\(\displaystyle \frac{dx}{ay} = \frac{dy}{bx} = \frac{dw}{cw^k}\)
\(\displaystyle \frac{dx}{ay} = \frac{dw}{cw^k}\)
From the previous post, we know that:
\(\displaystyle ay^2 - bx^2 = C_1\)
then
\(\displaystyle y = \pm\sqrt{\frac{C_1 + bx^2}{a}}\)
Let's take the positive square root, then we have:
\(\displaystyle \frac{c \ dx}{a\sqrt{\frac{C_1 + bx^2}{a}}} = \frac{dw}{w^k}\)
\(\displaystyle \frac{c}{\sqrt{a}}\int\frac{dx}{\sqrt{C_1 + bx^2}} = \int\frac{dw}{w^k}\)
\(\displaystyle \frac{c}{\sqrt{a}}\frac{\ln|\sqrt{C_1 + bx^2} + \sqrt{b}x|}{\sqrt{b}} + E = \frac{w^{1 - k}}{1 - k}\)
This solution is for \(\displaystyle k \neq 1\). Let us try to simplify it.
\(\displaystyle \frac{c}{\sqrt{ab}}\ln\left|\sqrt{ay^2 - bx^2 + bx^2} + \sqrt{b}x\right| + E = \frac{w^{1 - k}}{1 - k}\)
\(\displaystyle \frac{c}{\sqrt{ab}}\ln\left|\sqrt{a}y + \sqrt{b}x\right| + E = \frac{w^{1 - k}}{1 - k}\)
\(\displaystyle \frac{c}{\sqrt{ab}}\ln\left|\frac{ay + \sqrt{ab}x}{\sqrt{a}}\right| + E = \frac{w^{1 - k}}{1 - k}\)
\(\displaystyle \frac{c}{\sqrt{ab}}\bigg(\ln\left|\sqrt{ab}x + ay\right| - \ln\left|\sqrt{a}\right|\bigg) + E = \frac{w^{1 - k}}{1 - k}\)
\(\displaystyle \frac{c}{\sqrt{ab}}\ln\left|\sqrt{ab}x + ay\right| - \frac{c}{\sqrt{ab}}\ln\left|\sqrt{a}\right| + E = \frac{w^{1 - k}}{1 - k}\)
\(\displaystyle \frac{c}{\sqrt{ab}}\ln\left|\sqrt{ab}x + ay\right| + C_1 = \frac{w^{1 - k}}{1 - k}\)
\(\displaystyle \frac{c}{\sqrt{ab}}\ln\left|\sqrt{ab}x + ay\right| + \Phi(ay^2 - bx^2) = \frac{w^{1 - k}}{1 - k}\)
The solution for \(\displaystyle k = 1\) is:
\(\displaystyle \frac{c}{\sqrt{ab}}\ln\left|\sqrt{ab}x + ay\right| + \Phi(ay^2 - bx^2) = \ln|w|\)
Then, the general solution to the partial differential equation is:
\(\displaystyle \frac{c}{\sqrt{ab}}\ln\left|\sqrt{ab}x + ay\right| + \Phi(ay^2 - bx^2) =\begin{cases}\frac{w^{1 - k}}{1 - k} & \text{if} \ k \neq 1\\\ln|w| & \text{if} \ k = 1\end{cases} \)