Question

# The solution of the differential equation $$x \dfrac {dy}{dx} = \dfrac {y}{1 + \log x }$$ is :

A
y=logx+C
B
y=C1+logx
C
y=C(x+logx)
D
y=x+log(Cx)
E
y=C(1+logx)

Solution

## The correct option is D $$y = C ( 1 + \log x)$$$$x\dfrac { dy }{ dx } =\dfrac { y }{ 1+\log { x } } \\ \dfrac { dy }{ y } =\dfrac { dx }{ x(1+\log { x } ) }$$ Put $$t=\log x$$ $$\displaystyle{\dfrac { dt }{ dx } =\dfrac { 1 }{ x } \\ \Rightarrow dt=\dfrac { dx }{ x } \\ \Rightarrow \dfrac { dy }{ y } =\dfrac { dt }{ 1+t\\ } \\ \int { \dfrac { dy }{ y } } =\int { \dfrac { dt }{ 1+t\\ } } \\ \log { y } =\log { (1+t) } +\log { C } \\ \log { y } =\log { C(1+t) } \\ \Rightarrow y=C(1+t)}$$ Substituting $$t$$ $$\Rightarrow y=C(1+\log { x) }$$ So option $$E$$ is correct.Mathematics

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