partial differential equation - 8

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle ay\frac{\partial w}{\partial x} + bx\frac{\partial w}{\partial y} = cw^k\)

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle ay\frac{\partial w}{\partial x} + bx\frac{\partial w}{\partial y} = cw^k\)

Please show us what you have tried and exactly ()where you are stuck.

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[logistic_guy]

Did I say that we have become professionals in solving this type of partial differential equations? But happiness is always has limited time, so very soon we will get stuck as the equations are getting harder and harder!

\(\displaystyle \frac{dx}{ay} = \frac{dy}{bx} = \frac{dw}{cw^k}\)


\(\displaystyle \frac{dx}{ay} = \frac{dy}{bx}\)


\(\displaystyle bx \ dx = ay \ dy\)


\(\displaystyle b\int x \ dx = a\int y \ dy\)


\(\displaystyle b\frac{x^2}{2} + D = a\frac{y^2}{2}\)


\(\displaystyle a\frac{y^2}{2} - b\frac{x^2}{2} = D\)


\(\displaystyle ay^2 - bx^2 = 2D\)


\(\displaystyle ay^2 - bx^2 = C_1\)

Then,

\(\displaystyle \Phi(C_1) = \Phi(ay^2 - bx^2) = C_2\)

 

[logistic_guy]

\(\displaystyle \frac{dx}{ay} = \frac{dy}{bx} = \frac{dw}{cw^k}\)

\(\displaystyle \frac{dx}{ay} = \frac{dw}{cw^k}\)

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:

\(\displaystyle \frac{c}{\sqrt{ab}}\ln\left|\sqrt{ab}x + ay\right| + \Phi(ay^2 - bx^2) =\begin{cases}\frac{w^{1 - k}}{1 - k} & \text{if} \ k \neq 1\\\ln|w| & \text{if} \ k = 1\end{cases} \)