QJEEM20Maff2d: JEE Main 2020 Mathematics - Class 12 JEEM20Maff2d Solution: The solution curve of the differential equation, (1 + e -x )(1…

Question

Verified Solution • 2025–26 Session moderate

The solution curve of the differential equation,

(1 + e^-x)(1 + y²)

{{dy} \over {dx}}

= y²,

which passes through the point (0, 1), is :

A y2 + 1 = y $\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$
B y2 + 1 = y $\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$
C y2 = 1 + ${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$
D y2 = 1 + ${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$

Accepted Answer

Given (1 + e^-x)(1 + y²)

{{dy} \over {dx}}

= y²

\Rightarrow

\left( {{{{y^2} + 1} \over {{y^2}}}} \right)dy = {{{e^x}dx} \over {{e^x} + 1}}

Integrating both sides,

\int {\left( {{{{y^2} + 1} \over {{y^2}}}} \right)dy = } \int {{{{e^x}dx} \over {{e^x} + 1}}}

\Rightarrow

y - {1 \over y} = ln\left| {{e^x} + 1} \right| + c

It passes through (0, 1)

\Rightarrow

c =

- ln2

\Rightarrow

{y^2} - 1 = yln\left( {{{{e^x} + 1} \over 2}} \right)

\Rightarrow

{y^2} = 1 + yln\left( {{{{e^x} + 1} \over 2}} \right)

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