Question

The solution of the differential equation $\frac{dy}{dx}=\frac{x+y+7}{2x+2y+3}$ is

• A. $6(x+y)-11log\text{ (}3x+3y+10)=9x+c$
• B. $6(x+y)+11log\text{ (}3x+3y+10)=9x+c$
• C. $6(x+y)-\log \left(x+y+\frac{10}{3}\right)=9x+c$
• D. $6(x+y)\left(x+y+\frac{10}{3}\right)=9x+c$

Answer 1

Answer: (A)

Given, $\frac{dy}{dx}=\frac{x+y+7}{2x+2y+3}$ ...(i)
Let $x+y=v$ $\Rightarrow$ $\frac{dy}{dx}+1=\frac{dv}{dx}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{dv}{dx}-1$
On putting this value in Eq. (i), we get $\frac{dv}{dx}-1=\frac{v+7}{2v+3}$
$\Rightarrow$ $\frac{dv}{dx}=\frac{v+7}{2v+3}$ $\Rightarrow$ $\frac{dv}{dx}=\frac{v+7+2v+3}{2v+3}$
$\Rightarrow$ $\frac{dv}{dx}=\frac{3v+10}{2v+3}$
$\Rightarrow$ $\frac{2v+3}{3v+10}dv=dx$
$\Rightarrow$ $\left(\frac{2}{3}-\frac{11}{3(3v+10)}dv\right)=dx$
$\Rightarrow$ $\frac{2}{3}dv-\frac{11}{3(3v+10)}dv=dx$
On integrating, we get
$\frac{2}{3}v=-\frac{11}{3}\log (3v+10).\frac{1}{3}+c_1=x$
$\Rightarrow$ $6v-11\log (3v+10)9c_1=9x$
$\Rightarrow$ $6(x+y)-11\log \{3(x+y)+10\}=9x-9c_1$
$\Rightarrow$ $6(x+y)-11\log (3x+3y+10)=9x-c$
where $c=-9c_1$