Question

The solution for the differential equation $\frac{d^{2}x}{d{t}^2}=-9x$ with initial conditions $x\left(0\right)=1$ and ${\begin{array}{c}\frac{dx}{dt}\end{array}|}_{t=0}=1,$ is

• A. $t^2+t+1$
• B. $sin3t+\frac{1}{3}cos3t+\frac{2}{3}$
  
• C. $\frac{1}{3}sin3t+cos3t$
• D. $cos3t+t$

Answer 1

The correct option is C $\frac{1}{3}sin3t+cos3t$

$\frac{d^{2}x}{d{t}^2}=-9x$
$\frac{d^{2}x}{d{t}^2}+9x=0$
A.E. is $D^2+9=0$
$D^2=-9$
$D=\pm 3i$
$x=C_{1}cos3t+C_{2}sin3t$
$x\left(0\right)=1$
$\Rightarrow 1=C_1$
$\frac{dx}{dt}=-3C_{1}sin3t+3C_{2}cos3t$
${\begin{array}{c}\frac{dx}{dt}\end{array}|}_{t=0}=1$
$\Rightarrow$ $1=3C_2$
$\Rightarrow$ $C_2=\frac{1}{3}$
$x=cos3t+\frac{1}{3}sin3t$