Q119 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
The general solution of the differential equation $(9 x-3 y+5) d y=(3 x-y+1) d x$ is
- A$x-3 y-\log |12 x-4 y+7|=c$
- Correct answerB$4 x-12 y-\log |12 x-4 y+7|=c$
- C$4 x-12 y+\log |6 x-2 y+7|=c$
- D$2 x-6 y+\log |12 x-4 y+7|=c$
Solution
$\frac{d y}{d x}=\frac{(3 x-y+1)}{(9 x-3 y+5)}$
Let $3 x-y=v$
$\begin{aligned}
& \Rightarrow \frac{d y}{d x}=3-\frac{d v}{d x} \Rightarrow 3-\frac{d v}{d x}=\frac{v+1}{3 v+5} \\
& \Rightarrow \frac{d v}{d x}=\frac{9 v+15-v-1}{3 v+5} \Rightarrow \frac{d v}{d x}=\frac{8 v+14}{3 v+5} \\
& \Rightarrow \frac{3 v+5}{8 v+14} d v=d x \Rightarrow \frac{3}{8} \int \frac{24 v+40}{24 v+42} d v=d x \\
& \Rightarrow \frac{3}{8}\left[\left(1-\frac{2}{24 v+42}\right) d v\right]=d x \\
& \Rightarrow \frac{3}{8}\left[v-\frac{1}{12} \ln (24 v+42)\right]=x+C \\
& \Rightarrow \frac{3}{8}\left[3 x-y-\frac{1}{12} \ln (72 x-24 y+42)\right]=x+C \\
& \Rightarrow 36 x-12 y-\ln (72 x-24 y+42)=32 x+C \\
& \Rightarrow 4 x-12 y-\ln |12 x-4 y+7|=C .
\end{aligned}$
Hence, the correct answer is option: 'B'.