RAJASTHAN ­ PET Rajasthan PET Solved Paper-2008

  • question_answer
    The differential equation of the curve \[y=A{{e}^{x}}+B{{e}^{-x}}\]for different values of A and B, is

    A)  \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-2y=0\]

    B)  \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=y\]

    C)  \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=4y+3\]

    D)  \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+y=0\]

    Correct Answer: B

    Solution :

     Given, curve is \[y=A{{e}^{x}}+B{{e}^{-x}}\]                  ...(i) On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=A{{e}^{x}}-B{{e}^{-x}}\] Again, differentiating, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=A{{e}^{x}}+B{{e}^{-x}}\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=y\]                  [from Eq.(i)]


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