jee-main 2016 Q86

jee-main · India · 10apr Second order differential equations Second-order ODE with initial or boundary value conditions

The solution of the differential equation $\frac { d y } { d x } + \frac { y } { 2 } \sec x = \frac { \tan x } { 2 y }$, where $0 \leq x < \frac { \pi } { 2 }$ and $y ( 0 ) = 1$, is given by
(1) $y ^ { 2 } = 1 + \frac { x } { \sec x + \tan x }$
(2) $y = 1 + \frac { x } { \sec x + \tan x }$
(3) $y = 1 - \frac { x } { \sec x + \tan x }$
(4) $y ^ { 2 } = 1 - \frac { x } { \sec x + \tan x }$

The solution of the differential equation $\frac { d y } { d x } + \frac { y } { 2 } \sec x = \frac { \tan x } { 2 y }$, where $0 \leq x < \frac { \pi } { 2 }$ and $y ( 0 ) = 1$, is given by\\
(1) $y ^ { 2 } = 1 + \frac { x } { \sec x + \tan x }$\\
(2) $y = 1 + \frac { x } { \sec x + \tan x }$\\
(3) $y = 1 - \frac { x } { \sec x + \tan x }$\\
(4) $y ^ { 2 } = 1 - \frac { x } { \sec x + \tan x }$

Paper Questions

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