\(\displaystyle \frac{dx}{1} = \frac{dy}{a} = \frac{dw}{bx^nw + cx^mw^k}\)
\(\displaystyle \frac{dx}{1} = \frac{dw}{bx^nw + cx^mw^k}\)
\(\displaystyle \frac{dw}{dx} = bx^nw + cx^mw^k\)
Let \(\displaystyle v = w^{1-k}\)
Then,
\(\displaystyle \frac{dv}{dx} = \frac{dw^{1-k}}{dx} = \frac{dw^{1-k}}{dw}\frac{dw}{dx} = (1 - k)w^{-k}(bx^nw + cx^mw^k)\)
Simplify.
\(\displaystyle \frac{dv}{dx} = b(1 - k)x^nw^{1-k} + c(1 - k)x^m = b(1 - k)x^nv + c(1 - k)x^m\)
Or
\(\displaystyle \frac{dv}{dx} - b(1 - k)x^nv = c(1 - k)x^m\)
This is just an ordinary differential equation and can be solved by integrating factor.
Let \(\displaystyle F(x) = \text{exp}\left(\int b(1 - k)x^n \ dx\right) = \text{exp}\bigg(\frac{b(1 - k)x^{n+1}}{n + 1}\bigg)\) be the integrating factor.
Then,
\(\displaystyle \frac{1}{F(x)}\frac{dv}{dx} - \frac{b(1 - k)x^nv}{F(x)} = \frac{c(1 - k)x^m}{F(x)}\)
\(\displaystyle \frac{d}{dx}\left(\frac{v}{F(x)}\right) = \frac{c(1 - k)x^m}{F(x)}\)
\(\displaystyle \int \frac{d}{dx}\left(\frac{v}{F(x)}\right) \ dx = \int \frac{c(1 - k)x^m}{F(x)} \ dx\)
\(\displaystyle \frac{v}{F(x)} + D_1 = \int \frac{c(1 - k)x^m}{F(x)} \ dx\)
\(\displaystyle v = -D_1F(x) + F(x)\int \frac{c(1 - k)x^m}{F(x)} \ dx\)
\(\displaystyle v = C_2F(x) + F(x)\int \frac{c(1 - k)x^m}{F(x)} \ dx\)
\(\displaystyle v = \Phi(y - ax)F(x) + c(1 - k)F(x)\int \frac{x^m}{F(x)} \ dx\)
We know \(\displaystyle v = w^{1-k}\)
Then,
\(\displaystyle w = v^{\frac{1}{1-k}}\)
And the solution to the differential equation is:
\(\displaystyle w(x,y) = \left(\Phi(y - ax)F(x) + c(1 - k)F(x)\int \frac{x^m}{F(x)} \ dx\right)^{\frac{1}{1-k}}\)
where \(\displaystyle F(x) = \text{exp}\bigg(\frac{b(1 - k)x^{n+1}}{n + 1}\bigg)\)
