Question

The solution of differential equation

$\frac{dy}{dx}=\frac{3x-6y+7}{x-2y+4}$ is
(where $c$ is constant of integration and $\log$ is given with base ${}^{\prime }{e}^{\prime }$)

• A. $3x-2y+\log |x-y+2|=c$
• B. $3x+\log |x-2y+1|=c$
• C. $3x-y+\log |x-2y+2|=c$
• D. $2y+\log |x-2y|=c$

Answer 1

The correct option is C $3x-y+\log |x-2y+2|=c$

Given:
$\frac{dy}{dx}=\frac{3x-6y+7}{x-2y+4}=\frac{3(x-2y)+7}{x-2y+4}$
Put $x-2y=v\Rightarrow 1-2\frac{dy}{dx}=\frac{dv}{dx}$
$\text{Now differential equations reduces to}$

$1-\frac{dv}{dx}=2\left(\frac{3v+7}{v+4}\right)$
$\Rightarrow \frac{dv}{dx}=-5\left(\frac{v+2}{v+4}\right)\Rightarrow \int (1+\frac{2}{v+2})dv=-5\int dx$
$\Rightarrow v+2\log |v+2|=-5x+k$
On substituting back the initial variables and simplifying, we have:

$3x-y+\log |x-2y+2|=c$