Question

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

• A. $\log (2x+y-1)=C+x+y$
• B. $\log (4x+2y-1)=C+2x+y$
• C. $\log (2x+y+1)+x+2y=C$
• D. $\log (2x+y-1)+x+2y=C$

Answer 1

The correct option is D $\log (2x+y-1)+x+2y=C$

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

Let $t=2x+y$

$\Rightarrow \frac{dt}{dx}=2+\frac{dy}{dx}$

substituting in above equations;

$\Rightarrow 2-\frac{dt}{dx}=\frac{t+1}{2t-1}$

$\Rightarrow \frac{dt}{dx}=\frac{3t-3}{2t-1}$

$\Rightarrow (2t-1)dt=3(t-1)dx$

$\Rightarrow 2dt+\frac{dt}{t-1}=3dx$

Integrate on both sides;

$\Rightarrow 2t+log(t-1)=3x+C$

$\Rightarrow 2(2x+y)+log(2x+y-1)=3x+C$

$\Rightarrow log(2x+y-1)+x+2y=C$