Science:Math Exam Resources/Courses/MATH152/April 2022/Question B5 (c)/Solution 1

From the hint, we know that we may find a solution to a differential equation of the form ${\textstyle {\frac {d}{dt}}{\vec {y}}(t)=A{\vec {y}}(t)}$ as a function ${\textstyle {\vec {y}}(t)=e^{\lambda t}{\vec {v}}}$, where ${\textstyle (\lambda ,{\vec {v}})}$ is an eigenvalue-eigenvector pair of ${\textstyle A}$. A general solution to the differential equation is a linear combination of “eigensolutions” of this type.

If we want a non-oscillating solution, then we will need ${\textstyle \lambda }$ to be real-valued (see below for an explanation). From the given eigendecomposition of ${\textstyle A}$ we see that the third eigenvalue is real-valued. Thus, a non-oscillating solution is

$$\left[{\begin{array}{c}I(t)\\V_{1}(t)\\V_{2}(t)\end{array}}\right]=e^{-0.77t}\cdot \left[{\begin{array}{c}0.22\\0.85\\0.47\end{array}}\right].$$

To see why ${\textstyle \lambda }$ must be real-valued, consider an eigenvalue ${\textstyle \lambda =a+ib}$. Then ${\textstyle e^{\lambda t}=e^{at+ibt}=e^{at}(\cos(bt)+i\sin(bt))}$ has sine and cosine terms as long as ${\textstyle b\neq 0}$. Thus, to make sure there are no sine and cosine terms we have to take ${\textstyle b=0}$, meaning ${\textstyle \lambda }$ is real-valued.