jee-main 2020 Q68

jee-main · India · session2_04sep_shift2 Differential equations Solving Separable DEs with Initial Conditions

The solution of the differential equation $\frac { d y } { d x } - \frac { y + 3 x } { \log _ { e } ( y + 3 x ) } + 3 = 0$ is (where $C$ is a constant of integration)
(1) $x - \frac { 1 } { 2 } \left( \log _ { e } ( y + 3 x ) \right) ^ { 2 } = C$
(2) $x - \log _ { e } ( y + 3 x ) = C$
(3) $y + 3 x - \frac { 1 } { 2 } \left( \log _ { e } x \right) ^ { 2 } = C$
(4) $x - 2 \log _ { e } ( y + 3 x ) = C$

The solution of the differential equation $\frac { d y } { d x } - \frac { y + 3 x } { \log _ { e } ( y + 3 x ) } + 3 = 0$ is (where $C$ is a constant of integration)\\
(1) $x - \frac { 1 } { 2 } \left( \log _ { e } ( y + 3 x ) \right) ^ { 2 } = C$\\
(2) $x - \log _ { e } ( y + 3 x ) = C$\\
(3) $y + 3 x - \frac { 1 } { 2 } \left( \log _ { e } x \right) ^ { 2 } = C$\\
(4) $x - 2 \log _ { e } ( y + 3 x ) = C$

Paper Questions

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