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

  • question_answer
    The general solution the differential equation\[\frac{dy}{dx}-\frac{\tan \,\,y}{1+x}={{(1+x\,\,e)}^{x}}\sec \,\,y\] is

    A) \[\sin (1+x)=y({{e}^{x}}+c)\]

    B) \[y\sin (1+x)=c{{e}^{x}}\]

    C) \[(1+x)\sin \,\,y={{e}^{x}}+c\]

    D) \[\sin \,\,y=(1+x)({{e}^{x}}+c)\]

    Correct Answer: D

    Solution :

    [d] Divide the equation by sec y \[\cos y\frac{dy}{dx}-\frac{\sin y}{1+x}=(1+x){{e}^{x}}\] Put \[\sin y=z\Rightarrow \cos y\frac{dy}{dx}=\frac{dz}{dx}\] then \[\frac{dz}{dx}-\left( \frac{1}{1+x} \right)z=(1+x){{e}^{x}}\] I.E. \[={{e}^{-\int{\frac{1}{1+x}dx}}}={{e}^{-\log (1+x)}}=\frac{1}{1+x}\] The solution is \[z\left( \frac{1}{1+x} \right)={{e}^{x}}+c\Rightarrow \sin y=(1+x)({{e}^{x}}+c)\]

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