Q49 / 125 · BITSAT 2015 · Mathematics
If $\phi(\mathrm{x})$ is a differential function, then the solution of the differential equation $\mathrm{dy}+\left\{\mathrm{y} \phi^{\prime}(\mathrm{x})-\phi(\mathrm{x})\right.$
$\left.\phi^{\prime}(x)\right\} d x=0,$ is
- Correct answerA$y=\{\phi(x)-1\}+C e^{-\phi(x)}$
- B$y \phi(x)=\{\phi(x)\}^{2}+C$
- C$y e^{\varphi(x)}=\phi(x) e^{\phi(x)}+C$
- D$y-\phi(x)=\phi(x) e^{-\phi(x)}$
Solution
Given differential equation is
$\mathrm{dy}+\left\{y \phi^{\prime}(x)-\phi(x) \phi^{\prime}(x)\right\} d x=0$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\phi^{\prime}(\mathrm{x}) \mathrm{y}=\phi(\mathrm{x}) \phi^{\prime}(\mathrm{x})$
which is a linear differential equation with $\quad P=\phi^{\prime}(x), Q=\phi(x) \cdot \phi^{\prime}(x)$ and
$I \cdot F=e^{\int \phi^{\prime}(x) d x}=e^{\phi(x)}$
$\therefore$ Solution is $y \cdot e^{\phi(x)}=\int \phi(x) \cdot \phi^{\prime}(x) e^{\phi(x)} d x+C$
$\Rightarrow y \cdot e^{\phi(x)}=\int \phi(x) \cdot e^{\phi(x)} \phi^{\prime}(x) d x+C$
$\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-\int \phi^{\prime}(x) e^{\phi(x)} d x+C$
$\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-e^{\phi(x)}+C$
$\Rightarrow y=\lceil\phi(x)-1]+C e^{-\phi(x)}$
Hence, the correct answer is option: 'A'.