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

We start by substituting into the equation we found for in part b),

Set this equal to and solve for and

We can see that and solve the system. Thus, we have

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

From UBC Wiki

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

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