4
$\begingroup$

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

  • 0
    It seems that Kappa light off when leaving the chatroom :-) (the chatroom died for me!)2012-08-28

3 Answers 3

1

Both are not right.

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$

$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$

$\dfrac{dz}{dt}=(x+y)z=\left(-\dfrac{1}{t}-\dfrac{1}{t+y_0}\right)z$ , we have $z(x,y)=\dfrac{f(y_0)}{t(t+y_0)}=xy~f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$

  • 0
    @doraemonpaul its more accepted to write the general solution not as $f_1(x)\cdot f_2(x)$ but as $\phi(f_1(x),f_2(x))$2012-08-29
2

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\;. $

  • 0
    @Soham: Not a comment, an answer. I can't tell you where because the order in which the answers are displayed depends on your selections, but if you look through all the answers to the question, you should see three of them and one is by doraemonpaul.2012-08-29
1

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.

  • 0
    @Soham: It's not you, its the chatserver I think (I had the same problem). Ok to see the Cauchy problem another day (it's better to ask a question here...). Good night (I suppose it's late for you) ! Glad it helped a little,2012-08-28