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GATE 2026 EE – Question 50

Engineering Mathematics · Differential Equations: Initial and boundary value problems · 2 marks · Multiple choice

Consider the second-order differential equation

$$\frac{d^2y}{dx^2}+\frac{dy}{dx}+y=0$$

with initial conditions $y(0)=1,\ \dfrac{dy}{dx}\Big|_{x=0}=1$.

The solution is given by

  1. $y(x)=\exp(-x/2)\left(\cos\left(\frac{\sqrt3x}{2}\right)+\sqrt3\sin\left(\frac{\sqrt3x}{2}\right)\right)$
  2. $y(x)=\exp(-x/2)\left(\cos\left(\frac{\sqrt3x}{2}\right)+\frac1{\sqrt3}\sin\left(\frac{\sqrt3x}{2}\right)\right)$
  3. $y(x)=\exp(-x/2)\left(\cos\left(\frac{\sqrt3x}{2}\right)-\frac1{\sqrt3}\sin\left(\frac{\sqrt3x}{2}\right)\right)$
  4. $y(x)=\exp(-x/2)\left(\cos\left(\frac{\sqrt3x}{2}\right)-\sqrt3\sin\left(\frac{\sqrt3x}{2}\right)\right)$

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Show answer and explanation

Correct answer: (A) $y(x)=\exp(-x/2)\left(\cos\left(\frac{\sqrt3x}{2}\right)+\sqrt3\sin\left(\frac{\sqrt3x}{2}\right)\right)$

Explanation

The characteristic equation $m^2+m+1=0$ has roots $-\tfrac12\pm j\tfrac{\sqrt3}2$, so $y=e^{-x/2}\left(A\cos\tfrac{\sqrt3x}{2}+B\sin\tfrac{\sqrt3x}{2}\right)$. $y(0)=1$ gives $A=1$. $y'(0)=-\tfrac A2+\tfrac{\sqrt3}{2}B=1$ gives $B=\tfrac{3/2}{\sqrt3/2}=\sqrt3$.