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Question

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


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