Answer to Question #190507 in Differential Equations for randz

Question

4. (9x − 4y + 4)dx − (2x − y +) = 0

Answer 1

Given equation is-

$(9x-4y+4)dx-(2x-y+1)dy=0$

Let $x=u\Rightarrow dx=du$

$y=v+1\Rightarrow dy=dv$

$(9u-4v)du-(2u-v)dv=0$

Let $u=vz \Rightarrow du=vdz$

$\Rightarrow (9vz-4v)(zdv+vdz)-(2vz-v)dv=0\\[9pt] \Rightarrow (9z-4)(z dv+vdz)-(2z-1)dz=0 \\[9pt] \Rightarrow(9z^2-6z+1)dv+v(9z-4)dz=0 \\[9pt] \Rightarrow \dfrac{dv}{v}+\dfrac{9z-4}{(3z-1)^2}dz=0 \\[9pt] \Rightarrow \dfrac{dv}{v}+\dfrac{3dz}{3z-1}-\dfrac{dz}{(3z-1)^2}=0$

Integrating Both the sides-

$lnv+ln|3z-1|+\dfrac{1}{9z-3}+lnc=0\\[9pt] \Rightarrow 1=-3(3z-1)ln|c(3zv-v)|$

$v=-3(3vz-v)ln|c(3zv-v)|$

Putting $z=\dfrac{u}{v} , u=x \text{ and }v=y-1$

so we get-

$y=1+3(y-3x-1)ln|c(3x-y+1)|$