Question

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

• A. $ 6(x+y)-11\text{ }log\text{ (}3x+3y+10)=9x+c $
• B. $ 6(x+y)+11\text{ }log\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)9{{c}_{1}}=9x $
$ \Rightarrow $ $ 6(x+y)-11\log \{3(x+y)+10\}=9x-9{{c}_{1}} $
$ \Rightarrow $ $ 6(x+y)-11\log (3x+3y+10)=9x-c $
where $ c=-9{{c}_{1}} $