzoni Aug 28 '12 at 6:21
    
The only problems I see are that it should be $+k/(1-kx)$ (but the minus sign for the logarithm after integration is correct), and that you fudged over the point that the integral of $1/x$ is $\log|x|$, not $\log x$ (but I sort of did that, too). (I didn't downvote.) –  joriki Aug 28 '12 at 6:31
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You can't integrate $\mathrm dz/z$ to $\log z$ and treat $x$ and $y$ in the denominator as constant. Likewise, Raymond's answer treats $y$ as constant in integrating $y/x^2$. In the method of characteristics, since we're finding curves, there's only one independent variable at a time, and the others have to be expressed in terms of it when we want to integrate.

Thus, Raymond's answer is correct up to

$$ \frac{(x+y)\,\mathrm dx}{x^2}=\frac{\mathrm dz}z\;, $$

but then we have to express $y$ in terms of $x$,

$$ y=\left(\frac1x-k\right)^{-1}\;, $$

to obtain

$$ \left(\frac1x+\frac1x\frac1{1-kx}\right)\mathrm dx=\frac1z\mathrm dz\;. $$

Integrating this yields

$$ z=c\frac{x^2}{1-kx}\;, $$

and then substituting

$$ k=\frac1x-\frac1y $$

yields

$$ z=cxy\;, $$

which is readily confirmed to solve the given equation.

[Edit:]

As doraemonpaul rightly pointed out, that's not the general solution, since $c$ can be chosen independently for each value of $k$, so it should be

$$ z=c(k)xy=c\left(\frac1x-\frac1y\right)xy\;. $$

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Why cant we integrate it straight away keeping y,or z as constant? After all when we differentiate we do it in the same way.. –  Soham Aug 27 '12 at 17:38
    
@Soham: your second method is right you just wrote $xyz=c$ instead of $\frac {xy}z=c$ (see the minus sign). You had nearly Joriki's answer (+1 of course)... –  Raymond Manzoni Aug 27 '12 at 17:47
    
@Soham: Please elaborate; I'm not sure what form of differentiation you're referring to. Are you aware that the method of characteristics determines characteristic curves? –  joriki Aug 27 '12 at 18:00
    
@joriki I meant partial differentiation of say $xyz$ wrt $x$ yields $yz$ effectively implying $yz$ is a constant. So why cant we integrate (partially wrt x) as like this: xyz + f(y,z) Should be consistent isnt it? So in the similar way why cant we integrate the LHS in the first eqn which you wrote keeping y as a constant. –  Soham Aug 27 '12 at 18:39
    
@joriki No I was not aware of characteristic curves. Thanks will check it out –  Soham Aug 27 '12 at 18:40
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Both are not right.

In fact this PDE belongs to a PDE of the form http://eqworld.ipmnet.ru/en/solutions/fpde/fpde1309.pdf.

The general solution is $z(x,y)=e^{\int\frac{x}{x^2}dx+\int\frac{y}{y^2}dy}c\left(\int\dfrac{1}{x^2}dx-\int\dfrac{1}{y^2}dy\right)=e^{\int\frac{1}{x}dx+\int\frac{1}{y}dy}c\left(-\dfrac{1}{x}+\dfrac{1}{y}\right)=e^{\ln x+\ln y}C\left(\dfrac{1}{x}-\dfrac{1}{y}\right)=e^{\ln xy}C\left(\dfrac{1}{x}-\dfrac{1}{y}\right)=xy~C\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$

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If you consider the solution obtained : $\phi\left(\frac 1x -\frac 1y,\ \frac {xy}z\right)=D\ $ then for $\phi(a,b)=f(a)\,b\ $ you get : $f\left(\frac 1x-\frac 1y\right)\cdot \frac {xy}z=D\ $ i.e. $\ z=xy\ f\left(\frac 1x-\frac 1y\right)/D$ like in eqworld's solution ! –  Raymond Manzoni Aug 28 '12 at 7:06
    
@Raymond: I also first thought that your two solutions are equivalent, but actually yours allows $z$ to be any function at all, since you could let $\phi\equiv C$. The explicit form $\phi(a,b)=f(a)b$ is only one particular case of $\phi(a,b)=C$. –  joriki Aug 28 '12 at 7:13
    
@doraemonpaul: In any case, you were right about my answer; I edited it accordingly. –  joriki Aug 28 '12 at 7:15
    
@joriki: note that $z$ appears only in my second parameter, this limits the possibilities... (I only wrote that my solution implied eqworld's) –  Raymond Manzoni Aug 28 '12 at 7:20
    
@Raymond: Perhaps I misunderstand what you mean by $\phi\left(\frac 1x -\frac 1y,\ \frac {xy}z\right)=C$ -- why doesn't this allow for arbitrary $z$ if we choose $\phi\equiv C$? I don't see how this is related to whether $z$ occurs only in the second parameter. Are you making any assumptions about $\phi$? –  joriki Aug 28 '12 at 7:26
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