Let y(x) be the solution of the differential equation
\[\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right).\] Then y(e) is equal to :
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Official solution
Given, \[{{dy} \over {dx}} + \left( {{1 \over {x\,\log \,x}}} \right)y = 2\]
\[I.F. = {e^{\int {{1 \over {x\log x}}dx} }} = {e^{\log \left( {\log x} \right)}} = \log x\]
\[y.\log x = \int {2\,\log xdx + c} \]
\[y\log x = 2\left[ {x\log x - x} \right] + c\]
Put x=1,y.0=−2+c \[ \Rightarrow c = 2\]
Put x=e
\[y\log e = 2e\left( {\log e - 1} \right) + c \Rightarrow y\left( e \right) = c = 2\]