Question

(x-4y-9)dx+(4x+y-2)dy=0

Answer 1

$(x-4y-9)dx+(4x+y-2)dy=0$

Find the solution of system equation

$\left\{\begin{matrix} x-4y=9 \\ 4x+y=2 \end{matrix}\right.\\ \left\{\begin{matrix} x=1 \\ y=-2 \end{matrix}\right.$

Make a replacement

$\left\{\begin{matrix} x=t+1 \\ y=z(t)-2 \end{matrix}\right.$

and input in equation

$z'=-\frac{t+1-4(z-2)-9}{4(t+1)+z-2-2}\\ z'=-\frac{t-4z}{4t+z}$

Make a replacement

$z=t\cdot u(t)\\ tu'+u=-\frac{t-4tu}{4t+tu}\\ tu'=\frac{4u-1}{u+4}-u\\ tu'=\frac{-u^2-1}{u+4}\\ \frac{u+4}{u^2+1}du=-\frac{dt}{t}\\ \int{\frac{u}{u^2+1}+\frac{4}{u^2+1}}du=-\int\frac{dt}{t}\\ \frac{1}{2}\ln|u^2+1|+4\arctan u=-\ln|t|+\ln|C|\\ \sqrt{u^2+1}\cdot e^{4\arctan u}=\frac{C}{t}$

Then general solution is

$\sqrt{(\frac{y+2}{x-1})^2+1}\cdot e^{4\arctan \frac{y+2}{x-1}}=\frac{C}{x-1}$