Question

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

• A. $6(x+y)+11\log (3x+3y+10)=9x+c$
• B. $6(x+y)-11\log (3x+3y+10)=9x+c$
• C. 6 $(x+y)-11\log (x+y+3)=3x+c$
• D. 6 $(x+y)-11\log (x+y+3)=x+c$

Answer 1

The correct option is B $6(x+y)-11\log (3x+3y+10)=9x+c$

let $x+y=v$

$\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$

$\Rightarrow \frac{dv}{dx}-1=\frac{v+7}{2v+3}$

$\frac{dv}{dx}=\frac{v+7+2v+3}{2v+3}=\frac{3v+10}{2v+3}$

$\Rightarrow \frac{2v+3}{3v+10}dv=dx$

$\frac{2}{3}\left[\frac{v+\frac{3}{2}}{v+\frac{10}{3}}\right]=\frac{2}{3}\int [\frac{v+\frac{10}{3}}{v+\frac{10}{3}}+\frac{\frac{3}{2}-\frac{10}{3}}{v+\frac{10}{3}}]dv$

$=\frac{2}{3}[v+(\frac{3}{2}-\frac{10}{5})log(v+\frac{10}{3}\left)\right]=x+c$

$=\frac{2}{3}[v-\frac{11}{6}log(3x+3y+10\left)\right]=x+c$

$=\frac{2}{3}\left[\frac{6(x+y)-11log(3x+3y+10)}{6}\right]=x+c$

$\Rightarrow 6(x+y)-11log(3x+3y+10)=9x+3c$