JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 4)
The solution curve of the differential equation,
(1 + e-x)(1 + y2)$${{dy} \over {dx}}$$ = y2,
which passes through the point (0, 1), is :
(1 + e-x)(1 + y2)$${{dy} \over {dx}}$$ = y2,
which passes through the point (0, 1), is :
y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$$
y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$$
y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$$
y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$$
Explanation
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 = $$ - ln2$$
$$ \Rightarrow $$ $${y^2} - 1 = yln\left( {{{{e^x} + 1} \over 2}} \right)$$
$$ \Rightarrow $$ $${y^2} = 1 + yln\left( {{{{e^x} + 1} \over 2}} \right)$$
$$ \Rightarrow $$ $$\left( {{{{y^2} + 1} \over {{y^2}}}} \right)dy = {{{e^x}dx} \over {{e^x} + 1}}$$
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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