Question

The solution of the differential equation $(2x-y+2)dx+(4x-2y-1)dy=0$ is
(where $k$ is integration constant)

• A. $(2x-y)=k{e}^{|x-2y|}$
• B. $|2x+y|=k{e}^{-(2x-y)}$
• C. $|x+2y|=k{e}^{(2x-y)}$
• D. $|2x-y|=k{e}^{-(x+2y)}$

Answer 1

The correct option is D $|2x-y|=k{e}^{-(x+2y)}$

$(2x-y+2)dx+(4x-2y-1)dy=0\Rightarrow \frac{dy}{dx}=-\left(\frac{2x-y+2}{4x-2y-1}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{-(2x-y)-2}{2(2x-y)-1}\cdots \left(i\right)$

Put $2x-y=v$ so that $2-\frac{dy}{dx}=\frac{dv}{dx}$

Then, equation $\left(i\right)$ becomes: $2-\frac{dv}{dx}=\frac{-v-2}{2v-1}$
$\Rightarrow \frac{dv}{dx}=2-\left(\frac{-v-2}{2v-1}\right)\Rightarrow \frac{dv}{dx}=\frac{5v}{2v-1}$
$\Rightarrow \int \frac{2v-1}{v}dv=5\int dx\Rightarrow 2\int dv-\int \frac{dv}{v}=5\int dx$
$\Rightarrow 2v-\ln |v|=5x+\ln |c|\Rightarrow 2v-5x=\ln |vc|$
$\Rightarrow 2(2x-y)-5x=\ln |(2x-y)c|$
$\Rightarrow -(x+2y)=\ln |c(2x-y)|$
$\Rightarrow |2x-y|=k{e}^{-(x+2y)}$
where $k=\frac{1}{|c|}=$constant.