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

  • question_answer
    If \[\phi (x)\] is a differentiable function, then the solution of the differential equation\[dy+\{y\phi '(x)-\phi (x)\phi '(x)\}dx=0\] is

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

    B) \[y\phi (x)={{\{\phi (x)\}}^{2}}+c\]

    C) \[y{{e}^{\phi (x)}}=\phi (x){{e}^{\phi (x)}}+c\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] We have, \[dy+\{y\phi '(x)-\phi (x)\phi '(x)\}dx=0\]
    \[\Rightarrow \frac{dy}{dx}+\phi '(x)\cdot y=\phi (x)\phi '(x)\]                     ? (i)
    This is linear differential equation with
    I.F \[={{e}^{\int{\phi '(x)dx}}}={{e}^{\phi (x)}}\]
    Multiplying Eq. (i) by \[\phi (x)\] and integrating, we get
    \[y{{e}^{\phi (x)}}=\int{\phi (x)\phi '(x){{e}^{\phi (x)}}}dx\]
    \[\Rightarrow \,\,\,y{{e}^{\phi (x)}}=\int{{{e}^{\phi (x)}}\phi (x)\phi '(x)}dx\]
    \[\Rightarrow y{{e}^{\phi (x)}}=\int{\phi (x){{e}^{\phi (x)}}}\phi '(x)dx\]
    \[\Rightarrow y{{e}^{\phi (x)}}=\phi (x){{e}^{\phi (x)}}-\int{\phi '(x){{e}^{\phi (x)}}}dx\]
    \[\Rightarrow y{{e}^{\phi (x)}}=\phi (x){{e}^{\phi (x)}}-{{e}^{\phi (x)}}+c\]
    \[\Rightarrow y=\{\phi (x)-1\}+c{{e}^{-\phi (x)}}\]

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