0
$\begingroup$

I am trying to solve partial differential equation

x $\frac{\partial z}{\partial x}$ + y $\frac{\partial z}{\partial y}$ = $z$.

My approach involves using Lagrange's auxiliary equation given by

$\frac{dx}{x} = \frac{dy}{y} = \frac{dz}{z}$. Now grouping first two gives $u(x, y, z) = \frac{x}{y} = c_1$ and similarly grouping last two gives $v(x, y, z) = \frac{y}{z} = c_2$. So the general solution can be given by $F(u, v) = 0$.

My doubt: If instead of grouping first two I am grouping first and third then $u(x, y, z) = \frac{x}{z} = c$ and then will it not change to the solution $F(u, v) = 0$ obtained in the first approach? Kindly clarify my doubt.

Thank you

$\endgroup$
  • $\begingroup$ Is the derivative $\frac{\partial z}{\partial x}$ repeated twice? I think there is a typo there $\endgroup$ – minmax Sep 25 '17 at 0:15
  • $\begingroup$ @minimax I am sorry for the typo. I edited now. Thank you for pointing out. $\endgroup$ – mathscrazy Sep 25 '17 at 0:18
  • 1
    $\begingroup$ don't you have any boundary conditions? How about using the characteristics method. $\endgroup$ – minmax Sep 25 '17 at 1:00
  • $\begingroup$ Your problem has typos. I'm assuming you mean $dx/x = dy/y = dz/z$ not $dx/x = dx/y = dx/z$. Also, don't call both the equality to both $u$ and $v$ the letter '$c$'. Differentiate between them to avoid confusion by labelling them $c_{1}$ and $c_{2}$. $\endgroup$ – mattos Sep 25 '17 at 2:44
  • $\begingroup$ @Mattos I am extremely sorry for my ignorance. You are right. $\endgroup$ – mathscrazy Sep 25 '17 at 20:48
1
$\begingroup$

You system of ODE's is not correctly written. You should have : $$\frac{dx}{x} = \frac{dy}{y} = \frac{dz}{z}$$ A first family of characteristic curves comes from $\quad\frac{dx}{x} = \frac{dy}{y}\quad\to\quad \frac{x}{y}=c_1$

A second family of characteristic curves comes from $\quad \frac{dy}{y} = \frac{dz}{z}\quad\to\quad \frac{z}{y}=c_2$

A third family of characteristic curves comes from $\quad \frac{dx}{x} = \frac{dz}{z}\quad\to\quad \frac{z}{x}=c_3$

Take care that the three are not independent : $\quad\frac{c_2}{c_3}=\frac{\frac{z}{y}}{\frac{z}{x}}=\frac{x}{y}=c_1$

This means that the general solution on the form of implicit equation $\quad \Phi\left(\frac{x}{y} \:,\: \frac{z}{y} \:,\:\frac{z}{x} \right)=0\quad$ reduces to any one of $\quad \Phi_1\left(\frac{x}{y} \:,\: \frac{z}{y}\right)=0\quad$ or $\quad \Phi_2\left(\frac{x}{y} \:,\:\frac{z}{x} \right)=0\quad$ or $\quad \Phi_3\left( \frac{z}{y} \:,\:\frac{z}{x} \right)=0\quad$

where the $\Phi_1$ , $\Phi_2$ , $\Phi_3$ are arbitrary functions.

Equivalently, the general solution can be expressed on explicit form is solving them for $z$ (when it is possible, of course) :

$ \frac{z}{y}=F_1\left( \frac{x}{y} \right) \quad\to\quad z(x,y)=y\:F_1\left( \frac{x}{y}\right)$

$ \frac{z}{x}=F_2\left( \frac{x}{y} \right) \quad\to\quad z(x,y)=x\:F_2\left( \frac{x}{y}\right)$

This is two equivalent forms of the general solution since the functions $F_1$ and $F_2$ which are arbitrary, but are not independent :

$$F_1\left( \frac{x}{y} \right)=\frac{x}{y}F_2\left( \frac{x}{y} \right)$$

In conclusion: You can "group" the terms of the initial system of ODEs of any manners, you will obtain the general solution, expressed on different forms, but equivalent forms. For each form, the arbitrary function involved seems different at first sight, but in fact are related, so that the result is the same.

$\endgroup$
  • $\begingroup$ Thank you very much for clearing my doubt. $\endgroup$ – mathscrazy Sep 25 '17 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.