10.2.1: Basic Theory of Homogeneous Linear Systems (Exercises)

1. Prove: If \({\bf y}_1\), \({\bf y}_2\), …, \({\bf y}_n\) are solutions of \({\bf y}'=A(t){\bf y}\) on \((a,b)\), then any linear combination of \({\bf y}_1\), \({\bf y}_2\), …, \({\bf y}_n\) is also a solution of \({\bf y}'=A(t){\bf y}\) on \((a,b)\).

2. Let

\[A=\left[\begin{array}{cc}{2}&{4}\\{4}&{2}\end{array} \right], \quad {\bf y}_1=\left[\begin{array}{c} e^{6t} \\ e^{6t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r} e^{-2t} \\ -e^{-2t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r}-3 \\ 9\end{array}\right].\nonumber \]

1. Verify that \(\{{\bf y}_1,{\bf y}_2\}\) is a fundamental set of solutions for \({\bf y}'=A{\bf y}\).
2. Solve the initial value problem \[{\bf y}'=A{\bf y},\quad {\bf y}(0)={\bf k}. \tag{A}\]

3. Repeat Exercise 10.2.2 with

\[A=\left[\begin{array}{cc}{-2}&{-2}\\{-5}&{1}\end{array} \right], \quad {\bf y}_1=\left[\begin{array}{r} e^{-4t} \\ e^{-4t}\end{array}\right], \quad {\bf y}_2=\left[ \begin{array}{r}-2e^{3t} \\ 5e^{3t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 10 \\-4\end{array}\right].\nonumber \]

4. Repeat Exercise 10.2.2 with

\[A=\left[\begin{array}{cc}{-4}&{-10}\\{3}&{7}\end{array} \right], \quad {\bf y}_1=\left[\begin{array}{r}-5e^{2t} \\ 3e^{2t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r} 2e^t \\-e^t \end{array}\right], \quad {\bf k}=\left[\begin{array}{r}-19 \\ 11\end{array} \right ].\nonumber \]

5. Repeat Exercise 10.2.2 with

\[A=\left[\begin{array}{cc}{2}&{1}\\{1}&{2}\end{array} \right], \quad {\bf y}_1=\left[\begin{array}{r} e^{3t} \\ e^{3t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r}e^t \\ -e^t\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 2 \\ 8 \end{array}\right].\nonumber \]

6. Let

\[\begin{aligned} A&= \left[\begin{array}{ccc}{3}&{-1}&{-1}\\{-2}&{3}&{2}\\{4}&{-1}&{-2}\end{array} \right] , \\ {\bf y}_1&=\left[\begin{array}{c} e^{2t} \\ 0 \\ e^{2t}\end{array} \right], \quad {\bf y}_2=\left[\begin{array}{c} e^{3t} \\-e^{3t} \\ e^{3t}\end{array}\right], \quad {\bf y}_3=\left[\begin{array}{c} e^{-t} \\-3e^{-t} \\ 7e^{-t} \end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 2 \\-7 \\ 20\end{array}\right].\end{aligned}\nonumber \]

1. Verify that \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is a fundamental set of solutions for \({\bf y}'=A{\bf y}\).
2. Solve the initial value problem \[{\bf y}'=A{\bf y}, \quad {\bf y}(0)={\bf k}. \tag{A}\]

7. Repeat Exercise 10.2.6 with

\[\begin{aligned} A&=\left[\begin{array}{ccc}{0}&{2}&{2}\\{2}&{0}&{2}\\{2}&{2}&{0}\end{array} \right], \\ {\bf y}_1&=\left[\begin{array}{c}-e^{-2t} \\ 0 \\ e^{-2t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{c}-e^{-2t} \\ e^{-2t} \\ 0\end{array}\right], \quad {\bf y}_3=\left[\begin{array}{c} e^{4t} \\ e^{4t} \\ e^{4t}\end{array} \right], \quad {\bf k}=\left[\begin{array}{r} 0 \\-9 \\ 12\end{array} \right].\end{aligned}\nonumber \]

8. Repeat Exercise 10.2.6 with

\[\begin{aligned} A&=\left[\begin{array}{ccc}{-1}&{2}&{3}\\{0}&{1}&{6}\\{0}&{0}&{-2}\end{array} \right], \\ {\bf y}_1&=\left[\begin{array}{c} e^t \\ e^t \\ 0\end{array}\right], \quad {\bf y}_2=\left[\begin{array}{c} e^{-t} \\ 0 \\ 0\end{array}\right], \quad {\bf y}_3=\left[\begin{array}{c} e^{-2t} \\-2e^{-2t} \\ e^{-2t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 5 \\ 5 \\-1 \end{array}\right].\end{aligned}\nonumber \]

---