JEE MAIN - Mathematics (2008 - No. 7)
The solution of the differential equation
$${{dy} \over {dx}} = {{x + y} \over x}$$ satisfying the condition $$y(1)=1$$ is :
$${{dy} \over {dx}} = {{x + y} \over x}$$ satisfying the condition $$y(1)=1$$ is :
$$y = \ln x + x$$
$$y = x\ln x + {x^2}$$
$$y = x{e^{\left( {x - 1} \right)}}\,$$
$$y = x\,\ln x + x$$
Explanation
$${{dy} \over {dx}} = {{x + y} \over x} = 1 + {y \over x}$$
Putting $$y=$$ $$vx$$
and $${{dy} \over {dx}} = v + x{{dv} \over {dx}}$$
we get
$$v + x{{dv} \over {dx}} = 1 + v$$
$$ \Rightarrow \int {{{dx} \over x}} = \int {dv} $$
$$ \Rightarrow v = \ln {\mkern 1mu} x + c$$
$$ \Rightarrow y = x\ln x + cx$$
As $$\,\,\,\,y\left( 1 \right) = 1$$
$$\therefore$$ $$c=1$$
So solution is $$y=xlnx+x$$
Putting $$y=$$ $$vx$$
and $${{dy} \over {dx}} = v + x{{dv} \over {dx}}$$
we get
$$v + x{{dv} \over {dx}} = 1 + v$$
$$ \Rightarrow \int {{{dx} \over x}} = \int {dv} $$
$$ \Rightarrow v = \ln {\mkern 1mu} x + c$$
$$ \Rightarrow y = x\ln x + cx$$
As $$\,\,\,\,y\left( 1 \right) = 1$$
$$\therefore$$ $$c=1$$
So solution is $$y=xlnx+x$$
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