Question

The general solution of the differential equation $\frac{dy}{dx}=\frac{x+y+1}{2x+2y+1}$ is:

• A. ${\log }_e|3x+3y+2|+3x+6y=C$
• B. ${\log }_e|3x+3y+2|-3x+6y=C$
• C. ${\log }_e|3x+3y+2|-3x-6y=C$
• D. ${\log }_e|3x+3y+2|+3x-6y=C$

Answer 1

Answer: (D)

${\log }_e|3x+3y+2|+3x-6y=C$

Given differential equation,
$\frac{dy}{dx}=\frac{x+y+1}{2x+2y+1}$ .....(i)

Put $x+y=v$
$\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$$\therefore$ Eq. (i) becomes,

$\frac{dy}{dx}=\frac{v+1}{2v+1}$

$\Rightarrow \frac{dv}{dx}-1=\frac{v+1}{2v+1}\Rightarrow \frac{dv}{dx}=\frac{v+1}{2v+1}+1$

$\Rightarrow \frac{dv}{dx}=\frac{v+1+2v+1}{2v+1}\Rightarrow \frac{dv}{dx}=\frac{3v+2}{2v+1}$

$\Rightarrow \frac{2v+1}{3v+2}dv=dx\Rightarrow \frac{\left[\frac{2}{3}\right(3v+2)-\frac{1}{3}]}{3v+2}dv=dx$

$\Rightarrow [\frac{2}{3}-\frac{1}{3}\times \frac{1}{(3v+2)}]dv=dx$

Integrating both sides, we get

$\Rightarrow \int (\frac{2}{3}-\frac{1}{3}\times \frac{1}{(3v+2)})dv=\int dx+C^{\prime }$,

Where $C^{\prime }$ is a constant of integration.

$\Rightarrow \frac{2}{3}v-\frac{1}{3}\frac{\log (3v+2)}{3}=x+C^{\prime }$

$\Rightarrow \frac{2}{3}(x+y)-\frac{1}{9}\log (3x+3y+2)=x+C^{\prime }$

$\Rightarrow 6x+6y-\log (3x+3y+2)=9x+C^{\prime }$

$\Rightarrow 3x-6y+\log (3x+3y+2)=C$