Question

Solution of $\frac{dy}{dx}=\frac{2x-6y+7}{x-3y+4}$ is

• A. $2x-y+\frac{1}{5}\log (5x-15y+17)=C$
• B. $2x-6y+\frac{1}{5}\log (5x-15y+17)=C$
• C. $2x+y+\frac{1}{5}\log (5x-15y+17)=C$
• D. none of these.

Answer 1

Answer: (A)

Put $x-3y=z$
$\therefore 1-3\frac{dy}{dx}=\frac{dz}{dx}$
$\therefore \frac{1}{3}\left[1-\frac{dz}{dx}\right]=\frac{2z+7}{z+4}$
$\Rightarrow 1-\frac{dz}{dx}=\frac{6z+21}{z+4}$
$\Rightarrow \frac{dz}{dx}=1-\frac{6z+21}{z+4}$
$=\frac{z+4-6z-21}{z+4}=\frac{-5z-17}{z+4}$
$\therefore \frac{z+4}{5z+17}dz+dx=0$
$\Rightarrow \frac{5z+20}{5z+17}dz+5dx=0$
$\Rightarrow \left(1+\frac{3}{5z+17}\right)dz+5dx=0$
$\Rightarrow z+\frac{3}{5}log\left(5z+17\right)+5x=c$
$\Rightarrow x-3y+\frac{3}{5}log\left(5x-15y+117\right)+5x=C_1$
$\Rightarrow 6x-3y+\frac{3}{5}log\left(5x-15y+17\right)=C_1$
$\Rightarrow 2x-y+\frac{1}{5}log\left(5x-15y+17\right)=C$