Question
Let $$y\left( x \right)$$ be the solution of the differential equation $$\left( {x\,\log \,x} \right)\frac{{dy}}{{dx}} + y = 2x\,\log \,x,\,\left( {x \geqslant 1} \right).$$ Then $$y\left( e \right)$$ is equal to:
A.
$$2$$
B.
$$2e$$
C.
$$e$$
D.
$$0$$
Answer :
$$2$$
Solution :
Given, $$\frac{{dy}}{{dx}} + \left( {\frac{1}{{x\,\log \,x}}} \right)y = 2$$
$$\eqalign{
& {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{1}{{x\,\log \,x}}\,dx} }} = {e^{\log \left( {\log \,x} \right)}} = \log \,x \cr
& y.\log \,x = \int {2\,\log \,x\,dx + c} \cr
& y.\log \,x = 2\left[ {x\,\log \,x - x} \right] + c \cr
& {\text{Put }}\,x = 1,\,\,y.0 = - 2 + c\,\, \Rightarrow c = 2 \cr
& {\text{Put }}x = e \cr
& y\,\log \,e = 2e\left( {\log \,e - 1} \right) + c\,\,\, \Rightarrow y\left( e \right) = c = 2 \cr} $$