JEE Main & Advanced Mathematics Differential Equations Question Bank Self Evaluation Test - Differential Equations

  • question_answer
    The equation of the curve passing through the point \[\left( 0,\frac{\pi }{4} \right)\] whose differential equation is\[sin\text{ }x\text{ }cos\text{ }y\text{ }dx+cos\text{ }x\text{ }sin\text{ }y\text{ }dy=0\], is

    A) \[sec\,\,x\,\,sec\,\,y=\sqrt{2}\]

    B) \[cos\,\,x\,\,cos\,\,y=\sqrt{2}\]

    C) \[\sec \,\,x=\sqrt{2}\,\,\cos \,\,y\]

    D) \[cos\,\,y=\sqrt{2}\,\,\sec \,\,y\]

    Correct Answer: A

    Solution :

    [a] The given differential equation is \[\sin x\cos ydx+\cos x\sin ydy=0\] dividing by \[\cos x\cos y\Rightarrow \frac{\sin x}{\cos x}dx+\frac{\sin y}{\cos y}dy=0\] Integrating, \[\int{\tan xdx+\int{\tan ydy=\log c}}\] Or \[\log \sec x\sec y=\log cor\sec x\sec y=c\] curve passes through the point \[\left( 0,\frac{\pi }{4} \right)\] \[\sec 0\sec \frac{\pi }{4}=c=\sqrt{2}\] Hence. The required equation of the curve is \[\sec x\sec y=\sqrt{2}\]

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