Question

Solve: $\frac{dy}{dx}=\frac{2x-y+2}{2y-4x+1}$

• A. $x+2y+\log |2x-y|+c=0$
• B. $x+2y+\log |2x+y|+c=0$
• C. $x-2y+\log |2x+y|+c=0$
• D. $2x-y+\log |2x-y|+c=0$

Answer 1

The correct option is A $x+2y+\log |2x-y|+c=0$

$\frac{dy}{dx}=\frac{2x-y+2}{2y-4x+1}$
Substituting $v=2y-4x\Rightarrow \frac{dv}{dx}=2\frac{dy}{dx}-4$
$\frac{1}{2}(\frac{dv}{dx}+4)=\frac{2x+\frac{1}{2}(-4x-v)+2}{v+1}=-\frac{v-4}{2(v+1)}\Rightarrow \frac{dv}{dx}=-\frac{5v}{v+1}\Rightarrow \frac{\frac{dv}{dx}(v+1)}{v}=-5$
Integrating both sides w.r.t x, we get
$\int \frac{\frac{dv}{dx}(v+1)}{v}=\int -5dx\Rightarrow \log v+v=-5x+c\Rightarrow \log |2y-4x|+2y-4x=-5x+c\Rightarrow \log |y-2x|+2y+x+\log 2=cx+2y+\log |2x-y|=c$