10.6.1: Variation of Parameters for Nonhomogeneous Linear Systems (Exercises)

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In Exercises 1-20 find the general solution.

1. $\begin{array}{ccl}y'_1=3y_1 - 3y_2+4\\ y_2'=2y_1-2y_2-1\end{array}$

2. $\begin{array}{ccl}y'_1=2y_1 - y_2\\ y_2'=3y_1-2y_2+4t\end{array}$

3. ${\bf y}'=\left[\begin{array}{cc}{3}&{-5}\\{3\over 4}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\{-1} \end{array}\right]e^{t/2}$

4. ${\bf y}'=\left[\begin{array}{cc}{2}&{-1}\\{4}&{2}\end{array} \right]{\bf y}+\left[\begin{array}{c}{\sin 2t}\\{2\cos 2t} \end{array}\right]e^{2t}$

5. ${\bf y}'=\left[\begin{array}{cc}{0}&{2}\\{-1}&{3}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\{-1} \end{array}\right]e^{t}$

6. ${\bf y}'=\left[\begin{array}{cc}{0}&{2}\\{-1}&{3}\end{array} \right]{\bf y}+\left[\begin{array}{c}{2}\\{e^{-3t}} \end{array}\right]$

7. ${\bf y}'=\left[\begin{array}{cc}{1}&{8}\\{1}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{12}\\{12} \end{array}\right]t$

8. ${\bf y}'=\left[\begin{array}{cc}{1}&{8}\\{1}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{e^{-t}}\\{te^t} \end{array}\right]$

9. ${\bf y}'=\left[\begin{array}{cc}{3}&{2}\\{-2}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{2}\\{1} \end{array}\right]e^{-t}$

10. ${\bf y}'=\left[\begin{array}{cc}{3}&{2}\\{-2}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\{1} \end{array}\right]$

11. ${\bf y}'=\left[\begin{array}{cc}{0}&{-1}\\{1}&{0}\end{array} \right]{\bf y}+\left[\begin{array}{c}{\sec t}\\{0} \end{array}\right]$

12. ${\bf y}'=\left[\begin{array}{cc}{1}&{-1}\\{1}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{3}\\{3} \end{array}\right]e^{t}$

13. ${\bf y}'=\left[\begin{array}{cc}{1}&{-1}\\{1}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{\cos t}\\{\sin t} \end{array}\right]e^{t}$

14. ${\bf y}'=\left[\begin{array}{cc}{2}&{-2}\\{8}&{-6}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\{3} \end{array}\right]{e^{-2t}\over t}$

15. ${\bf y}'=\left[\begin{array}{cc}{0}&{1}\\{-1}&{0}\end{array} \right]{\bf y}+\left[\begin{array}{c}{0}\\{\sec t\tan t} \end{array}\right]$

16. ${\bf y}'=\left[\begin{array}{cc}{0}&{1}\\{-1}&{0}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\{\cot t} \end{array}\right]$

17. ${\bf y}'=\left[\begin{array}{cc}{1}&{2}\\-{1\over 2}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{\csc t}\\{\sec t} \end{array}\right]e^{t}$

18. ${\bf y}'=\left[\begin{array}{cc}{1}&{-2}\\{1}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{\tan t}\\{1} \end{array}\right]$

19. ${\bf y}'=\left[\begin{array}{ccc}{1}&{1}&{0}\\{1}&{1}&{0}\\{0}&{0}&{3}\end{array} \right]{\bf y}+\left[\begin{array}{c}{e^t}\\{e^{2t}}\\{te^{3t}} \end{array}\right]$

20. ${\bf y}'=\left[\begin{array}{ccc}{3}&{-1}&{-1}\\{1}&{1}&{-1}\\{1}&{-1}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{0}\\{t}\\{2e^{t}} \end{array}\right]$

In Exercises 21-22 solve the initial value problem.

21. ${\bf y}'=\left[\begin{array}{cc}{3}&{-1}\\{-1}&{3}\end{array} \right]{\bf y}+\left[\begin{array}{c}{4e^{2t}}\\{4e^{4t}} \end{array}\right] ,\quad {\bf y}(0)=\left[\begin{array}{c}{1}\\{1}\end{array} \right] $

22. ${\bf y}'=\left[\begin{array}{cc}{1}&{-1}\\{1}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1/t}\\{1/t} \end{array}\right],\quad {\bf y}(1)=\left[\begin{array}{c}{2}\\{-1}\end{array} \right] $

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