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I was presented with the equation:

$\begin{equation} \begin{split} \frac{dx}{x(x+y)} = \frac{dy}{y(x+y)} = \frac{dz}{(x-y)(2x+2y+z)} \end{split} \end{equation}$

So, we immediately know that the solution must be in this form:

$F(c_1,c_2) =0$.

Taking the the first two terms:

$\begin{equation} \begin{split} \frac{dy}{dx} = \frac{y(x+y)}{x(x+y)} = \frac{y}{x} \end{split} \end{equation}$

$\begin{equation} \begin{split} \int_{}{\frac{1}{x}} \ dx =\int_{}{\frac{1}{y}} \ dy \end{split} \end{equation}$

Then,

$\begin{equation} \begin{split} \ln {\frac{x}{y}} => \frac{x}{y} = c_1 \end{split} \end{equation}$

The solution: $\frac{x}{y} = c_1$

Next,

$\begin{equation} \begin{split} \frac{dx - dy}{(x-y)(x+y)}= \frac{dz}{(x-y)(2x+2y+z)} \end{split} \end{equation}$

Finally subtracting and multiplying by a factor of $(x-y)$ (on both sides),

$\begin{equation} \begin{split} \frac{dx-dy-dz}{(-x-y+z)} = 0 \end{split} \end{equation}$

The solution is: $\frac{x}{yz^2} =c_2$ So, The solution is $F(\frac{x}{y},\frac{x}{yz^2}) =0$

Am I correct?

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  • $\begingroup$ How did you get $x-y$ in the argument of $F$ from $c_{1} = x/y$? $\endgroup$ – Mattos Apr 14 '18 at 7:49
  • $\begingroup$ @Mattos sorry about that $\endgroup$ – DwightD Apr 14 '18 at 10:58
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    $\begingroup$ Please tell again why $z^2y/x$ is a constant of the characteristic curves. The present equations seem not to support this. Is there a typo in the third fraction? $\endgroup$ – LutzL Apr 14 '18 at 15:24
  • $\begingroup$ @LutzL Because in general, when taking the integral of some function, there exist a constant. Although the equation equals zero, by subtracting the constant on both sides, equals a still equals a constant. Hopefully you get what I’m saying lol $\endgroup$ – DwightD Apr 14 '18 at 15:30
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    $\begingroup$ Along a characteristic curve $x$ can be chosen as parameter, you found $y=ax$. Then $$ \frac{dz}{dx}=\frac{(1−a)(2x+2ax+z)}{x(1+a)}=2(1-a)+\frac{1-a}{1+a}\frac zx $$ and that can be solved as 1st order linear ODE, but with an ugly solution. Are you sure the original equations are correctly derived from the PDE? $\endgroup$ – LutzL Apr 14 '18 at 16:00

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