JEE MAIN - Mathematics (2015 (Offline) - No. 6)
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 :
$$\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right).$$ Then $$y(e)$$ is equal to :
$$2$$
$$2e$$
$$e$$
$$0$$
Explanation
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 $$x=e$$
$$y\log e = 2e\left( {\log e - 1} \right) + c \Rightarrow y\left( e \right) = c = 2$$
$$I.F. = {e^{\int {{1 \over {x\log x}}dx} }} = {e^{\log \left( {\log x} \right)}} = \log x$$
$$y.\log x = \int {2\,\log xdx + c} $$
$$y\log x = 2\left[ {x\log x - x} \right] + c$$
Put $$x=1,y.0=-2+c$$ $$ \Rightarrow c = 2$$
Put $$x=e$$
$$y\log e = 2e\left( {\log e - 1} \right) + c \Rightarrow y\left( e \right) = c = 2$$
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