JEE MAIN - Mathematics (2005 - No. 27)
If $$x{{dy} \over {dx}} = y\left( {\log y - \log x + 1} \right),$$ then the solution of the equation is :
$$y\log \left( {{x \over y}} \right) = cx$$
$$x\log \left( {{y \over x}} \right) = cy$$
$$\log \left( {{y \over x}} \right) = cx$$
$$\log \left( {{x \over y}} \right) = cy$$
Explanation
$${{xdy} \over {dx}} = y\left( {\log y - \log x + 1} \right)$$
$${{dy} \over {dx}} = {y \over x}\left( {\log \left( {{y \over x}} \right) + 1} \right)$$
Put $$\,\,\,\,y = vx$$
$${{dy} \over {dx}} = v + {{xdv} \over {dx}}$$
$$\,\,\,\,\,\,\,\,\,\,\, \Rightarrow v + {{xdv} \over {dx}} = v\left( {\log v + 1} \right)$$
$${{xdv} \over {dx}} = v\,\log \,v$$
$$\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{dv} \over {v\,\log \,v}} = {{dx} \over x}$$
Put $$\,\,\,\,\log \,v = z$$
$$ \Rightarrow {1 \over v}dv = dz$$
$$ \Rightarrow {{dz} \over x} = {{dx} \over x}$$
$$ \Rightarrow \ln \,z = \ln x + \ln \,c$$
$$x = cx\,\,\,\,$$ or $$\,\,\,\,\,\log v = cx\,\,\,$$
or $$\,\,\,\,$$ $$\log \left( {{y \over x}} \right) = cx.$$
$${{dy} \over {dx}} = {y \over x}\left( {\log \left( {{y \over x}} \right) + 1} \right)$$
Put $$\,\,\,\,y = vx$$
$${{dy} \over {dx}} = v + {{xdv} \over {dx}}$$
$$\,\,\,\,\,\,\,\,\,\,\, \Rightarrow v + {{xdv} \over {dx}} = v\left( {\log v + 1} \right)$$
$${{xdv} \over {dx}} = v\,\log \,v$$
$$\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{dv} \over {v\,\log \,v}} = {{dx} \over x}$$
Put $$\,\,\,\,\log \,v = z$$
$$ \Rightarrow {1 \over v}dv = dz$$
$$ \Rightarrow {{dz} \over x} = {{dx} \over x}$$
$$ \Rightarrow \ln \,z = \ln x + \ln \,c$$
$$x = cx\,\,\,\,$$ or $$\,\,\,\,\,\log v = cx\,\,\,$$
or $$\,\,\,\,$$ $$\log \left( {{y \over x}} \right) = cx.$$
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