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JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    The solution of the differential equation\[\frac{dy}{dx}+y\frac{y}{2}\sec \,x=\frac{\tan x}{2y}\], where\[\le \,x<\frac{\pi }{2}\] and y(0) = 1, is given by   JEE Main Online Paper (Held On 10 April 2016)

    A) \[{{y}^{2}}=1-\frac{x}{\sec x\,+\tan x}\]  

    B) \[{{y}^{2}}=1+\frac{x}{\sec x\,+\tan x}\]

    C) \[y=1+\frac{x}{\sec x\,+\tan x}\]

    D) \[y=1-\frac{x}{\sec x\,+\tan x}\]

    Correct Answer: D

    Solution :

                 \[2y\frac{dy}{dx}+{{y}^{2}}\sec x=\tan x\]              put \[{{y}^{2}}=t\]              \[\Rightarrow \]               \[2y=\frac{dy}{dx}=\frac{dt}{dx}\]              \[\frac{dt}{dx}+t\,\sec =\tan x\]              \[I.F.\,\,{{e}^{\int{\sec \,xdx}}}={{e}^{\ell n(\sec x+\tan x)}}=\sec x+\tan x\]              \[t.(\sec x+\tan x)=\int{(\sec x+\tan x)\tan x\,\,dx}\]                                 \[=\int{\sec x\ \tan x+}\int{{{\tan }^{2}}xdx}\]                \[{{y}^{2}}(\sec x+\tan x)=\sec x+\tan x-x+c\]              \[y(0)=1\]                               \[\Rightarrow \]                               \[c=0\]                                 \[\Rightarrow \]               \[{{y}^{2}}=1-\frac{x}{\sec x\,+\tan x}\]


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