NIMCET 2017 Question Paper with solutions

Question : 113 of 120

Marks: +1, -0

The solution of

(e^x + 1)

y dy = (y +1)

e^x

dx is

Solution:

Concept:

\log x=y ⇒ x=e^{y}

Calculation:
Given:

(e^{x}+1) y d y=(y+1) e^{x} d x

⇒ \;{ {ydy}}/{( {y}+1)}=\;{ {e}^{ {x}}}/{( {e}^{ {x}}+1)} {dx}

⇒ \;{( {y}+1-1) {dy}}/{( {y}+1)}=\;{( {e}^{ {x}})}/{( {e}^{ {x}}+1)} {dx}

⇒ {dy}-\;{ {dy}}/{( {y}+1)}=\;{ {e}^{ {x}}}/{( {e}^{ {x}}+1)} {dx}

Integrating both sides, we get

⇒ {y}-\log | {y}+1|=\log ( {e}^{ {x}}+1)+\log {c}

⇒ {y}=\log |( {y}+1)( {e}^{ {x}}+1)|+\log {c}

⇒ {c}( {y}+1)( {e}^{ {x}}+1)= {e}^{ {y}}

∴ {e}^{ {y}}= {c}( {e}^{ {x}}+1)( {y}+1)

Hence, option (1) is correct.

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