jee-main 2020 Q67

jee-main · India · session2_03sep_shift1 Differential equations Solving Separable DEs with Initial Conditions
The solution curve of the differential equation, $\left( 1 + e ^ { - x } \right) \left( 1 + y ^ { 2 } \right) \frac { d y } { d x } = y ^ { 2 }$ which passes through the point $( 0,1 )$, is
(1) $y ^ { 2 } + 1 = y \left( \log _ { e } \left( \frac { 1 + e ^ { - x } } { 2 } \right) + 2 \right)$
(2) $y ^ { 2 } + 1 = y \left( \log _ { e } \left( \frac { 1 + e ^ { x } } { 2 } \right) + 2 \right)$
(3) $y ^ { 2 } = 1 + y \log _ { e } \left( \frac { 1 + e ^ { x } } { 2 } \right)$
(4) $y ^ { 2 } = 1 + y \log _ { e } \left( \frac { 1 + e ^ { - x } } { 2 } \right)$
The solution curve of the differential equation, $\left( 1 + e ^ { - x } \right) \left( 1 + y ^ { 2 } \right) \frac { d y } { d x } = y ^ { 2 }$ which passes through the point $( 0,1 )$, is

(1) $y ^ { 2 } + 1 = y \left( \log _ { e } \left( \frac { 1 + e ^ { - x } } { 2 } \right) + 2 \right)$\\
(2) $y ^ { 2 } + 1 = y \left( \log _ { e } \left( \frac { 1 + e ^ { x } } { 2 } \right) + 2 \right)$\\
(3) $y ^ { 2 } = 1 + y \log _ { e } \left( \frac { 1 + e ^ { x } } { 2 } \right)$\\
(4) $y ^ { 2 } = 1 + y \log _ { e } \left( \frac { 1 + e ^ { - x } } { 2 } \right)$