Chapter 10

Solving $$\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=\left|\begin{array}{cc} 2-\lambda & 3 \\\\-1 & -2-\lambda\end{array}\right|=\lambda^{2}-1=(\lambda-1)(\lambda+1)=0$$ we obtain eigenvalues \(\lambda_{1}=-1\) and \(\lambda_{2}=1 .\) Corresponding eigenvectors are $$\mathbf{K}_{1}=\left(\begin{array}{r}-1 \\\1\end{array}\right) \quad \text { and } \quad \mathbf{K}_{2}=\left(\begin{array}{r}-3 \\\1\end{array}\right).$$ Thus $$\mathbf{X}_{c}=c_{1}\left(\begin{array}{r}-1 \\\1\end{array}\right) e^{-t}+c_{2}\left(\begin{array}{r} -3 \\\1\end{array}\right) e^{t}.$$ Substituting $$\mathbf{X}_{p}=\left(\begin{array}{l}a_{1} \\\b_{1}\end{array}\right)$$ into the system yields $$\begin{aligned}2 a_{1}+3 b_{1} &=7 \\\\-a_{1}-2 b_{1} &=-5,\end{aligned}$$ $$\mathbf{X}(t)=c_{1}\left(\begin{array}{r}-1 \\\1\end{array}\right) e^{-t}+c_{2}\left(\begin{array}{r}-3 \\\1\end{array}\right) e^{t}+\left(\begin{array}{r}-1 \\\3\end{array}\right).$$

For \(\mathbf{A}=\left(\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right)\) we have $$\begin{aligned} &\mathbf{A}^{2}=\left(\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right)\left(\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right)=\left(\begin{array}{ll} 1 & 0 \\ 0 & 4 \end{array}\right),\\\ &\begin{array}{l} \mathbf{A}^{3}=\mathbf{A} \mathbf{A}^{2}=\left(\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right)\left(\begin{array}{ll} 1 & 0 \\ 0 & 4 \end{array}\right)=\left(\begin{array}{ll} 1 & 0 \\ 0 & 8 \end{array}\right), \\ \mathbf{A}^{4}=\mathbf{A} \mathbf{A}^{3}=\left(\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right)\left(\begin{array}{ll} 1 & 0 \\ 0 & 8 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 16 \end{array}\right), \end{array} \end{aligned}$$ and so on. In general $$\mathbf{A}^{k}=\left(\begin{array}{cc} 1 & 0 \\ 0 & 2^{k} \end{array}\right) \quad \text { for } \quad k=1,2,3, \ldots$$ Thus $$\begin{aligned} e^{\mathbf{A} t} &=\mathbf{I}+\frac{\mathbf{A}}{1 !} t+\frac{\mathbf{A}^{2}}{2 !} t^{2}+\frac{\mathbf{A}^{3}}{3 !} t^{3}+\cdots \\ &=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)+\frac{1}{1 !}\left(\begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array}\right) t+\frac{1}{2 !}\left(\begin{array}{cc} 1 & 0 \\ 0 & 4 \end{array}\right) t^{2}+\frac{1}{3 !}\left(\begin{array}{cc} 1 & 0 \\ 0 & 8 \end{array}\right) t^{3}+\cdots \\ &=\left(\begin{array}{ccc} 1+t+\frac{t^{2}}{2 !}+\frac{t^{3}}{3 !}+\cdots & & 0 \\ 0 & & \left.1+t+\frac{(2 t)^{2}}{2 !}+\frac{(2 t)^{3}}{3 !}+\cdots\right) & \left(\begin{array}{cc} e^{t} & 0 \\ 0 & e^{2 t} \end{array}\right) \end{array}\right. \end{aligned}$$ and $$e^{-\mathbf{A} t}=\left(\begin{array}{cc} e^{-t} & 0 \\ 0 & e^{-2 t} \end{array}\right).$$

For \(\mathbf{A}=\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\) we have $$\begin{array}{l} \mathbf{A}^{2}=\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)=\mathbf{I} \\ \mathbf{A}^{3}=\mathbf{A} \mathbf{A}^{2}=\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) \mathbf{I}=\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)=\mathbf{A} \\ \mathbf{A}^{4}=\left(\mathbf{A}^{2}\right)^{2}=\mathbf{I} \\ \mathbf{A}^{5}=\mathbf{A} \mathbf{A}^{4}=\mathbf{A} \mathbf{I}=\mathbf{A}, \end{array}$$ and so on. In general, $$\mathbf{A}^{k}=\left\\{\begin{array}{ll} \mathbf{A}, & k=1,3,5, \ldots \\ \mathbf{I}, & k=2,4,6, \ldots \end{array}\right.$$ Thus $$\begin{aligned} e^{\mathbf{A} t} &=\mathbf{I}+\frac{\mathbf{A}}{1 !} t+\frac{\mathbf{A}^{2}}{2 !} t^{2}+\frac{\mathbf{A}^{3}}{3 !} t^{3}+\cdots \\ &=\mathbf{I}+\mathbf{A} t+\frac{1}{2 !} \mathbf{I} t^{2}+\frac{1}{3 !} \mathbf{A} t^{3}+\cdots \\ &=\mathbf{I}\left(1+\frac{1}{2 !} t^{2}+\frac{1}{4 !} t^{4}+\cdots\right)+\mathbf{A}\left(t+\frac{1}{3 !} t^{3}+\frac{1}{5 !} t^{5}+\cdots\right) \\ &=\mathbf{I} \cosh t+\mathbf{A} \sinh t=\left(\begin{array}{cc} \cosh t & \sinh t \\ \sinh t & \cosh t \end{array}\right) \end{aligned}$$ and $$e^{-\mathbf{A} t}=\left(\begin{array}{cc} \cosh (-t) & \sinh (-t) \\ \sinh (-t) & \cosh (-t) \end{array}\right)=\left(\begin{array}{rr} \cosh t & -\sinh t \\ -\sinh t & \cosh t \end{array}\right).$$

The system is $$\mathbf{X}^{\prime}=\left(\begin{array}{ll} -4 & 2 \\ -\frac{5}{2} & 2 \end{array}\right) \mathbf{x}$$ and \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=(\lambda-1)(\lambda+3)=0 .\) For \(\lambda_{1}=1\) we obtain $$\left(\begin{array}{cc|c} -5 & 2 & 0 \\ -\frac{5}{2} & 1 & 1 \end{array}\right) \Longrightarrow\left(\begin{array}{rr|r} -5 & 2 & 0 \\ 0 & 0 & 0 \end{array}\right) \quad \text { so that } \quad \mathbf{K}_{1}=\left(\begin{array}{c} 2 \\ 5 \end{array}\right)$$ For \(\lambda_{2}=-3\) we obtain \left(\begin{array}{cc|c} -1 & 2 & 0 \\ -\frac{5}{2} & 5 & | 0 \end{array}\right) \Longrightarrow\left(\begin{array}{rr|r} -1 & 2 & 0 \\ 0 & 0 & 0 \end{array}\right) \quad \text { so that } \quad \mathbf{K}_{2}=\left(\begin{array}{l} 2 \\ 1 \end{array}\right) Then $$\mathbf{X}=c_{1}\left(\begin{array}{l} 2 \\ 5 \end{array}\right) e^{t}+c_{2}\left(\begin{array}{l} 2 \\ 1 \end{array}\right) e^{-3 t}$$

For $$\mathbf{A}=\left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 1 \\ -2 & -2 & -2 \end{array}\right)$$ we have $$\mathbf{A}^{2}=\left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 1 \\ -2 & -2 & -2 \end{array}\right)\left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 1 \\ -2 & -2 & -2 \end{array}\right)=\left(\begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right).$$ Thus, \(\mathbf{A}^{3}=\mathbf{A}^{4}=\mathbf{A}^{5}=\cdots=\mathbf{0}\) and $$e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t=\left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)+\left(\begin{array}{rrr} t & t & t \\ t & t & t \\ -2 t & -2 t & -2 t \end{array}\right)=\left(\begin{array}{ccc} t+1 & t & t \\ t & t+1 & t \\ -2 t & -2 t & -2 t+1 \end{array}\right).$$

For $$\mathbf{A}=\left(\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 5 & 1 & 0 \end{array}\right)$$ we have $$\begin{aligned} &\mathbf{A}^{2}=\left(\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 5 & 1 & 0 \end{array}\right)\left(\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 5 & 1 & 0 \end{array}\right)=\left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 3 & 0 & 0 \end{array}\right)\\\ &\mathbf{A}^{3}=\mathbf{A} \mathbf{A}^{2}=\left(\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 5 & 1 & 0 \end{array}\right)\left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 3 & 0 & 0 \end{array}\right)=\left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right). \end{aligned}$$ Thus, \(\mathbf{A}^{4}=\mathbf{A}^{5}=\mathbf{A}^{6}=\cdots=\mathbf{0}\) and $$\begin{aligned} e^{\mathbf{A} t} &=\mathbf{I}+\mathbf{A} t+\frac{1}{2} \mathbf{A}^{2} t^{2} \\\ &=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)+\left(\begin{array}{lll} 0 & 0 & 0 \\ 3 t & 0 & 0 \\ 5 t & t & 0 \end{array}\right)+\left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \frac{3}{2} t^{2} & 0 & 0 \end{array}\right)=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 3 t & 1 & 0 \\ \frac{3}{2} t^{2}+5 t & t & 1 \end{array}\right). \end{aligned}$$

\(\operatorname{Let} \mathbf{X}=\left(\begin{array}{l}x \\ y \\\ z\end{array}\right) .\) Then \(\mathbf{X}^{\prime}=\left(\begin{array}{rrr}1 & -1 & 0 \\ 1 & 0 & 2 \\ -1 & 0 & 1\end{array}\right) \mathbf{X}\)

The system is $$\mathbf{X}^{\prime}=\left(\begin{array}{rr} 10 & -5 \\ 8 & -12 \end{array}\right) \mathbf{X}$$ and \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=(\lambda-8)(\lambda+10)=0 .\) For \(\lambda_{1}=8\) we obtain $$\left(\begin{array}{cc|c} 2 & -5 & 0 \\ 8 & -20 & 0 \end{array}\right) \Longrightarrow\left(\begin{array}{rr|r} 1 & -\frac{5}{2} & 0 \\ 0 & 0 & 0 \end{array}\right) \quad \text { so that } \quad \mathbf{K}_{1}=\left(\begin{array}{l} 5 \\ 2 \end{array}\right)$$ For \(\lambda_{2}=-10\) we obtain $$\left(\begin{array}{rr} 20 & -5 & 0 \\ 8 & -2 & 0 \end{array}\right) \Longrightarrow\left(\begin{array}{rr|r} 1 & -\frac{1}{4} & 0 \\ 0 & 0 & 0 \end{array}\right) \quad \text { so that } \quad \mathbf{K}_{2}=\left(\begin{array}{l} 1 \\ 4 \end{array}\right)$$ Then $$\mathbf{X}=c_{1}\left(\begin{array}{l} 5 \\ 2 \end{array}\right) e^{8 t}+c_{2}\left(\begin{array}{l} 1 \\ 4 \end{array}\right) e^{-10 t}$$

The system is $$\mathbf{X}^{\prime}=\left(\begin{array}{ll} -6 & 2 \\ -3 & 1 \end{array}\right) \mathbf{X}$$ and \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=\lambda(\lambda+5)=0 .\) For \(\lambda_{1}=0\) we obtain $$\left(\begin{array}{rr|r} -6 & 2 & 0 \\ -3 & 1 & 0 \end{array}\right) \Longrightarrow\left(\begin{array}{rr|r} 1 & -\frac{1}{3} & 0 \\ 0 & 0 & 0 \end{array}\right) \quad \text { so that } \quad \mathbf{K}_{1}=\left(\begin{array}{l} 1 \\ 3 \end{array}\right)$$ For \(\lambda_{2}=-5\) we obtain $$\left(\begin{array}{rr|r} -1 & 2 & 0 \\ -3 & 6 & 0 \end{array}\right) \Longrightarrow\left(\begin{array}{rr|r} 1 & -2 & 0 \\ 0 & 0 & 0 \end{array}\right) \quad \text { so that } \quad \mathbf{K}_{2}=\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ Then $$\mathbf{X}=c_{1}\left(\begin{array}{l} 1 \\ 3 \end{array}\right)+c_{2}\left(\begin{array}{l} 2 \\ 1 \end{array}\right) e^{-5 t}$$

Solving $$\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=\left|\begin{array}{ccc}1-\lambda & 1 & 1 \\ 0 & 2-\lambda & 3 \\\0 & 0 & 5-\lambda\end{array}\right|=(1-\lambda)(2-\lambda)(5-\lambda)=0$$ we obtain the eigenvalues \(\lambda_{1}=1, \lambda_{2}=2,\) and \(\lambda_{3}=5 .\) Corresponding eigenvectors are $$\mathbf{K}_{1}=\left(\begin{array}{l}1 \\\0 \\\0\end{array}\right), \quad \mathbf{K}_{2}=\left(\begin{array}{l}1 \\\1 \\\0\end{array}\right) \quad \text { and } \quad \mathbf{K}_{3}=\left(\begin{array}{l}1 \\\2 \\\2\end{array}\right).$$ Thus $$\mathbf{X}_{c}=C_{1}\left(\begin{array}{l}1 \\\0 \\\0\end{array}\right) e^{t}+C_{2}\left(\begin{array}{l} 1 \\\1 \\\0\end{array}\right) e^{2 t}+C_{3}\left(\begin{array}{l}1 \\\2 \\\2\end{array}\right) e^{5 t}$$ Substituting $$\mathbf{X}_{p}=\left(\begin{array}{l}a_{1} \\\b_{1} \\\c_{1}\end{array}\right) e^{4 t}$$ into the system yields $$\begin{aligned}-3 a_{1}+b_{1}+c_{1} &=-1 \\\\-2 b_{1}+3 c_{1} &=1 \\\c_{1} &=-2 \end{aligned}.$$

The partial differential equation is \(k \frac{\partial^{2} u}{\partial x^{2}}-h u=\frac{\partial u}{\partial t}\). Substituting \(u(x, t)=v(x, t)+\psi(x)\) gives \(k \frac{\partial^{2} v}{\partial x^{2}}+k \psi^{\prime \prime}-h v-h \psi=\frac{\partial v}{\partial t}\). This equation will be homogeneous provided \(\psi\) satisfies \(k \psi^{\prime \prime}-h \psi=0\). Assuming \(h>0\) and \(k>0,\) we have \(\psi=c_{1} e^{\sqrt{h / k} x}+c_{2} e^{-\sqrt{h / k} x}\), where we have used the exponential form of the solution since the rod is infinite. Now, in order that the steady-state temperature \(\psi(x)\) be bounded as \(x \rightarrow \infty,\) we require \(c_{1}=0 .\) Then \(\psi(x)=c_{2} e^{-\sqrt{h / k} x}\) and \(\psi(0)=u_{0}\) implies \(c_{2}=u_{0} .\) Thus \(\psi(x)=u_{0} e^{-\sqrt{h / k} x}\).

The system is $$\mathbf{X}^{\prime}=\left(\begin{array}{rrr} 2 & -7 & 0 \\ 5 & 10 & 4 \\ 0 & 5 & 2 \end{array}\right) \mathbf{X}$$ and \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=(2-\lambda)(\lambda-5)(\lambda-7)=0 .\) For \(\lambda_{1}=2, \lambda_{2}=5,\) and \(\lambda_{3}=7\) we obtain $$\mathbf{K}_{1}=\left(\begin{array}{r} 4 \\ 0 \\ -5 \end{array}\right), \quad \mathbf{K}_{2}=\left(\begin{array}{r} -7 \\ 3 \\ 5 \end{array}\right), \quad \text { and } \quad \mathbf{K}_{3}=\left(\begin{array}{r} -7 \\ 5 \\ 5 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{r} 4 \\ 0 \\ -5 \end{array}\right) e^{2 t}+c_{2}\left(\begin{array}{r} -7 \\ 3 \\ 5 \end{array}\right) e^{5 t}+c_{3}\left(\begin{array}{r} -7 \\ 5 \\ 5 \end{array}\right) e^{7 t}$$

Substituting \(u(x, t)=v(x, t)+\psi(x)\) into the partial differential equation gives \(a^{2} \frac{\partial^{2} v}{\partial x^{2}}+a^{2} \psi^{\prime \prime}+A x=\frac{\partial^{2} v}{\partial t^{2}}\). This equation will be homogeneous provided \(\psi\) satisfies \(a^{2} \psi^{\prime \prime}+A x=0\). The solution of this differential equation is \(\psi(x)=-\frac{A}{6 a^{2}} x^{3}+c_{1} x+c_{2}\). From \(\psi(0)=0\) we obtain \(c_{2}=0,\) and from \(\psi(1)=0\) we obtain \(c_{1}=A / 6 a^{2} .\) Hence \(\psi(x)=\frac{A}{6 a^{2}}\left(x-x^{3}\right)\). Now the new problem is \(a^{2} \frac{\partial^{2} v}{\partial x^{2}}=\frac{\partial^{2} v}{\partial t^{2}}\) \(v(0, t)=0, \quad v(1, t)=0, \quad t>0\), \(v(x, 0)=-\psi(x), \quad v_{t}(x, 0)=0, \quad 0

We solve\(a^{2} \frac{\partial^{2} u}{\partial x^{2}}-g=\frac{\partial^{2} u}{\partial t^{2}}, \quad 00\) \(u(0, t)=0, \quad u(1, t)=0, \quad t>0\) \(u(x, 0)=0,\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=0, \quad 00\) \(v(0, t)=0, \quad v(1, t)=0\) \(v(x, 0)=\frac{g}{2 a^{2}}\left(x-x^{2}\right),\left.\quad \frac{\partial v}{\partial t}\right|_{t=0}=0\). Substituting \(v=X T\) we find in the usual manner \(X^{\prime \prime}+\alpha^{2} X=0\) \(T^{\prime \prime}+a^{2} \alpha^{2} T=0\) with solutions \(\begin{aligned} X(x) &=c_{3} \cos \alpha x+c_{4} \sin \alpha x \\ T(t) &=c_{5} \cos a \alpha t+c_{6} \sin a \alpha t \end{aligned}\). The conditions \(X(0)=0\) and \(X(1)=0\) imply in turn that \(c_{3}=0\) and \(\alpha=n \pi\) for \(n=1,2,3, \ldots .\) The condition \(T^{\prime}(0)=0\) implies \(c_{6}=0 .\) Hence, by the superposition principle \(v(x, t)=\sum_{n=1}^{\infty} A_{n} \cos (a n \pi t) \sin (n \pi x)\). At \(t=0\), \(\frac{g}{2 a^{2}}\left(x-x^{2}\right)=\sum_{n=1}^{\infty} A_{n} \sin (n \pi x)\) and so \(A_{n}=\frac{g}{a^{2}} \int_{0}^{1}\left(x-x^{2}\right) \sin (n \pi x) d x=\frac{2 g}{a^{2} n^{3} \pi^{3}}\left[1-(-1)^{n}\right]\). Thus the solution to the original problem is \(u(x, t)=\psi(x)+v(x, t)=\frac{g}{2 a^{2}}\left(x^{2}-x\right)+\frac{2 g}{a^{2} \pi^{3}} \sum_{n=1}^{\infty} \frac{1-(-1)^{n}}{n^{3}} \cos (a n \pi t) \sin (n \pi x)\).

We have \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=-\lambda(\lambda-1)(\lambda-2)=0 .\) For \(\lambda_{1}=0, \lambda_{2}=1,\) and \(\lambda_{3}=2\) we obtain $$\mathbf{K}_{1}=\left(\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right), \quad \mathbf{K}_{2}=\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right), \quad \text { and } \quad \mathbf{K}_{3}=\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right)+c_{2}\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) e^{t}+c_{3}\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right) e^{2 t}$$

(a) since \(\mathbf{M}=\left(\begin{array}{cc}m_{1} & 0 \\ 0 & m_{2}\end{array}\right)\) is a diagonal matrix with nonzero diagonal entries, it has an inverse. Writing the system in the form $$\begin{array}{c} m_{1} x_{1}^{\prime \prime}+\left(k_{1}+k_{2}\right) x_{1}-k_{2} x_{2}=0 \\ m_{2} x_{2}^{\prime \prime}-k_{2} x_{1}+k_{2} x_{2}=0 \end{array}$$ we see that \(\mathbf{K}=\left(\begin{array}{cc}k_{1}+k_{2} & -k_{2} \\ -k_{2} & k_{2}\end{array}\right)\). (b) since \(\mathbf{M}\) has an inverse, \(\mathbf{M X}^{\prime \prime}+\mathbf{K X}=\mathbf{0}\) can be written as \(\mathbf{X}^{\prime \prime}+\mathbf{M}^{-1} \mathbf{K} \mathbf{X}=\mathbf{0}\) or \(\mathbf{X}^{\prime \prime}+\mathbf{B} \mathbf{X}=\mathbf{0}\) where $$\mathbf{B}=\mathbf{M}^{-1} \mathbf{K}=\left(\begin{array}{cc} \frac{1}{m_{1}} & 0 \\ 0 & \frac{1}{m_{2}} \end{array}\right)\left(\begin{array}{cc} k_{1}+k_{2} & -k_{2} \\ -k_{2} & k_{2} \end{array}\right)=\left(\begin{array}{cc} \frac{k_{1}+k_{2}}{m_{1}} & -\frac{k_{2}}{m_{1}} \\ -\frac{k_{2}}{m_{2}} & \frac{k_{2}}{m_{2}} \end{array}\right)$$ (c) With \(m_{1}=1, m_{2}=1, k_{1}=3,\) and \(k_{2}=2\) we have \(\mathbf{B}=\left(\begin{array}{rr}5 & -2 \\ -2 & 2\end{array}\right) .\) The eigenvalues of \(\mathbf{B}\) are \(\lambda_{1}=1\) and \(\lambda_{2}=6\) with corresponding eigenvectors \(\left(\begin{array}{c}1 \\ 2\end{array}\right)\) and \(\left(\begin{array}{r}-2 \\ 1\end{array}\right) .\) Letting \(\mathbf{X}=\mathbf{P} \mathbf{Y}\) the system can be written \(\mathbf{P Y}^{\prime \prime}+\mathbf{B P Y}=\mathbf{0}\) or \(\mathbf{Y}^{\prime \prime}+\mathbf{P}^{-1} \mathbf{B} \mathbf{P} \mathbf{Y}=\mathbf{0}\) where \(\left(\begin{array}{rr}1 & -2 \\ 2 & 1\end{array}\right)\) and \(\mathbf{P}^{-1} \mathbf{B} \mathbf{P}=\left(\begin{array}{ll}1 & 0 \\ 0 & 6\end{array}\right) .\) The system is then \(\mathbf{Y}^{\prime \prime}+\left(\begin{array}{ll}1 & 0 \\ 0 & 6\end{array}\right) \mathbf{Y}=\mathbf{0},\) which is uncoupled and equivalent to \(y_{1}^{\prime \prime}+y_{1}=0\) and \(y_{2}^{\prime \prime}+6 y_{2}=0 .\) The solutions are \(y_{1}=c_{1} \cos t+c_{2} \sin t\) and \(y_{2}=c_{3} \cos \sqrt{6} t+c_{4} \sin \sqrt{6} t\) (d) From $$\mathbf{X}=\mathbf{P} \mathbf{Y}=\left(\begin{array}{rr} 1 & -2 \\ 2 & 1 \end{array}\right)\left(\begin{array}{l} y_{1} \\ y_{2} \end{array}\right)=\left(\begin{array}{l} y_{1}-2 y_{2} \\ 2 y_{1}+y_{2} \end{array}\right)$$ we have $$\begin{array}{l} x_{1}=c_{1} \cos t+c_{2} \sin t-2 c_{3} \cos \sqrt{6} t-2 c_{4} \sin \sqrt{6} t \\ x_{1}=2 c_{1} \cos t+2 c_{2} \sin t+c_{3} \cos \sqrt{6} t+c_{4} \sin \sqrt{6} t \end{array}$$ which is the same as $$\mathbf{X}=c_{1}\left(\begin{array}{l} 1 \\ 2 \end{array}\right) \cos t+c_{2}\left(\begin{array}{l} 1 \\ 2 \end{array}\right) \sin t+c_{3}\left(\begin{array}{r} -2 \\ 1 \end{array}\right) \cos \sqrt{6} t+c_{4}\left(\begin{array}{r} -2 \\ 1 \end{array}\right) \sin \sqrt{6} t$$.

since $$\mathbf{X}^{\prime}=\left(\begin{array}{c} -5 \\ -10 \end{array}\right) e^{-5 t} \quad \text { and } \quad\left(\begin{array}{rr} 3 & -4 \\ 4 & -7 \end{array}\right) \mathbf{X}=\left(\begin{array}{c} -5 \\ -10 \end{array}\right) e^{-5 t}$$ we see that $$\mathbf{X}^{\prime}=\left(\begin{array}{ll} 3 & -4 \\ 4 & -7 \end{array}\right) \mathbf{X}$$

since \\[ \mathbf{X}^{\prime}=\left(\begin{array}{c} 5 \cos t-5 \sin t \\ 2 \cos t-4 \sin t \end{array}\right) e^{t} \quad \text { and } \quad\left(\begin{array}{rr} -2 & 5 \\ -2 & 4 \end{array}\right) \mathbf{X}=\left(\begin{array}{c} 5 \cos t-5 \sin t \\ 2 \cos t-4 \sin t \end{array}\right) e^{t} \\] we see that \\[ \mathbf{X}^{\prime}=\left(\begin{array}{ll} -2 & 5 \\ -2 & 4 \end{array}\right) \mathbf{X} \\]

Identifying \(k=1\) and \(L=\pi\) we see that the eigenfunctions of \(X^{\prime \prime}+\lambda X=0, X(0)=0, X(\pi)=0\) are \(\sin n x\), \(n=1,2,3, \ldots .\) Assuming that \(u(x, t)=\sum_{n=1}^{\infty} u_{n}(t) \sin n x,\) the formal partial derivatives of \(u\) are \(\frac{\partial^{2} u}{\partial x^{2}}=\sum_{n=1}^{\infty} u_{n}(t)\left(-n^{2}\right) \sin n x \quad\) and \(\quad \frac{\partial u}{\partial t}=\sum_{n=1}^{\infty} u_{n}^{\prime}(t) \sin n x\). Assuming that \(x e^{-3 t}=\sum_{n=1}^{\infty} F_{n}(t) \sin n x\) we have \(F_{n}(t)=\frac{2}{\pi} \int_{0}^{\pi} x e^{-3 t} \sin n x d x=\frac{2 e^{-3 t}}{\pi} \int_{0}^{\pi} x \sin n x d x=\frac{2 e^{-3 t}(-1)^{n+1}}{n}\). Then \(x e^{-3 t}=\sum_{n=1}^{\infty} \frac{2 e^{-3 t}(-1)^{n+1}}{n} \sin n x\) and \(u_{t}-u_{x x}=\sum_{n=1}^{\infty}\left[u_{n}^{\prime}(t)+n^{2} u_{n}(t)\right] \sin n x=x e^{-3 t}=\sum_{n=1}^{\infty} \frac{2 e^{-3 t}(-1)^{n+1}}{n} \sin n x\). Equating coefficients we obtain \(u_{n}^{\prime}(t)+n^{2} u_{n}(t)=\frac{2 e^{-3 t}(-1)^{n+1}}{n}\). This is a linear first-order differential equation whose solution is \(u_{n}(t)=\frac{2(-1)^{n+1}}{n\left(n^{2}-3\right)} e^{-3 t}+C_{n} e^{-n^{2} t}\). Thus \(u(x, t)=\sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n\left(n^{2}-3\right)} e^{-3 t} \sin n x+\sum_{n=1}^{\infty} C_{n} e^{-n^{2} t} \sin n x\) and \(u(x, 0)=0\) implies \(\sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n\left(n^{2}-3\right)} \sin n x+\sum_{n=1}^{\infty} C_{n} \sin n x=0\) so that \(C_{n}=2(-1)^{n} / n\left(n^{2}-3\right) .\) Therefore \(u(x, t)=2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n\left(n^{2}-3\right)} e^{-3 t} \sin n x+2 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n\left(n^{2}-3\right)} e^{-n^{2} t} \sin n x\).

We have \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=(\lambda+1 / 2)(\lambda-1 / 2)=0 .\) For \(\lambda_{1}=-1 / 2\) and \(\lambda_{2}=1 / 2\) we obtain $$\mathbf{K}_{1}=\left(\begin{array}{l} 0 \\ 1 \end{array}\right) \quad \text { and } \quad \mathbf{K}_{2}=\left(\begin{array}{l} 1 \\ 1 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{l} 0 \\ 1 \end{array}\right) e^{-t / 2}+c_{2}\left(\begin{array}{l} 1 \\ 1 \end{array}\right) e^{t / 2}$$ If $$\mathbf{X}(0)=\left(\begin{array}{l} 3 \\ 5 \end{array}\right)$$ then \(c_{1}=2\) and \(c_{2}=3\)

We have $$\mathbf{X}(0)=c_{1}\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right)+c_{2}\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)+c_{3}\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right)=\left(\begin{array}{l} c_{1} \\ c_{2} \\ c_{3} \end{array}\right)=\left(\begin{array}{r} 1 \\ -4 \\ 6 \end{array}\right).$$ Thus, the solution of the initial-value problem is $$\mathbf{X}=\left(\begin{array}{c} t+1 \\ t \\ -2 t \end{array}\right)-4\left(\begin{array}{c} t \\ t+1 \\ -2 t \end{array}\right)+6\left(\begin{array}{c} t \\ t \\ -2 t+1 \end{array}\right).$$

since \\[ \mathbf{X}^{\prime}=\left(\begin{array}{r} \frac{3}{2} \\ -3 \end{array}\right) e^{-3 t / 2} \quad \text { and } \quad\left(\begin{array}{rr} -1 & \frac{1}{4} \\ 1 & -1 \end{array}\right) \mathbf{X}=\left(\begin{array}{r} \frac{3}{2} \\ -3 \end{array}\right) e^{-3 t / 2} \\] we see that \\[ \mathbf{X}^{\prime}=\left(\begin{array}{rr} -1 & 1 / 4 \\ 1 & -1 \end{array}\right) \mathbf{X} \\]

We have \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=(2-\lambda)(\lambda-3)(\lambda+1)=0 .\) For \(\lambda_{1}=2, \lambda_{2}=3,\) and \(\lambda_{3}=-1\) we obtain $$\mathbf{K}_{1}=\left(\begin{array}{r} 5 \\ -3 \\ 2 \end{array}\right), \quad \mathbf{K}_{2}=\left(\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right), \quad \text { and } \quad \mathbf{K}_{3}=\left(\begin{array}{r} -2 \\ 0 \\ 1 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{r} 5 \\ -3 \\ 2 \end{array}\right) e^{2 t}+c_{2}\left(\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right) e^{3 t}+c_{3}\left(\begin{array}{r} -2 \\ 0 \\ 1 \end{array}\right) e^{-t}$$ If $$\mathbf{X}(0)=\left(\begin{array}{l} 1 \\ 3 \\ 0 \end{array}\right)$$ then \(c_{1}=-1, c_{2}=5 / 2,\) and \(c_{3}=-1 / 2\)

Identifying \(k=1\) and \(L=1\) we see that the eigenfunctions of \(X^{\prime \prime}+\lambda X=0, X(0)=0, X(1)=0\) are \(\sin n \pi x\), \(n=1,2,3, \ldots .\) Assuming that \(u(x, t)=\sum_{n=1}^{\infty} u_{n}(t) \sin n \pi x,\) the formal partial derivatives of \(u\) are \\[\frac{\partial^{2} u}{\partial x^{2}}=\sum_{n=1}^{\infty} u_{n}(t)\left(-n^{2} \pi^{2}\right) \sin n \pi x \quad \text { and } \quad \frac{\partial u}{\partial t}=\sum_{n=1}^{\infty} u_{n}^{\prime}(t) \sin n \pi x\\]. Assuming that \(-1+x-x \cos t=\sum_{n=1}^{\infty} F_{n}(t) \sin n \pi x\) we have\\[ F_{n}(t)=\frac{2}{1} \int_{0}^{1}(-1+x-x \cos t) \sin n \pi x d x=\frac{2\left[-1+(-1)^{n} \cos t\right]}{n \pi}\\]. Then \\[-1+x-x \cos t=\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{-1+(-1)^{n} \cos t}{n} \sin n \pi x\\] and\\[u_{t}-u_{x x}=\sum_{n=1}^{\infty}\left[u_{n}^{\prime}(t)+n^{2} \pi^{2} u_{n}(t)\right] \sin n \pi x\\] \(=-1+x-x \cos t=\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{-1+(-1)^{n} \cos t}{n} \sin n \pi x\). Equating coefficients we obtain\\[u_{n}^{\prime}(t)+n^{2} \pi^{2} u_{n}(t)=\frac{2\left[-1+(-1)^{n} \cos t\right]}{n \pi}\\]. This is a linear first-order differential equation whose solution is\\[u_{n}(t)=\frac{2}{n \pi}\left[-\frac{1}{n^{2} \pi^{2}}+(-1)^{n} \frac{n^{2} \pi^{2} \cos t+\sin t}{n^{4} \pi^{4}+1}\right]+C_{n} e^{-n^{2} \pi^{2} t}\\] .Thus\\[u(x, t)=\sum_{n=1}^{\infty} \frac{2}{n \pi}\left[-\frac{1}{n^{2} \pi^{2}}+(-1)^{n} \frac{n^{2} \pi^{2} \cos t+\sin t}{n^{4} \pi^{4}+1}\right] \sin n \pi x+\sum_{n=1}^{\infty} C_{n} e^{-n^{2} \pi^{2} t} \sin n \pi x\\] and \(u(x, 0)=x(1-x)\) implies\\[\sum_{n=1}^{\infty} \frac{2}{n \pi}\left[-\frac{1}{n^{2} \pi^{2}}+(-1)^{n} \frac{n^{2} \pi^{2}}{n^{4} \pi^{4}+1}+C_{n}\right] \sin n \pi x=x(1x)\\]. Hence\\[\frac{2}{n \pi}\left[-\frac{1}{n^{2} \pi^{2}}+(-1)^{n} \frac{n^{2} \pi^{2}}{n^{4} \pi^{4}+1}+C_{n}\right]=\frac{2}{1} \int_{0}^{1} x(1-x) \sin n \pi x d x=2\left[\frac{1-(-1)^{n}}{n^{3} \pi^{3}}\right] \\] and\\[C_{n}=\frac{4-2(-1)^{n}}{n^{3} \pi^{3}}-(-1)^{n} \frac{2 n \pi}{n^{4} \pi^{4}+1} \\] Therefore\\[\begin{aligned}u(x, t)=\sum_{n=1}^{\infty} & \frac{2}{n \pi}\left[-\frac{1}{n^{2} \pi^{2}}+(-1)^{n} \frac{n^{2} \pi^{2} \cos t+\sin t}{n^{4} \pi^{4}+1}\right] \sin n \pi x \\\&+\sum_{n=1}^{\infty}\left[\frac{4-2(-1)^{n}}{n^{3} \pi^{3}}-(-1)^{n} \frac{2 n \pi}{n^{4} \pi^{4}+1}\right] e^{-n^{2} \pi^{2} t} \sin n \pi x\end{aligned}\\].

Identifying \(k=1\) and \(L=\pi\) we see that the eigenfunctions of \(X^{\prime \prime}+\lambda X=0, X(0)=0, X(\pi)=0\) are \(\sin n x\), \(n=1,2,3, \ldots .\) Assuming that \(u(x, t)=\sum_{n=1}^{\infty} u_{n}(t) \sin n x,\) the formal partial derivatives of \(u\) are \(\frac{\partial^{2} u}{\partial x^{2}}=\sum_{n=1}^{\infty} u_{n}(t)\left(-n^{2}\right) \sin n x \quad\) and \(\quad \frac{\partial^{2} u}{\partial t^{2}}=\sum_{n=1}^{\infty} u_{n}^{\prime \prime}(t) \sin n x\). Then\\[u_{t t}-u_{x x}=\sum_{n=1}^{\infty}\left[u_{n}^{\prime \prime}(t)+n^{2} u_{n}(t)\right] \sin n x=\cos t \sin x\\]. Equating coefficients, we obtain \(u_{1}^{\prime \prime}(t)+u_{1}(t) \cos t\) and \(u_{n}^{\prime \prime}(t)+n^{2} u_{n}(t)=0\) for \(n=2,3,4, \ldots .\) Solving the first differential equation we obtain \(u_{1}(t)=A_{1} \cos t+B_{1} \sin t+\frac{1}{2} t \sin t .\) From the second differential equation we obtain \(u_{n}(t)=A_{n} \cos n t+B_{n} \sin n t\) for \(n=2,3,4, \ldots .\) Thus \(u(x, t)=\left(A_{1} \cos t+B_{1} \sin t+\frac{1}{2} t \sin t\right) \sin x+\sum_{n=2}^{\infty}\left(A_{n} \cos n t+B_{n} \sin n t\right) \sin n x\). From \(u(x, 0)=A_{1} \sin x+\sum_{n=2}^{\infty} A_{n} \sin n x=0\) we see that \(A_{n}=0\) for \(n=1,2,3, \ldots .\) Thus \(u(x, t)=\left(B_{1} \sin t+\frac{1}{2} t \sin t\right) \sin x+\sum_{n=2}^{\infty} B_{n} \sin n t \sin n x\) and \(\frac{\partial u}{\partial t}=\left(B_{1} \cos t+\frac{1}{2} t \cos t+\frac{1}{2} \sin t\right) \sin x+\sum_{n=2}^{\infty} n B_{n} \cos n t \sin n x\), so \(\left.\frac{\partial u}{\partial t}\right|_{t=0}=B_{1} \sin x+\sum_{n=2}^{\infty} n B_{n} \sin n x=0\). We see that \(B_{n}=0\) for all \(n\) so \(u(x, t)=\frac{1}{2} t \sin t \sin x\).

since \\[ \mathbf{X}^{\prime}=\left(\begin{array}{c} \cos t \\ \frac{1}{2} \sin t-\frac{1}{2} \cos t \\ -\cos t-\sin t \end{array}\right) \quad \text { and } \quad\left(\begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 0 \\ -2 & 0 & -1 \end{array}\right) \mathbf{X}=\left(\begin{array}{c} \cos t \\ \frac{1}{2} \sin t-\frac{1}{2} \cos t \\ -\cos t-\sin t \end{array}\right) \\] we see that \\[ \mathbf{X}^{\prime}=\left(\begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 0 \\ -2 & 0 & -1 \end{array}\right) \mathbf{X} \\]

In the general case the associated Sturm-Liouville problem is \(X^{\prime \prime}+\lambda X=0, \quad X^{\prime}(0)=0, \quad X^{\prime}(L)=0\) with eigenvalues and eigenfunctions \(\lambda_{0}=0, X_{0}=1,\) and \(\lambda_{n}=n^{2} \pi^{2} / L^{2}, X_{n}=\cos n \pi x / L, n=1,2,3, \ldots .\) The entire set of eigenfunctions can be written as \(X_{n}=\cos n \pi x / L, n=0,1,2, \ldots,\) which serves as the basis for the Fourier cosine series. Hence, we assume in this problem that \(u(x, t)=\frac{1}{2} u_{0}(t)+\sum_{n=1}^{\infty} u_{n}(t) \cos \frac{n \pi}{L} x\) and \(F(x, t)=\frac{1}{2} F_{0}(t)+\sum_{n=1}^{\infty} F_{n}(t) \cos \frac{n \pi}{L} x\). Taking \(k=1, L=1, F(x, t)=t x,\) and \(f(x)=0\) we have \(u(x, t)=\frac{1}{2} u_{0}(t)+\sum_{n=1}^{\infty} u_{n}(t) \cos n \pi x\) and \(F(x, t)=\frac{1}{2} F_{0}(t)+\sum_{n=1}^{\infty} F_{n}(t) \cos n \pi x\). By treating \(t\) as a parameter, the coefficients \(F_{n}\) can be computed: \(F_{0}(t)=2 \int_{0}^{1} t x d x=t\) \(F_{n}(t)=2 \int_{0}^{1} t x \cos n \pi x d x=2 t \int_{0}^{1} x \cos n \pi x d x=2 t \frac{(-1)^{n}-1}{n^{2} \pi^{2}}\). Hence \(t x=\frac{1}{2} t+2 t \sum_{n=1}^{\infty} \frac{(-1)^{n}-1}{n^{2} \pi^{2}} \cos n \pi x\). Now, using the series representation for \(u(x, t),\) we have \(\frac{\partial^{2} u}{\partial x^{2}}=\sum_{n=1}^{\infty} u_{n}(t)\left(-n^{2} \pi^{2}\right) \cos n \pi x \quad\) and \(\quad \frac{\partial u}{\partial t}=\frac{1}{2} u_{0}^{\prime}(t)+\sum_{n=1}^{\infty} u_{n}^{\prime}(t) \cos n \pi x\). Writing the partial differential equation as \(u_{t}-u_{x x}=t x\) and using the above results we have \(\frac{1}{2} u_{0}^{\prime}(t)+\sum_{n=1}^{\infty}\left[u_{n}^{\prime}(t)+n^{2} \pi^{2} u_{n}(t)\right] \cos n \pi x=\frac{1}{2} t+2 t \sum_{n=1}^{\infty} \frac{(-1)^{n}-1}{n^{2} \pi^{2}} \cos n \pi x\). Equating coefficients we get \(u_{0}^{\prime}(t)=t \quad\) and \(\quad u_{n}^{\prime}(t)+n^{2} \pi^{2} u_{n}(t)=2 t \frac{(-1)^{n}-1}{n^{2} \pi^{2}}\). From the first equation we obtain \(u_{0}(t)=\frac{1}{2} t^{2}+C_{0} .\) The second equation is a linear, first-order differential equation whose general solution is \(u_{n}(t)=2\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right)\left(\frac{n^{2} \pi^{2} t-1}{n^{4} \pi^{4}}\right)+C_{n} e^{-n^{2} \pi^{2} t}\). Thus \(u(x, t)=\frac{1}{4} t^{2}+\frac{1}{2} C_{0}+\sum_{n=1}^{\infty}\left[2\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right)\left(\frac{n^{2} \pi^{2} t-1}{n^{4} \pi^{4}}\right)+C_{n} e^{-n^{2} \pi^{2} t}\right] \cos n \pi x\). The initial condition \(u(x, 0)=0\) implies \(\frac{1}{2} C_{0}+\sum_{n=1}^{\infty}\left[2\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right)\left(\frac{-1}{n^{4} \pi^{4}}\right)+C_{n}\right] \cos n \pi x=\frac{1}{2} C_{0}+\sum_{n=1}^{\infty}\left[2 \frac{1-(-1)^{n}}{n^{6} \pi^{6}}+C_{n}\right] \cos n \pi x=0\) so that \(C_{0}=0 \quad\) and \(\quad C_{n}=2 \frac{(-1)^{n}-1}{n^{6} \pi^{6}}\). Therefore \(\begin{aligned} u(x, t) &=\frac{1}{4} t^{2}+\sum_{n=1}^{\infty}\left[2\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right)\left(\frac{n^{2} \pi^{2} t-1}{n^{4} \pi^{4}}\right)+2 \frac{(-1)^{n}-1}{n^{6} \pi^{6}} e^{-n^{2} \pi^{2} t}\right] \cos n \pi x \\\ &=\frac{1}{4} t^{2}+\frac{2}{\pi^{6}} \sum_{n=1}^{\infty}\left[\frac{\left[(-1)^{n}-1\right]\left[n^{2} \pi^{2} t-1+e^{-n^{2} \pi^{2} t}\right]}{n^{6}}\right] \cos n \pi x \end{aligned}\).

Yes, since \(W\left(\mathbf{X}_{1}, \mathbf{X}_{2}, \mathbf{X}_{3}\right)=-84 e^{-t} \neq 0\) the set \(\mathbf{X}_{1}, \mathbf{X}_{2}, \mathbf{X}_{3}\) is linearly independent on \(-\infty

The eigenvalues are \(\lambda_{1}=\frac{1}{4}\) and \(\lambda_{2}=\frac{1}{2} .\) This leads to the system $$\begin{array}{l} e^{t / 4}=b_{0}+\frac{1}{4} b_{1} \\ e^{t / 2}=b_{0}+\frac{1}{2} b_{1} \end{array}$$ which has the solution \(b_{0}=2 e^{t / 4}+e^{t / 2}\) and \(b_{1}=-4 e^{t / 4}+4 e^{t / 2} .\) Then $$e^{\mathbf{A} t}=b_{0} \mathbf{I}+b_{1} \mathbf{A}=\left(\begin{array}{cc} -2 e^{t / 4}+3 e^{t / 2} & 6 e^{t / 4}-6 e^{t / 2} \\ -e^{t / 4}+e^{t / 2} & 3 e^{t / 4}-2 e^{t / 2} \end{array}\right)$$ The general solution of the system is then $$\begin{aligned} \mathbf{X}=e^{\mathbf{A} t} \mathbf{C} &=\left(\begin{array}{cc} -2 e^{t / 4}+3 e^{t / 2} & 6 e^{t / 4}-6 e^{t / 2} \\ -e^{t / 4}+e^{t / 2} & 3 e^{t / 4}-2 e^{t / 2} \end{array}\right)\left(\begin{array}{c} c_{1} \\ c_{2} \end{array}\right) \\ &=c_{1}\left(\begin{array}{c} -2 \\ -1 \end{array}\right) e^{t / 4}+c_{1}\left(\begin{array}{c} 3 \\ 1 \end{array}\right) e^{t / 2}+c_{2}\left(\begin{array}{c} 6 \\ 3 \end{array}\right) e^{t / 4}+c_{2}\left(\begin{array}{c} -6 \\ -2 \end{array}\right) e^{t / 2} \\ &=\left(-c_{1}+3 c_{2}\right)\left(\begin{array}{c} 2 \\ 1 \end{array}\right) e^{t / 4}+\left(c_{1}-2 c_{2}\right)\left(\begin{array}{c} 3 \\ 1 \end{array}\right) e^{t / 2} \\ &=c_{3}\left(\begin{array}{c} 2 \\ 1 \end{array}\right) e^{t / 4}+c_{4}\left(\begin{array}{c} 3 \\ 1 \end{array}\right) e^{t / 2}. \end{aligned}$$

since \\[ \mathbf{X}_{p}^{\prime}=\left(\begin{array}{l} 0 \\ 0 \end{array}\right) \quad \text { and } \quad\left(\begin{array}{rr} 2 & 1 \\ 1 & -1 \end{array}\right) \mathbf{X}_{p}+\left(\begin{array}{r} -5 \\ 2 \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \end{array}\right) \\] we see that \\[ \mathbf{X}_{p}^{\prime}=\left(\begin{array}{rr} 2 & 1 \\ 1 & -1 \end{array}\right) \mathbf{X}_{p}+\left(\begin{array}{r} -5 \\ 2 \end{array}\right) \\]

since \\[ \mathbf{X}_{p}^{\prime}=\left(\begin{array}{c} 3 \cos 3 t \\ 0 \\ -3 \sin 3 t \end{array}\right) \quad \text { and } \quad\left(\begin{array}{rrr} 1 & 2 & 3 \\ -4 & 2 & 0 \\ -6 & 1 & 0 \end{array}\right) \mathbf{X}_{p}+\left(\begin{array}{r} -1 \\ 4 \\ 3 \end{array}\right) \sin 3 t=\left(\begin{array}{c} 3 \cos 3 t \\ 0 \\ -3 \sin 3 t \end{array}\right) \\] we see that \\[ \mathbf{X}_{p}^{\prime}=\left(\begin{array}{rrr} 1 & 2 & 3 \\ -4 & 2 & 0 \\ -6 & 1 & 0 \end{array}\right) \mathbf{X}_{p}+\left(\begin{array}{r} -1 \\ 4 \\ 3 \end{array}\right) \sin 3 t \\]

We have det \((\mathbf{A}-\lambda \mathbf{I})=-\lambda(5-\lambda)^{2}=0 .\) For \(\lambda_{1}=0\) we obtain $$\mathbf{K}_{1}=\left(\begin{array}{r} -4 \\ -5 \\ 2 \end{array}\right)$$ For \(\lambda_{2}=5\) we obtain $$\mathbf{K}=\left(\begin{array}{r} -2 \\ 0 \\ 1 \end{array}\right)$$ A solution of \(\left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{P}=\mathbf{K}\) is $$\mathbf{P}=\left(\begin{array}{c} \frac{5}{2} \\ \frac{1}{2} \\ 0 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{r} -4 \\ -5 \\ 2 \end{array}\right)+c_{2}\left(\begin{array}{r} -2 \\ 0 \\ 1 \end{array}\right) e^{5 t}+c_{3}\left[\left(\begin{array}{r} -2 \\ 0 \\ 1 \end{array}\right) t e^{5 t}+\left(\begin{array}{c} \frac{5}{2} \\ \frac{1}{2} \\ 0 \end{array}\right) e^{5 t}\right]$$

Let \\[ \begin{array}{l} \mathbf{X}_{1}=\left(\begin{array}{c} 1 \\ -1-\sqrt{2} \end{array}\right) e^{\sqrt{2} t} \\ \mathbf{X}_{2}=\left(\begin{array}{c} 1 \\ -1+\sqrt{2} \end{array}\right) e^{-\sqrt{2} t} \end{array} \\] $$\mathbf{X}_{p}=\left(\begin{array}{l} 1 \\ 0 \end{array}\right) t^{2}+\left(\begin{array}{r} -2 \\ 4 \end{array}\right) t+\left(\begin{array}{l} 1 \\ 0 \end{array}\right)$$ and \\[ \mathbf{A}=\left(\begin{array}{rr} -1 & -1 \\ -1 & 1 \end{array}\right) \\] Then \\[ \begin{array}{l} \mathbf{X}_{1}^{\prime}=\left(\begin{array}{c} \sqrt{2} \\ -2-\sqrt{2} \end{array}\right) e^{\sqrt{2} t}=\mathbf{A} \mathbf{X}_{1} \\ \mathbf{X}_{2}^{\prime}=\left(\begin{array}{c} -\sqrt{2} \\ -2+\sqrt{2} \end{array}\right) e^{-\sqrt{2} t}=\mathbf{A} \mathbf{X}_{2} \\ \mathbf{X}_{p}^{\prime}=\left(\begin{array}{c} 2 \\ 0 \end{array}\right) t+\left(\begin{array}{r} -2 \\ 4 \end{array}\right)=\mathbf{A} \mathbf{X}_{p}+\left(\begin{array}{c} 1 \\ 1 \end{array}\right) t^{2}+\left(\begin{array}{r} 4 \\ -6 \end{array}\right) t+\left(\begin{array}{r} -1 \\ 5 \end{array}\right) \end{array} \\] and \(W\left(\mathbf{X}_{1}, \mathbf{X}_{2}\right)=2 \sqrt{2} \neq 0\) so that \(\mathbf{X}_{p}\) is a particular solution and \(\mathbf{X}_{1}\) and \(\mathbf{X}_{2}\) form a fundamental set on \(-\infty

(a) The following commands can be used in Mathematica: \(\mathbf{A}=\\{\\{4,2\\},\\{3,3\\}\\}\) \(\mathbf{c}=\\{\mathbf{c} 1, \mathbf{c} 2\\}\) \(\mathrm{m}=\) MatrixExp \([\mathrm{A} \mathrm{t}]\) sol = Expand [m.c] Collect[sol, \(\\{\mathrm{c} 1, \mathrm{c} 2\\}] /\) /MatrixForm The output gives $$\begin{array}{l} x(t)=c_{1}\left(\frac{2}{5} e^{t}+\frac{3}{5} e^{6 t}\right)+c_{2}\left(-\frac{2}{5} e^{t}+\frac{2}{5} e^{6 t}\right) \\ y(t)=c_{1}\left(-\frac{3}{5} e^{t}+\frac{3}{5} e^{6 t}\right)+c_{2}\left(\frac{3}{5} e^{t}+\frac{2}{5} e^{6 t}\right). \end{array}$$ The eigenvalues are 1 and 6 with corresponding eigenvectors $$\left(\begin{array}{r} -2 \\ 3 \end{array}\right) \quad \text { and } \quad\left(\begin{array}{l} 1 \\ 1 \end{array}\right),$$ so the solution of the system is $$\mathbf{X}(t)=b_{1}\left(\begin{array}{r} -2 \\ 3 \end{array}\right) e^{t}+b_{2}\left(\begin{array}{l} 1 \\ 1 \end{array}\right) e^{6 t}$$ or $$\begin{array}{l} x(t)=-2 b_{1} e^{t}+b_{2} e^{6 t} \\ y(t)=3 b_{1} e^{t}+b_{2} e^{6 t}. \end{array}$$ If we replace \(b_{1}\) with \(-\frac{1}{5} c_{1}+\frac{1}{5} c_{2}\) and \(b_{2}\) with \(\frac{3}{5} c_{1}+\frac{2}{5} c_{2},\) we obtain the solution found using the matrix exponential. (b) \(x(t)=c_{1} e^{-2 t} \cos t-\left(c_{1}+c_{2}\right) e^{-2 t} \sin t\) \(y(t)=c_{2} e^{-2 t} \cos t+\left(2 c_{1}+c_{2}\right) e^{-2 t} \sin t\)

We have det \((\mathbf{A}-\lambda \mathbf{I})=(\lambda-4)^{3}=0 .\) For \(\lambda_{1}=4\) we obtain $$\mathbf{K}=\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right)$$ Solutions of \(\left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{P}=\mathbf{K}\) and \(\left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{Q}=\mathbf{P}\) are $$\mathbf{P}=\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) \quad \text { and } \quad \mathbf{Q}=\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right) e^{4 t}+c_{2}\left[\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right) t e^{4 t}+\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) e^{4 t}\right]+c_{3}\left[\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right) \frac{t^{2}}{2} e^{4 t}+\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) t e^{4 t}+\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right) e^{4 t}\right]$$

We have det \((\mathbf{A}-\lambda \mathbf{I})=(\lambda-4)^{2}=0 .\) For \(\lambda_{1}=4\) we obtain $$\mathbf{K}=\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ A solution of \(\left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{P}=\mathbf{K}\) is $$\mathbf{P}=\left(\begin{array}{l} 1 \\ 1 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{l} 2 \\ 1 \end{array}\right) e^{4 t}+c_{2}\left[\left(\begin{array}{l} 2 \\ 1 \end{array}\right) t e^{4 t}+\left(\begin{array}{l} 1 \\ 1 \end{array}\right) e^{4 t}\right]$$ If$$\mathbf{X}(0)=\left(\begin{array}{r} -1 \\ 6 \end{array}\right)$$ then \(c_{1}=-7\) and \(c_{2}=13\)

Since \(\mathbf{A}^{3}=\mathbf{0}, \mathbf{A}\) is nilpotent. Since $$e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t+\mathbf{A}^{2} \frac{t^{2}}{2 !}+\cdots+\mathbf{A}^{k} \frac{t^{k}}{k !}+\cdots,$$ if \(\mathbf{A}\) is nilpotent and \(\mathbf{A}^{m}=\mathbf{0},\) then \(\mathbf{A}^{k}=\mathbf{0}\) for \(k \geq m\) and $$e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t+\mathbf{A}^{2} \frac{t^{2}}{2 !}+\cdots+\mathbf{A}^{m-1} \frac{t^{m-1}}{(m-1) !}.$$ In this problem \(\mathbf{A}^{3}=\mathbf{0},\) so $$\begin{aligned} e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t+\mathbf{A}^{2} \frac{t^{2}}{2} &=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)+\left(\begin{array}{rrr} -1 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right) t+\left(\begin{array}{rrr} -1 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{array}\right) \frac{t^{2}}{2} \\ &=\left(\begin{array}{ccc} 1-t-t^{2} / 2 & t & t+t^{2} / 2 \\ -t & 1 & t \\ -t-t^{2} / 2 & t & 1+t+t^{2} / 2 \end{array}\right) \end{aligned}$$ and the solution of \(\mathbf{X}^{\prime}=\mathbf{A X}\) is $$\mathbf{X}(t)=e^{\mathbf{A} t} \mathbf{C}=e^{\mathbf{A} t}\left(\begin{array}{c} c_{1} \\ c_{2} \\ c_{3} \end{array}\right)=\left(\begin{array}{c} c_{1}\left(1-t-t^{2} / 2\right)+c_{2} t+c_{3}\left(t+t^{2} / 2\right) \\ -c_{1} t+c_{2}+c_{3} t \\ c_{1}\left(-t-t^{2} / 2\right)+c_{2} t+c_{3}\left(1+t+t^{2} / 2\right) \end{array}\right).$$

We have \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=-(\lambda+1)(\lambda-1)^{2}=0 .\) For \(\lambda_{1}=-1\) we obtain $$\mathbf{K}_{1}=\left(\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right)$$ For \(\lambda_{2}=1\) we obtain $$\mathbf{K}_{2}=\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right) \quad \text { and } \quad \mathbf{K}_{3}=\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right) e^{-t}+c_{2}\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right) e^{t}+c_{3}\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) e^{t}$$ If $$\mathbf{X}(0)=\left(\begin{array}{l} 1 \\ 2 \\ 5 \end{array}\right)$$ then \(c_{1}=2, c_{2}=3,\) and \(c_{3}=2\)

We have \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=\lambda^{2}-8 \lambda+17=0 .\) For \(\lambda_{1}=4+i\) we obtain $$\mathbf{K}_{1}=\left(\begin{array}{c} -1-i \\ 2 \end{array}\right)$$ so that $$\mathbf{X}_{1}=\left(\begin{array}{c} -1-i \\ 2 \end{array}\right) e^{(4+i) t}=\left(\begin{array}{c} \sin t-\cos t \\ 2 \cos t \end{array}\right) e^{4 t}+i\left(\begin{array}{c} -\sin t-\cos t \\ 2 \sin t \end{array}\right) e^{4 t}$$ Then $$\mathbf{X}=c_{1}\left(\begin{array}{c} \sin t-\cos t \\ 2 \cos t \end{array}\right) e^{4 t}+c_{2}\left(\begin{array}{c} -\sin t-\cos t \\ 2 \sin t \end{array}\right) e^{4 t}$$

We have \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=\lambda^{2}-10 \lambda+34=0 .\) For \(\lambda_{1}=5+3 i\) we obtain $$\mathbf{K}_{1}=\left(\begin{array}{c} 1-3 i \\ 2 \end{array}\right)$$ so that $$\mathbf{X}_{1}=\left(\begin{array}{c} 1-3 i \\ 2 \end{array}\right) e^{(5+3 i) t}=\left(\begin{array}{c} \cos 3 t+3 \sin 3 t \\ 2 \cos 3 t \end{array}\right) e^{5 t}+i\left(\begin{array}{c} \sin 3 t-3 \cos 3 t \\ 2 \sin 3 t \end{array}\right) e^{5 t}$$ Then $$\mathbf{X}=c_{1}\left(\begin{array}{c} \cos 3 t+3 \sin 3 t \\ 2 \cos 3 t \end{array}\right) e^{5 t}+c_{2}\left(\begin{array}{c} \sin 3 t-3 \cos 3 t \\ 2 \sin 3 t \end{array}\right) e^{5 t}$$

From \(x=2 \cos 2 t-2 \sin 2 t, y=-\cos 2 t\) we find \(x+2 y=-2 \sin 2 t .\) Then \((x+2 y)^{2}=4 \sin ^{2} 2 t=4\left(1-\cos ^{2} 2 t\right)=4-4 \cos ^{2} 2 t=4-4 y^{2}\) and \(x^{2}+4 x y+4 y^{2}=4-4 y^{2} \quad\) or \(\quad x^{2}+4 x y+8 y^{2}=4\) This is a rotated conic section and, from the discriminant \(b^{2}-4 a c=16-32 < 0,\) we see that the curve is an ellipse.

(a) The given system can be written as $$x_{1}^{\prime \prime}=-\frac{k_{1}+k_{2}}{m_{1}} x_{1}+\frac{k_{2}}{m_{1}} x_{2}, \quad x_{2}^{\prime \prime}=\frac{k_{2}}{m_{2}} x_{1}-\frac{k_{2}}{m_{2}} x_{2}$$ In terms of matrices this is \(\mathbf{X}^{\prime \prime}=\mathbf{A X}\) where $$\mathbf{X}=\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) \text { and } \mathbf{A}=\left(\begin{array}{cc} -\frac{k_{1}+k_{2}}{m_{1}} & \frac{k_{2}}{m_{1}} \\ \frac{k_{2}}{m_{2}} & -\frac{k_{2}}{m_{2}} \end{array}\right)$$ (b) If \(\mathbf{X}=\mathbf{K} e^{\omega t}\) then \(\mathbf{X}^{\prime \prime}=\omega^{2} \mathbf{K} e^{\omega t}\) and \(\mathbf{A X}=\mathbf{A K} e^{\omega t}\) so that \(\mathbf{X}^{\prime \prime}=\mathbf{A} \mathbf{X}\) becomes \(\omega^{2} \mathbf{K} e^{\omega t}=\mathbf{A} \mathbf{K} e^{\omega t}\) or \(\left(\mathbf{A}-\omega^{2} \mathbf{I}\right) \mathbf{K}=0 .\) Now let \(\omega^{2}=\lambda\) (c) When \(m_{1}=1, m_{2}=1, k_{1}=3,\) and \(k_{2}=2\) we obtain \(\mathbf{A}=\left(\begin{array}{rr}-5 & 2 \\ 2 & -2\end{array}\right) .\) The eigenvalues and corresponding eigenvectors of \(\mathbf{A}\) are \(\lambda_{1}=-1, \lambda_{2}=-6, \mathbf{K}_{1}=\left(\begin{array}{l}1 \\\ 2\end{array}\right), \mathbf{K}_{2}=\left(\begin{array}{r}-2 \\\ 1\end{array}\right) .\) since \(\omega_{1}=i, \omega_{2}=-i, \omega_{3}=\sqrt{6} i,\) and \(\omega_{4}=-\sqrt{6} i\) a solution is $$\mathbf{X}=c_{1}\left(\begin{array}{l} 1 \\ 2 \end{array}\right) e^{i t}+c_{2}\left(\begin{array}{l} 1 \\ 2 \end{array}\right) e^{-i t}+c_{3}\left(\begin{array}{r} -2 \\ 1 \end{array}\right) e^{\sqrt{6} i t}+c_{4}\left(\begin{array}{r} -2 \\ 1 \end{array}\right) e^{-\sqrt{6} i t}$$ (d) Using \(e^{i t}=\cos t+i \sin t\) and \(e^{\sqrt{6} i t}=\cos \sqrt{6} t+i \sin \sqrt{6} t\) the preceding solution can be rewritten as $$\begin{aligned} \mathbf{X}=c_{1}\left(\begin{array}{c} 1 \\ 2 \end{array}\right)(\cos t+i \sin t)+c_{2}\left(\begin{array}{c} 1 \\ 2 \end{array}\right)(\cos t-i \sin t) \\ +c_{3}\left(\begin{array}{c} -2 \\ 1 \end{array}\right)(\cos \sqrt{6} t+i \sin \sqrt{6} t)+c_{4}\left(\begin{array}{c} -2 \\ 1 \end{array}\right)(\cos \sqrt{6} t+i \sin \sqrt{6} t) \\ =\left(c_{1}+c_{2}\right)\left(\begin{array}{c} 1 \\ 2 \end{array}\right) \cos t+i\left(c_{1}-c_{2}\right)\left(\begin{array}{c} 1 \\ 2 \end{array}\right) \sin t \\ +\left(c_{3}+c_{4}\right)\left(\begin{array}{c} -2 \\ 1 \end{array}\right) \cos \sqrt{6} t+i\left(c_{3}-c_{4}\right)\left(\begin{array}{c} -2 \\ 1 \end{array}\right) \sin \sqrt{6} t \\ =b_{1}\left(\begin{array}{c} 1 \\ 2 \end{array}\right) \cos t+b_{2}\left(\begin{array}{c} 1 \\ 2 \end{array}\right) \sin t+b_{3}\left(\begin{array}{c} -2 \\ 1 \end{array}\right) \cos \sqrt{6} t+b_{4}\left(\begin{array}{c} -2 \\ 1 \end{array}\right) \sin \sqrt{6} t \end{aligned}$$ where \(b_{1}=c_{1}+c_{2}, b_{2}=i\left(c_{1}-c_{2}\right), b_{3}=c_{3}+c_{4},\) and \(b_{4}=i\left(c_{3}-c_{4}\right)\)