Question

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

• A. $lo{g}_e|3x+3y+2|+3x+6y=C$
• 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

The correct option is D

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

Find the general solution of the differential equation

Given, $\frac{dy}{dx}=\frac{x+y+1}{2x+2y+1}$

Let, $x+y=v$

$$\begin{gathered} \Rightarrow \frac{dv}{dx}=\frac{dy}{dx}+1 \\ \Rightarrow \frac{dv}{dx}-1=\frac{dy}{dx} \\ \Rightarrow \frac{dv}{dx}-1=\frac{v+1}{2v+1} \\ \Rightarrow \frac{dv}{dx}=\frac{3v+2}{2v+1} \\ \Rightarrow \frac{3v+2}{2v+1}dv=dx \end{gathered}$$

Integrating both sides, we get

$$\begin{gathered} \Rightarrow \int \left[\frac{2}{3}-\frac{1}{3(3v+2)}\right]dv=\int dx \\ \Rightarrow \frac{2}{3}v-\frac{1}{9}\log \left(3v+2\right)=x+C_1 \\ \Rightarrow \frac{2}{3}\left(x+y\right)-\frac{1}{9}\log \left(3x+3y+2\right)=x+C_1 \\ \Rightarrow 6x+6y-\log \left(3x+3y+2\right)=9x+C_1 \\ \Rightarrow \log \left(3x+3y+2\right)+3x-6y=C \\ \Rightarrow {\log }_e\left|3x+3y+2\right|+3x-6y=C \end{gathered}$$

Hence, option (D) is the correct answer.