KVPY Sample Paper KVPY Stream-SX Model Paper-24

  • question_answer
    Solution of the equation \[{{x}^{3}}\frac{dy}{dx}+4{{x}^{2}}\tan y={{e}^{x}}\sec \,y\] when \[y(1)=0\]is -

    A) \[\sin y-{{e}^{x}}(x-1){{x}^{-4}}\]

    B) \[\sin y-{{e}^{x}}(x-1){{x}^{-3}}\]

    C) \[\tan y-{{e}^{x}}(x-1){{x}^{-3}}\]

    D) \[\tan y-{{e}^{x}}(x-2)logx\]

    Correct Answer: A

    Solution :

    Given equation can be written as
    \[{{x}^{4}}\cos y\frac{dy}{dx}+4{{x}^{3}}\sin y=x{{e}^{x}}\]\[\Rightarrow \frac{d}{dx}({{x}^{4}}\sin y)=x{{e}^{x}}\]\[\Rightarrow {{x}^{4}}\sin y=\int{x{{e}^{x}}\,\,\,dx}\]\[\Rightarrow {{x}^{4}}\sin y=(x-1){{e}^{x}}+c\]
    But when\[x=1,y=0\,\,so\,\,c=0\], Hence required solution is \[\sin y={{x}^{-4}}(x-1){{e}^{x}}\]


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