Question
If $$x\frac{{dy}}{{dx}} = y\left( {\log \,y - \log \,x + 1} \right),$$ then the solution of the equation is-
A.
$$y\,\log \left( {\frac{x}{y}} \right) = cx$$
B.
$$x\,\log \left( {\frac{y}{x}} \right) = cy$$
C.
$$\log \left( {\frac{y}{x}} \right) = cx$$
D.
$$\log \left( {\frac{x}{y}} \right) = cy$$
Answer :
$$\log \left( {\frac{y}{x}} \right) = cx$$
Solution :
$$\eqalign{
& \frac{{xdy}}{{dx}} = y\left( {\log \,y - \log \,x + 1} \right) \cr
& \frac{{dy}}{{dx}} = \frac{y}{x}\left( {\log \left( {\frac{y}{x}} \right) + 1} \right) \cr
& {\text{Put }}y = vx \cr
& \frac{{dy}}{{dx}} = v + \frac{{xdv}}{{dx}} \Rightarrow v + \frac{{xdv}}{{dx}} = v\left( {\log \,v + 1} \right) \cr
& \frac{{xdv}}{{dx}} = v\,\log \,v \Rightarrow \frac{{dv}}{{v\,\log \,v}} = \frac{{dx}}{x} \cr
& {\text{Put }}\log \,v = z \Rightarrow \frac{1}{v}dv = dz \Rightarrow \frac{{dz}}{z} = \frac{{dx}}{x} \cr
& \Rightarrow \ln \,z = \ln \,x + \ln \,c \cr
& x = cx\,\,\,\,{\text{or, }}\log \,v = cx\,\,\,{\text{or, }}\log \left( {\frac{y}{x}} \right) = cx \cr} $$