The solution of the differential equation ${{dy} \over {dx}} = {{x + y} \over x}$ satisfying the condition $y(1)=1$ is :

Question

The solution of the differential equation

${{dy} \over {dx}} = {{x + y} \over x}$ satisfying the condition $y(1)=1$ is :

$y = \ln x + x$

$y = x\ln x + {x^2}$

$y = x{e^{\left( {x - 1} \right)}}\,$

$y = x\,\ln x + x$

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