Q44 / 100 · BITSAT 2023 (Memory Based Paper 1) · Mathematics
The number of solutions of $\frac{d y}{d x}=\frac{y+1}{x-1}$, when $y(1)=2$ is
- Anone
- Correct answerBone
- Ctwo
- Dinfinite
Solution
Since, $\frac{d y}{d x}=\frac{y+1}{x-1}$
$\Rightarrow \quad \frac{d y}{y+1}=\frac{d x}{x-1}$
After integrating on both sides, we have
$\log (y+1)=\log (x-1)-\log C$
$C(y+1)=(x-1)$
$C=\frac{x-1}{y+1}$
If $x=1$, then $y=2$, so $C=0$
Therefore, $x-1=0$
Hence, there is only one solution.
$\Rightarrow \quad \frac{d y}{y+1}=\frac{d x}{x-1}$
After integrating on both sides, we have
$\log (y+1)=\log (x-1)-\log C$
$C(y+1)=(x-1)$
$C=\frac{x-1}{y+1}$
If $x=1$, then $y=2$, so $C=0$
Therefore, $x-1=0$
Hence, there is only one solution.
Hence, the correct answer is option: 'B'.