Science:Math Exam Resources/Courses/MATH152/April 2022/Question B4 (c)/Solution 1

We start by substituting ${\textstyle t=0}$ into the equation we found for ${\textstyle {\vec {y}}(t)}$ in part b),

$${\begin{aligned}{\vec {y}}(0)&=c_{1}e^{2(0)}\left[{\begin{array}{c}\cos(0)\\-\cos(0)-\sin(0)\end{array}}\right]+c_{2}e^{2(0)}\left[{\begin{array}{c}\sin(0)\\\cos(0)-\sin(0)\end{array}}\right]\\&=\left[{\begin{array}{c}-c_{1}\\c_{1}-c_{2}\end{array}}\right].\end{aligned}}$$

Set this equal to ${\textstyle [1\ \ -1]^{T}}$ and solve for ${\textstyle c_{1}}$ and ${\textstyle c_{2}}$

$${\begin{aligned}\left[{\begin{array}{c}-c_{1}\\c_{1}-c_{2}\end{array}}\right]=\left[{\begin{array}{c}1\\-1\end{array}}\right].\end{aligned}}$$

We can see that ${\textstyle c_{1}=-1}$ and ${\textstyle c_{2}=0}$ solve the system. Thus, we have

$${\vec {y}}(t)=-e^{2t}\left[{\begin{array}{c}\cos(t)\\-\cos(t)-\sin(t)\end{array}}\right].$$