ODE Non linear Non separable

Question

$$ \frac{dy} {dx} = \frac{3x-4y-2}{3x-4y-3}$$

I don't know how to solve this. Searched about the equation to know about the non separable one. I know about the separable one. Any hint will be helpful.

Answer 1

Given $\dfrac{dy}{dx}=\dfrac{3x-4y-2}{3x-4y-3}.....(1)$

Take $t=3x-4y-2$

$$\dfrac{dt}{dx}=3-4\dfrac{dy}{dx}$$

$$\dfrac{dy}{dx}=\dfrac{3-4\dfrac{dy}{dx}}{4}$$

$$3-\dfrac{dt}{dx}=\dfrac{4t}{t-1}$$

$$\dfrac{1-t}{t+3}dt=dx$$

$$\int\dfrac{1-t}{t+3}dt=\int dx$$

Can you take it from here?

Answer 2

With the Substitution $$z=4y-3x$$ we get $$y'=\frac{z'+3}{4}$$ and $z'$ is equal to $$z'=\frac{z-1}{z+3}$$

Answer 3

$$\frac{dy} {dx} = \frac{3x-4y-2}{3x-4y-3}$$ $$\frac{dy}{3x-4y-2} = \frac {dx}{3x-4y-3}$$ $$\frac{d(3x-4y)}{-3x+4y-1} = \frac {d(y-x)}{1}$$ After integration $$ -\ln (3x-4y+1)=(y-x)+C$$ Finally $$\boxed {(y-x)+\ln (3x-4y+1)=K}$$