partial differential equation - 6

[logistic_guy]

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

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

 

[khansaheb]

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

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

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Please share your work/thoughts about this problem

 

[logistic_guy]

We start as usual.

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


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


\(\displaystyle b \ dx = a \ dy\)


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


\(\displaystyle bx = ay + C_1\)


\(\displaystyle bx - ay = C_1\)

Then,

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

 

[logistic_guy]

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

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


\(\displaystyle \frac{c}{a}\int \frac{dx}{x} = \int \frac{dw}{w^k}\)


\(\displaystyle \frac{c}{a}\ln|x| + C_2 = \frac{w^{1 - k}}{1 - k}\)


\(\displaystyle \frac{c}{a}\ln|x| + \Phi(bx - ay) = \frac{w^{1 - k}}{1 - k}\)

It seems that this solution is not valid when \(\displaystyle k = 1\). Therefore, there is a second solution when \(\displaystyle k = 1\).

We start from here.

\(\displaystyle \frac{c}{a}\int \frac{dx}{x} = \int \frac{dw}{w}\)


\(\displaystyle \frac{c}{a}\ln|x| + \Phi(bx - ay) = \ln|w|\)

Then, the combined general solution is:

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