Question 4 of 8: General Solution of a Resonantly-Forced Linear System

Question 4: General Solution of a Resonantly-Forced Linear System (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. $\mathbf X'=A\mathbf X+\mathbf Fe^{-5t}$ with $A=\begin{pmatrix}4&-18\\-3&1\end{pmatrix}$, $\mathbf F=(0,1)^T$.

Find. The general solution $x(t),y(t)$.

Approach. Diagonalize the homogeneous system first. Because the forcing rate $e^{-5t}$ turns out to match one of the eigenvalues (resonance), the particular solution needs the extended trial $\mathbf X_p=(t\mathbf a+\mathbf b)e^{-5t}$ rather than a bare $\mathbf a e^{-5t}$.

1. Eigenvalues. $\det(A-\lambda I)=(4-\lambda)(1-\lambda)-54=\lambda^2-5\lambda-50=0\Rightarrow\lambda=\dfrac{5\pm15}2=10,-5.$
2. Eigenvectors. For $\lambda=10$: $(A-10I)\mathbf v=0\Rightarrow\begin{pmatrix}-6&-18\\-3&-9\end{pmatrix}\mathbf v=0\Rightarrow v_1=-3v_2$, take $\mathbf v_1=(-3,1)$. For $\lambda=-5$: $(A+5I)\mathbf v=0\Rightarrow\begin{pmatrix}9&-18\\-3&6\end{pmatrix}\mathbf v=0\Rightarrow v_1=2v_2$, take $\mathbf v_2=(2,1)$. $$\mathbf X_h(t)=C_1\begin{pmatrix}-3\\1\end{pmatrix}e^{10t}+C_2\begin{pmatrix}2\\1\end{pmatrix}e^{-5t}.$$
3. Resonance check. The forcing rate $-5$ coincides with the eigenvalue $\lambda=-5$, so the plain trial $\mathbf a\,e^{-5t}$ would solve the homogeneous system and cannot balance the forcing. Use $\mathbf X_p=(t\mathbf a+\mathbf b)e^{-5t}$ instead, which requires $(A+5I)\mathbf a=0$ (so $\mathbf a=k(2,1)$, a multiple of the $\lambda=-5$ eigenvector) and $(A+5I)\mathbf b=\mathbf a-\mathbf F$.
4. Solve the consistency condition for $k$. With $\mathbf a=k(2,1)$, solving $(A+5I)\mathbf b=\mathbf a-\mathbf F=(2k,\,k-1)$ means the two scalar equations $9b_1-18b_2=2k$ and $-3b_1+6b_2=k-1$ must be consistent, since $A+5I=\begin{pmatrix}9&-18\\-3&6\end{pmatrix}$ is singular (row 1 $=-3\times$row 2). Multiplying the second equation by $-3$ gives $9b_1-18b_2=-3k+3$; matching this to the first equation's right side, $2k=-3k+3\Rightarrow k=\tfrac35$. Then $\mathbf a=\left(\tfrac65,\tfrac35\right)$.
5. Solve for $\mathbf b$. $9b_1-18b_2=\tfrac65$; choosing the free component $b_2=0$ (its complement is absorbed into the homogeneous $C_2$ term) gives $b_1=\tfrac2{15}$, so $\mathbf b=\left(\tfrac2{15},0\right)$ and $$\mathbf X_p(t)=\left[t\begin{pmatrix}6/5\\3/5\end{pmatrix}+\begin{pmatrix}2/15\\0\end{pmatrix}\right]e^{-5t}.$$

$$x(t)=\boxed{-3C_1e^{10t}+2C_2e^{-5t}+\left(\tfrac65t+\tfrac2{15}\right)e^{-5t}},\qquad y(t)=\boxed{C_1e^{10t}+C_2e^{-5t}+\tfrac35t\,e^{-5t}}$$