GATE 2023 IN Q.20: solution and answer

Official answer: CSource: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Write the equation as $\ddot{x}+k\dot{x}=0$. It is linear with constant coefficients, so its auxiliary equation is $m^2+km=0$, i.e. $m(m+k)=0$, giving $m=0$ and $m=-k$.

2. General solution: $x(t)=A+B{e}^{-kt}$, so $\dot{x}(t)=-kB{e}^{-kt}$.

3. Apply $\dot{x}(0)=0$: $-kB=0$. Since $k>0$, $B=0$.

4. Apply $x(0)=1$: $A+B=1$, so $A=1$. Hence $x(t)=1$ for all $t\ge 0$. Check: $\dot{x}=0$ and $\ddot{x}=0$, so $\ddot{x}=-k\dot{x}$ holds, and both initial conditions hold.

5. Quicker route: put $v=\dot{x}$. Then $\dot{v}=-kv$ with $v(0)=0$, so $v(t)=v(0)e^{-kt}=0$. The velocity is zero at every instant, so $x$ stays at its starting value $x(0)=1$.

6. Check the other options. (B): $\dot{x}=-2k{e}^{-kt}$, so $\dot{x}(0)=-2k\ne 0$. (D): $\dot{x}=-2k{e}^{-kt}-k$, so $\dot{x}(0)=-3k\ne 0$. (A) satisfies both initial conditions, since $x(0)=2-1=1$ and $\dot{x}=-2k{e}^{-kt}+2k$ gives $\dot{x}(0)=0$, but it fails the equation itself: $\ddot{x}=2k^2{e}^{-kt}$, while $-k\dot{x}=2k^2{e}^{-kt}-2k^2$, and these differ by $2k^2\ne 0$.

7. So the answer is (C), $x(t)=1$.