Question

Consider two solution $x\left(t\right)=x_1\left(t\right)$ and $x\left(t\right)=x_2\left(t\right)$ of the differential equation $\frac{d^{2}x\left(t\right)}{d{t}^2}+x\left(t\right)=0,t>0,$ such that $x_1\left(0\right)=1,{\begin{array}{c}\frac{d{x}_1\left(t\right)}{dt}\end{array}|}_{t=0}=0,x_2\left(0\right)=0,{\begin{array}{c}\frac{d{x}_2\left(t\right)}{dt}\end{array}|}_{t=0}=1$.
The Wronskian $W\left(t\right)=\left|\begin{array}{cc}{x}_1\left(t\right)& x_2\left(t\right)\\ \frac{d{x}_1\left(t\right)}{dt}& \frac{d{x}_2\left(t\right)}{dt}\end{array}\right|$ at $t=\pi /2$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $\pi /2$

Answer 1

The correct option is A $1$

$\frac{d^{2}x\left(t\right)}{d{t}^2}+x\left(t\right)=0$
$D^2+1=0$
$D=\pm i$
$x=c_{1}cost+c_{2}sint$
$x_1=cost$ and $x_2=sint$ satisfies the given condition.
$W\left(t\right)={\left|\begin{array}{cc}cost& sint\\ -sint& cost\end{array}\right|}_{t=\pi /2}$
$=\left|\begin{array}{cc}0& 1\\ -1& 0\end{array}\right|$