Question
Solution of the differential equation $$\frac{{dx}}{{dy}} - \frac{{x\,\log x}}{{1 + \log \,x}} = \frac{{{e^y}}}{{1 + \log \,x}},{\text{ if }}y\left( 1 \right) = 0,{\text{ is :}}$$
A.
$${x^x} = {e^{y{e^y}}}$$
B.
$${e^y} = {x^{{e^y}}}$$
C.
$${x^x} = y{e^y}$$
D.
none of these
Answer :
$${x^x} = {e^{y{e^y}}}$$
Solution :
$$\eqalign{
& \left( {1 + \log \,x} \right)\frac{{dx}}{{dy}} - x\,\log \,x = {e^y} \cr
& {\text{putting }}x\,\log \,x = t \Rightarrow \left( {1 + \log \,x} \right)dx = dt \cr
& \therefore \,\frac{{dt}}{{dy}} - t = {e^y}\,\,\,\,\,{\text{Now, I}}{\text{.F}}{\text{.}} = {e^{\int { - 1\,dy} }} = {e^{ - y}} \cr
& \Rightarrow t{e^{ - y}} = \int {{e^{ - y}}{e^y}dy + C} \cr
& \Rightarrow t = C{e^y} + y{e^y} \cr
& \Rightarrow x\,\log \,x = \left( {C + y} \right){e^y}, \cr
& {\text{Since, }}y\left( 1 \right) = 0,\,{\text{then}} \cr
& 0 = \left( {C + 0} \right)1 \Rightarrow C = 0 \cr
& \therefore \,y{e^y} = x\,\log \,x \Rightarrow {x^x} = {e^{y{e^y}}} \cr} $$