Question 4 of 8: Initial Value Problem for a 2×2 Linear System with Complex Eigenvalues

Question 4: Initial Value Problem for a 2×2 Linear System with Complex Eigenvalues 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A homogeneous linear system $\mathbf x'=A\mathbf x$ with $A=\begin{pmatrix}3&1\\-2&1\end{pmatrix}$ and initial condition $\mathbf x(0)=(1,0)^T$.

Find. $x(t)$ and $y(t)$.

Approach. Find the eigenvalues of $A$; since they turn out complex, build the real general solution from the real/imaginary parts of one complex eigenvector, then fit the initial conditions.

1. Eigenvalues. $\operatorname{tr}A=4$, $\det A=3(1)-1(-2)=5$, so $\lambda^2-4\lambda+5=0\Rightarrow\lambda=\dfrac{4\pm\sqrt{16-20}}{2}=2\pm i$.
2. Eigenvector for $\lambda=2+i$. $(A-\lambda I)\mathbf v=0$: row 1 gives $(1-i)v_1+v_2=0\Rightarrow v_2=(i-1)v_1$. Taking $v_1=1$: $\mathbf v=(1,\,-1+i)=\underbrace{(1,-1)}_{\mathbf p}+i\underbrace{(0,1)}_{\mathbf q}$.
3. Build the real general solution. With $a=2,\,b=1$ (from $\lambda=a+ib$), $$\mathbf x(t)=e^{at}\Big[C_1(\mathbf p\cos bt-\mathbf q\sin bt)+C_2(\mathbf p\sin bt+\mathbf q\cos bt)\Big].$$ Component-wise this gives $$x(t)=e^{2t}\big(C_1\cos t+C_2\sin t\big),\qquad y(t)=e^{2t}\big[C_1(-\cos t-\sin t)+C_2(\cos t-\sin t)\big].$$
4. Apply the initial conditions. At $t=0$: $x(0)=C_1=1$. $y(0)=-C_1+C_2=0\Rightarrow C_2=1$.
5. Final solution. $$\boxed{x(t)=e^{2t}(\cos t+\sin t)},\qquad\boxed{y(t)=-2e^{2t}\sin t}$$ (the $y(t)$ expression simplifies since the $\cos t$ terms from $C_1,C_2$ cancel). (Check: $x'=e^{2t}(3\cos t+\sin t)=3x+y$ and $y'=e^{2t}(-4\sin t-2\cos t)=-2x+y$.)