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Joint Entrance Examination
Mathematics
Differential Equations
Question 378
Question 378 of 490
Medium
MCQ
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 + \[{y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}\]
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Official solution
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 =
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l
n
2
- ln2
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l
n
2
\[ \Rightarrow \] \[{y^2} - 1 = yln\left( {{{{e^x} + 1} \over 2}} \right)\] \[ \Rightarrow \] \[{y^2} = 1 + yln\left( {{{{e^x} + 1} \over 2}} \right)\]
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