Question
If $$\phi \left( x \right)$$ is a differentiable function, then the solution of the differential equation $$dy + \left\{ {y\phi '\left( x \right) - \phi \left( x \right)\phi '\left( x \right)} \right\}dx = 0$$ is :
A.
$$y = \left\{ {\phi \left( x \right) - 1} \right\} + c{e^{ - \phi \left( x \right)}}$$
B.
$$y\phi \left( x \right) = {\left\{ {\phi \left( x \right)} \right\}^2} + c$$
C.
$$y{e^{\phi \left( x \right)}} = \phi \left( x \right){e^{\phi \left( x \right)}} + c$$
D.
none of these
Answer :
$$y = \left\{ {\phi \left( x \right) - 1} \right\} + c{e^{ - \phi \left( x \right)}}$$
Solution :
$$\eqalign{
& {\text{We have, }}dy + \left\{ {y\phi '\left( x \right) - \phi \left( x \right)\phi '\left( x \right)} \right\}dx = 0 \cr
& \Rightarrow \frac{{dy}}{{dx}} + \phi '\left( x \right).y = \phi \left( x \right)\phi '\left( x \right)......\left( {\text{i}} \right) \cr} $$
This is a linear differential equation with $${\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\phi '\left( x \right)} dx}} = {e^{\phi \left( x \right)}}$$
Multiplying Equation $$\left( {\text{i}} \right)$$ by $$\phi \left( x \right)$$ and integrating, we get
$$\eqalign{
& y{e^{\phi \left( x \right)}} = \int {\phi \left( x \right)\phi '\left( x \right){e^{\phi \left( x \right)}}dx} \cr
& \Rightarrow y{e^{\phi \left( x \right)}} = \int {{e^{\phi \left( x \right)}}\phi \left( x \right)\phi '\left( x \right)dx} \cr
& \Rightarrow y{e^{\phi \left( x \right)}} = \int {\phi \left( x \right){e^{\phi \left( x \right)}}\phi '\left( x \right)dx} \cr
& \Rightarrow y{e^{\phi \left( x \right)}} = \phi \left( x \right){e^{\phi \left( x \right)}} - \int {\phi '\left( x \right){e^{\phi \left( x \right)}}dx} \cr
& \Rightarrow y{e^{\phi \left( x \right)}} = \phi \left( x \right){e^{\phi \left( x \right)}} - {e^{\phi \left( x \right)}} + c \cr
& \Rightarrow y = \left\{ {\phi \left( x \right) - 1} \right\} + c{e^{ - \phi \left( x \right)}} \cr} $$