partial differential equation - 5

[logistic_guy]

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

\(\displaystyle \frac{\partial w}{\partial x} + a\frac{\partial w}{\partial y} = bx^ny^mw^k\)

 

[logistic_guy]

\(\displaystyle \frac{dx}{1} = \frac{dy}{a} = \frac{dw}{bx^ny^mw^k}\)


\(\displaystyle \int \frac{dx}{1} = \int \frac{dy}{a}\)


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


\(\displaystyle ax + C_1 = y\)


\(\displaystyle y - ax = C_1\)

Then,

\(\displaystyle \Phi(C_1) = \Phi(y - ax) = C_2\)

 

[logistic_guy]

\(\displaystyle \frac{dx}{1} = \frac{dy}{a} = \frac{dw}{bx^ny^mw^k}\)

\(\displaystyle \frac{dx}{1} = \frac{dw}{bx^ny^mw^k}\)


\(\displaystyle bx^ny^m \ dx = \frac{dw}{w^k}\)


\(\displaystyle \int bx^ny^m \ dx = \int \frac{dw}{w^k}\)

It seems that we are in trouble as we have two variables on the left side. But from the last post we know that:

\(\displaystyle y = ax + C_1\)

Then,

\(\displaystyle \int bx^n(ax + C_1)^m \ dx = \int \frac{dw}{w^k}\)


\(\displaystyle \int bx^n(ax + C_1)^m \ dx = \frac{w^{1 - k}}{1 - k} + D\)


\(\displaystyle w^{1 - k} = (1 - k)\int bx^n(ax + C_1)^m \ dx - (1 - k)D\)


\(\displaystyle w^{1 - k} = (1 - k)\int bx^n(ax + C_1)^m \ dx + (1 - k)C_2\)


\(\displaystyle w^{1 - k} = (1 - k)\int bx^n(ax + C_1)^m \ dx + (1 - k)\Phi(y - ax)\)

Then, the general solution to the partial differential equation is:

\(\displaystyle w(x,y) = \left(b(1 - k)\int x^n(ax + C_1)^m \ dx + (1 - k)\Phi(y - ax)\right)^{\frac{1}{1 - k}}\)

where \(\displaystyle C_1 = y - ax\)