• question_answer If $\phi (x)$ is a differentiable function, then the solution of the differential equation$dy+\{y\phi '(x)-\phi (x)\phi '(x)\}dx=0$ is A) $y=\{\phi (x)-1\}+c{{e}^{-\phi (x)}}$ B) $y\phi (x)={{\{\phi (x)\}}^{2}}+c$ C) $y{{e}^{\phi (x)}}=\phi (x){{e}^{\phi (x)}}+c$ D) None of these

 [a] We have, $dy+\{y\phi '(x)-\phi (x)\phi '(x)\}dx=0$ $\Rightarrow \frac{dy}{dx}+\phi '(x)\cdot y=\phi (x)\phi '(x)$                     ? (i) This is linear differential equation with I.F $={{e}^{\int{\phi '(x)dx}}}={{e}^{\phi (x)}}$ Multiplying Eq. (i) by $\phi (x)$ and integrating, we get $y{{e}^{\phi (x)}}=\int{\phi (x)\phi '(x){{e}^{\phi (x)}}}dx$ $\Rightarrow \,\,\,y{{e}^{\phi (x)}}=\int{{{e}^{\phi (x)}}\phi (x)\phi '(x)}dx$ $\Rightarrow y{{e}^{\phi (x)}}=\int{\phi (x){{e}^{\phi (x)}}}\phi '(x)dx$ $\Rightarrow y{{e}^{\phi (x)}}=\phi (x){{e}^{\phi (x)}}-\int{\phi '(x){{e}^{\phi (x)}}}dx$ $\Rightarrow y{{e}^{\phi (x)}}=\phi (x){{e}^{\phi (x)}}-{{e}^{\phi (x)}}+c$ $\Rightarrow y=\{\phi (x)-1\}+c{{e}^{-\phi (x)}}$