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JEE Main & Advanced Mathematics Differential Equations Question Bank Critical Thinking

  • question_answer
    The solution of the differential equation \[{{x}^{4}}\frac{dy}{dx}+{{x}^{3}}y+\text{cosec}\,(xy)=0\] is equal to [Pb. CET 2004]

    A) \[2\cos \,(xy)+{{x}^{-2}}=c\]    

    B) \[2\cos \,(xy)+{{y}^{-2}}=c\]

    C) \[2\sin \,(xy)+{{x}^{-2}}=c\]     

    D) \[2\sin \,(xy)+{{y}^{-2}}=c\]

    Correct Answer: A

    Solution :

    • \[{{x}^{4}}\frac{dy}{dx}+{{x}^{3}}y+\text{cosec}(xy)=0\]                   
    • \[{{x}^{4}}dy+{{x}^{3}}y\,dx+\text{cosec }(xy)\,dx=0\]                   
    • \[{{x}^{3}}(x\,dy+y\,dx)+\text{cosec }(xy)\,dx=0\]                   
    • \[{{x}^{3}}d(xy)+\text{cosec }(xy)dx=0\]                   
    • \[\frac{d(xy)}{\text{cosec }(xy)}+\frac{dx}{{{x}^{3}}}=0\]                   
    • Integrating both sides, \[\int_{{}}^{{}}{\frac{d\,(xy)}{\text{cosec }(xy)}}+\int_{{}}^{{}}{\frac{dx}{{{x}^{3}}}}=0\]                   
    • \[\int_{{}}^{{}}{\sin (xy)\,d(xy)+\int_{{}}^{{}}{{{x}^{-3}}dx=0}}\]                   
    • \[-\cos (xy)+\left( \frac{{{x}^{-2}}}{-2} \right)=c\]; \[2\cos (xy)+{{x}^{-2}}=c\].


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