Question

The solution $x(t)$, $t\ge 0$, to $\ddot{x}=-k\dot{x}$ ($k>0$) with $x(0)=1$ and $\dot{x}(0)=0$ is:

• A. $x(t)=2e^{-kt}+2kt-1$
• B. $x(t)=2e^{-kt}-1$
• C. $x(t)=1$
• D. $x(t)=2e^{-kt}-kt-1$

Hint:

When the equation is $\ddot{x}=-k\dot{x}$, solve the first-order decay for $\dot{x}$ first; a zero initial velocity makes the velocity (and hence change in $x$) identically zero.

Answer 1

The Correct Option is C

Step 1: Reduce order. Let $v(t)=\dot{x}(t)$. Then $\dot{v}=-kv$ with solution \[ v(t)=v(0)\,e^{-kt}. \] Step 2: Apply initial condition. $v(0)=\dot{x}(0)=0 \Rightarrow v(t)\equiv 0$ for all $t$.
Step 3: Integrate for $x(t)$. Since $\dot{x}(t)=0$, $x(t)$ is constant; using $x(0)=1$, we get $x(t)\equiv 1$.