Question

Find and graph the vector function determined $\mathbf{r}\left(t\right)=⟨x\left(t\right),y\left(t\right)⟩$by the differential equation

$x^{\mathrm{\prime }}\left(t\right)=-y,y^{\mathrm{\prime }}\left(t\right)=x,x\left(0\right)=1,y\left(0\right)=0$. ( HINT: What familiar pair of functions have the given properties ?)

Short answer

Ans: The vector function is $r\left(t\right)=<\cos t,-\sin t>$

and the graph is,

Step-by-step solution

Step 1. Given information.

given

$x^{\mathrm{\prime }}\left(t\right)=-y,y^{\mathrm{\prime }}\left(t\right)=x,x\left(0\right)=1,y\left(0\right)=0$

Step 2. The objective of this problem is to find and graph the vector r(t)=⟨x(t),y(t)⟩ determined by the differential equations

$\begin{array}{r}{x}^{\mathrm{\prime }}\left(t\right)=-y\dots \dots (1\\ y^{\mathrm{\prime }}\left(t\right)=x\dots \dots \left(2\right)\\ x\left(0\right)=1,y\left(0\right)=0\end{array}$

Differentiate equation (1) with respect to $t$

$x\text{'}\text{'}\left(t\right)=-y\text{'}\left(t\right)....\left(3\right)$

Substitute (2) in (3)

$\begin{array}{r}{x}^{\prime \prime }\left(t\right)=-x\\ x^{\prime \prime }\left(t\right)+x=0\\ D^{2}x+x=0\\ \left(D^2+1\right)x=0\end{array}$

Auxilary equation is

$$\begin{gathered} m^2+1=0 \\ m=\pm i \end{gathered}$$

Therefore,

$x\left(t\right)=c_1\cos t+c_2\sin t$

At $t=0,x\left(0\right)=1$

$\begin{array}{r}x\left(0\right)=c_1\cos \left(0\right)+c_2\sin \left(0\right)\\ 1=c_1\end{array}$

Then

$x\left(t\right)=\cos t+c_2\sin t$

Differentiate with respect to $t$

role="math" localid="1649589790829" $\begin{array}{r}{x}^{\mathrm{\prime }}\left(t\right)=-\sin t+c_2\cos t\\ y\left(t\right)=-\sin t+c_2\cos t\end{array}$

Substitute$y\left(0\right)=0$

role="math" localid="1649589878483" $$\begin{gathered} 0=-\sin \left(0\right)+c_2\cos \left(0\right) \\ {c}_2=0 \end{gathered}$$

Therefore,

$x\left(t\right)=\cos t,y\left(t\right)=-\sin t$

Thus, the vector function is

role="math" localid="1649590028804" $r\left(t\right)=<\cos t,-\sin t>$

Step 3. Draw the graph,