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GATE 2023 AG – Question 14

Engineering Mathematics · Differential Equations · 1 mark · Multiple choice

$y=ae^{mx}+be^{-mx}$ solves which ODE?

  1. $y^{\prime}-my=0$
  2. $y^{\prime}+my=0$
  3. $y^{\prime\prime}+m^2y=0$
  4. $y^{\prime\prime}-m^2y=0$

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Correct answer: (D) $y^{\prime\prime}-m^2y=0$

Explanation

Both exponentials have second derivative m² times themselves, so $y^{\prime\prime}=m^2y$ for arbitrary constants a, b. Rearranging gives **D**. Either first-order alternative would retain only one of the two independent exponential components.

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