Linear System of ODEs (Eigenvalue Method with Resonant Forcing): 04-BS-1, May 2016, Q3

Question 3: Linear System of ODEs (Eigenvalue Method with Resonant Forcing) 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 linear system $\mathbf X'=A\mathbf X+\mathbf F(t)$ with $A=\begin{pmatrix}4&2\\3&-1\end{pmatrix}$ and $\mathbf F(t)=(0,e^{-2t})^T$.

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

Approach. Diagonalize $A$ for the homogeneous solution, then find a particular solution for the forcing $e^{-2t}$. Since $-2$ turns out to be an eigenvalue of $A$, this is a resonant case and the particular solution needs a $t\,e^{-2t}$ term.

Eigenvalues of $A$. $$\det(A-\lambda I)=(4-\lambda)(-1-\lambda)-6=\lambda^2-3\lambda-10=0 \;\Rightarrow\; \lambda=\frac{3\pm\sqrt{9+40}}{2}=\frac{3\pm7}{2}=5,\,-2.$$
Eigenvectors. For $\lambda=5$: $(4-5)v_1+2v_2=0\Rightarrow v_1=2v_2$, so $\mathbf v_5=(2,1)$. For $\lambda=-2$: $(4+2)v_1+2v_2=0\Rightarrow v_2=-3v_1$, so $\mathbf v_{-2}=(1,-3)$. The homogeneous solution is $$\mathbf X_h(t)=C_1e^{5t}(2,1)+C_2e^{-2t}(1,-3).$$
Resonant particular solution — set-up. Since $-2$ is an eigenvalue and the forcing is $e^{-2t}$, try $\mathbf X_p=(\mathbf a\,t+\mathbf b)\,e^{-2t}$. Substituting into $\mathbf X_p'=A\mathbf X_p+\mathbf F$ and matching the $t\,e^{-2t}$ terms gives $-2\mathbf a=A\mathbf a$, i.e. $\mathbf a$ must itself be the $\lambda=-2$ eigenvector: $\mathbf a=c(1,-3)$.
Resonant particular solution — constant terms. Matching the plain $e^{-2t}$ terms gives $(A+2I)\mathbf b=\mathbf a-(0,1)$. With $A+2I=\begin{pmatrix}6&2\\3&1\end{pmatrix}$ (rank 1, rows proportional $2\times$), solvability requires the right side's components to be in the same $2\!:\!1$ ratio: $c=2(-3c-1)\Rightarrow c=-\tfrac27$, so $\mathbf a=-\tfrac27(1,-3)=(-\tfrac27,\tfrac67)$. Taking the free component $b_1=0$, row 2 gives $b_2=-3c-1=-\tfrac17$, so $\mathbf b=(0,-\tfrac17)$ (check row 1: $6(0)+2(-\tfrac17)=-\tfrac27=c$ ✓).
Assemble the general solution. $$\mathbf X_p(t)=\left(-\tfrac27t,\ \tfrac67t-\tfrac17\right)e^{-2t}.$$ Adding to the homogeneous solution: $$x(t)=\boxed{2C_1e^{5t}+C_2e^{-2t}-\tfrac27t\,e^{-2t}}$$ $$y(t)=\boxed{C_1e^{5t}-3C_2e^{-2t}+\tfrac17(6t-1)e^{-2t}}$$ (Verified by direct substitution into both original equations — residuals vanish identically for all $C_1,C_2$.)

Quantity	Result
Eigenvalues	$\lambda=5,\,-2$
Eigenvectors	$(2,1)$ for $\lambda=5$; $(1,-3)$ for $\lambda=-2$
$x(t)$	$2C_1e^{5t}+C_2e^{-2t}-\tfrac27t\,e^{-2t}$
$y(t)$	$C_1e^{5t}-3C_2e^{-2t}+\tfrac17(6t-1)e^{-2t}$

Concept & Theory

Topic. Linear systems of ODEs $\mathbf X'=A\mathbf X+\mathbf F(t)$ solved by the eigenvalue/eigenvector method, including resonant forcing.

Key relations. The homogeneous solution is $\sum C_ie^{\lambda_it}\mathbf v_i$. When the forcing $\mathbf F(t)=\mathbf f_0e^{\mu t}$ has $\mu$ coinciding with an eigenvalue of $A$ (resonance), the usual trial $\mathbf X_p=\mathbf a\,e^{\mu t}$ fails (it solves the homogeneous equation with zero right side), so one instead tries $\mathbf X_p=(\mathbf a\,t+\mathbf b)e^{\mu t}$.

Why this works. Matching powers of $t\,e^{\mu t}$ forces $\mathbf a$ to be the eigenvector for $\lambda=\mu$; matching the plain $e^{\mu t}$ terms then gives a rank-deficient linear system for $\mathbf b$ (singular because $\mu$ is an eigenvalue), which is solvable precisely because the forcing vector was consistent with that same eigenvector's null-space condition — exactly analogous to the scalar resonance case $y''+\omega^2y=\cos\omega t\Rightarrow y_p=t(A\cos\omega t+B\sin\omega t)$.

Common pitfall. Missing that $-2$ is an eigenvalue and trying the non-resonant ansatz $\mathbf X_p=\mathbf a\,e^{-2t}$, which leads to an inconsistent (unsolvable) linear system. Sign errors in $(A+2I)$ or forgetting that $\mathbf b$ is only determined up to an additive multiple of the $\lambda=-2$ eigenvector (any choice of the free component is valid, since it is absorbed into $C_2$).

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 4 (§4.4 Systems of ODEs: General Theory; §4.6 Nonhomogeneous Linear Systems).

Question 3 of 8: Linear System of ODEs (Eigenvalue Method with Resonant Forcing)

Question 3: Linear System of ODEs (Eigenvalue Method with Resonant Forcing) 20 marks