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

  • question_answer
    If \[y={{e}^{4x}}+2{{e}^{-x}}\] satisfies the relation \[\frac{{{d}^{3}}y}{d{{x}^{3}}}+A\frac{dy}{dx}+By=0,\] then values of A and B respectively are:

    A) -13, 14

    B) -13, -12

    C) -13, 12

    D) 12, -13

    Correct Answer: B

    Solution :

    [b] Given \[y={{e}^{4x}}+2{{e}^{-x}}\] Differentiating we get \[\frac{dy}{dx}=4{{e}^{4x}}-2{{e}^{-x}}\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=16{{e}^{14x}}+2{{e}^{-x}}\] \[\Rightarrow \frac{{{d}^{3}}y}{d{{x}^{3}}}=64{{e}^{4x}}-2{{e}^{-x}}\] Putting these values in \[\frac{{{d}^{3}}y}{d{{x}^{3}}}+A\frac{dy}{dx}+By=0\] We have, \[(64+4A+B){{e}^{4x}}+(-2-2A+2B){{e}^{-x}}=0\] \[\Rightarrow \,\,\,64+4A+B=0,\,\,\,-2-2A+2B=0\] Solving these eqs. we get \[A=-13,B=-12\]


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