# 10.7.1: Variation of Parameters for Nonhomogeneous Linear Systems (Exercises)

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$$\newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\$1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill\displaystyle#1\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}}$$ ## Q10.7.1 In Exercises 10.7.1-10.7.10 find a particular solution. 1. $${\bf y}'=\left[\begin{array}{cc}{-1}&{-4}\\{-1}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{21e^{4t}}\\{8e^{-3t}} \end{array}\right]$$ 2. $${\bf y}'=\frac{1}{5}\left[\begin{array}{cc}{-4}&{3}\\{-2}&{-11}\end{array} \right]{\bf y}+\left[\begin{array}{c}{50e^{3t}}\\{10e^{-3t}} \end{array}\right]$$ 3. $${\bf y}'=\left[\begin{array}{cc}{1}&{2}\\{2}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\{t} \end{array}\right]$$ 4. $${\bf y}'=\left[\begin{array}{cc}{-4}&{-3}\\{6}&{5}\end{array} \right]{\bf y}+\left[\begin{array}{c}{2}\\{-2e^{t}} \end{array}\right]$$ 5. $${\bf y}'=\left[\begin{array}{cc}{-6}&{-3}\\{1}&{-2}\end{array} \right]{\bf y}+\left[\begin{array}{c}{4e^{-3t}}\\{4e^{-5t}} \end{array}\right]$$ 6. $${\bf y}'=\left[\begin{array}{cc}{0}&{1}\\{-1}&{0}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\{t} \end{array}\right]$$ 7. $${\bf y}'=\left[\begin{array}{ccc}{3}&{1}&{-1}\\{3}&{5}&{1}\\{-6}&{2}&{4}\end{array} \right]{\bf y}+\left[\begin{array}{c}{3}\\{6}\\{3} \end{array}\right]$$ 8. $${\bf y}'=\left[\begin{array}{ccc}{3}&{-1}&{-1}\\{-2}&{3}&{2}\\{4}&{-1}&{-2}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\{e^{t}}\\{e^{t}} \end{array}\right]$$ 9. $${\bf y}'=\left[\begin{array}{ccc}{-3}&{2}&{2}\\{2}&{-3}&{2}\\{2}&{2}&{-3}\end{array} \right]{\bf y}+\left[\begin{array}{c}{e^{t}}\\{e^{-5t}}\\{e^{t}} \end{array}\right]$$ 10. $${\bf y}'=\frac{1}{3}\left[\begin{array}{ccc}{1}&{1}&{-3}\\{-4}&{-4}&{3}\\{-2}&{1}&{0}\end{array} \right]{\bf y}+\left[\begin{array}{c}{e^{t}}\\{e^{t}}\\{e^{t}} \end{array}\right]$$ ## Q10.7.2 In Exercises 10.7.11-10.7.20 find a particular solution, given that $$Y$$ is a fundamental matrix for the complementary system. 11. $${\bf y}'=\frac{1}{t}\left[\begin{array}{cc}{1}&{t}\\{-t}&{1}\end{array} \right]{\bf y}+t\left[\begin{array}{c}{\cos t}\\{\sin t}\end{array} \right];\quad Y=t\left[\begin{array}{cc}{\cos t}&{\sin t}\\{-\sin t}&{\cos t}\end{array} \right]$$ 12. $${\bf y}'=\frac{1}{t}\left[\begin{array}{cc}{1}&{t}\\{t}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{t}\\{t^{2}}\end{array} \right];\quad Y=t\left[\begin{array}{cc}{e^{t}}&{e^{-t}}\\{e^{t}}&{-e^{-t}}\end{array} \right]$$ 13. $${\bf y}'=\frac{1}{t^{2}-1}\left[\begin{array}{cc}{t}&{-1}\\{-1}&{t}\end{array} \right]{\bf y}+t\left[\begin{array}{c}{1}\\{-1}\end{array} \right];\quad Y=\left[\begin{array}{cc}{t}&{1}\\{1}&{t}\end{array} \right]$$ 14. $${\bf y}'=\frac{1}{3}\left[\begin{array}{cc}{1}&{-2e^{-t}}\\{2e^{t}}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{e^{2t}}\\{e^{-2t}}\end{array} \right];\quad Y=\left[\begin{array}{cc}{2}&{e^{-t}}\\{e^{t}}&{2}\end{array} \right]$$ 15. $${\bf y}'=\frac{1}{2t^{4}}\left[\begin{array}{cc}{3t^{3}}&{t^{6}}\\{1}&{-3t^{3}}\end{array} \right]{\bf y}+\frac{1}{t}\left[\begin{array}{c}{t^{2}}\\{1}\end{array} \right];\quad Y=\frac{1}{t^{2}}\left[\begin{array}{cc}{t^{3}}&{t^{4}}\\{-1}&{t}\end{array} \right]$$ 16. $${\bf y}'=\left[\begin{array}{cc}{\frac{1}{t-1}}&{-\frac{e^{-t}}{t-1}}\\{\frac{e^{t}}{t+1}}&{\frac{1}{t+1}}\end{array} \right]{\bf y}+\left[\begin{array}{c}{t^{2}-1}\\{t^{2}-1}\end{array} \right];\quad Y=t\left[\begin{array}{cc}{t}&{e^{-t}}\\{e^{t}}&{t}\end{array} \right]$$ 17. $${\bf y}' = \frac{1}{t}\left[\begin{array}{ccc}{1}&{1}&{0}\\{0}&{2}&{1}\\{-2}&{2}&{2}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\{2}\\{1}\end{array} \right];\quad Y=\left[\begin{array}{ccc}{t^{2}}&{t^{3}}&{1}\\{t^{2}}&{2t^{3}}&{-1}\\{0}&{2t^{3}}&{2}\end{array} \right]$$ 18. $${\bf y}' = \left[\begin{array}{ccc}{3}&{e^{t}}&{e^{2t}}\\{e^{-t}}&{2}&{e^{t}}\\{e^{-2t}}&{e^{-t}}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{e^{3t}}\\{0}\\{0}\end{array} \right];\quad Y=\left[\begin{array}{ccc}{e^{5t}}&{e^{2t}}&{0}\\{e^{4t}}&{0}&{e^{t}}\\{e^{3t}}&{-1}&{-1}\end{array} \right]$$ 19. $${\bf y}' = \frac{1}{t}\left[\begin{array}{ccc}{1}&{t}&{0}\\{0}&{1}&{t}\\{0}&{-t}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{t}\\{t}\\{t}\end{array} \right];\quad Y=t\left[\begin{array}{ccc}{1}&{\cos t}&{\sin t}\\{0}&{-\sin t}&{\cos t}\\{0}&{-\cos t}&{-\sin t}\end{array} \right]$$ 20. $${\bf y}' = -\frac{1}{t}\left[\begin{array}{ccc}{e^{-t}}&{-t}&{1-e^{-t}}\\{e^{-t}}&{1}&{-t-e^{-t}}\\{e^{-t}}&{-t}&{1-e^{-t}}\end{array} \right]{\bf y}+\frac{1}{t}\left[\begin{array}{c}{e^{t}}\\{0}\\{e^{t}}\end{array} \right];\quad Y=\frac{1}{t}\left[\begin{array}{ccc}{e^{t}}&{e^{-t}}&{t}\\{e^{t}}&{-e^{-t}}&{e^{-t}}\\{e^{t}}&{e^{-t}}&{0}\end{array} \right]$$ ## Q10.7.3 21. Prove Theorem 10.7.1. 22. 1. Convert the scalar equation \[P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=F(t) \tag{A}$ into an equivalent $$n\times n$$ system ${\bf y}'=A(t){\bf y}+{\bf f}(t). \tag{B}$
2. Suppose (A) is normal on an interval $$(a,b)$$ and $$\{y_1,y_2,\dots,y_n\}$$ is a fundamental set of solutions of $P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=0 \tag{C}$ on $$(a,b)$$. Find a corresponding fundamental matrix $$Y$$ for ${\bf y}'=A(t){\bf y} \tag{D}$ on $$(a,b)$$ such that $y=c_1y_1+c_2y_2+\cdots+c_ny_n\nonumber$ is a solution of (C) if and only if $${\bf y}=Y{\bf c}$$ with ${\bf c}=\left[\begin{array}{c}c_1\\c_2\\\vdots\\c_n\end{array}\right]\nonumber$ is a solution of (D).
3. Let $$y_p=u_1y_1+u_1y_2+\cdots+u_ny_n$$ be a particular solution of (A), obtained by the method of variation of parameters for scalar equations as given in Section 9.4, and define ${\bf u}=\left[\begin{array}{c}u_1\\u_2\\\vdots\\u_n\end{array}\right].\nonumber$ Show that $${\bf y}_p=Y{\bf u}$$ is a solution of (B).
4. Let $${\bf y}_p=Y{\bf u}$$ be a particular solution of (B), obtained by the method of variation of parameters for systems as given in this section. Show that $$y_p=u_1y_1+u_1y_2+\cdots+u_ny_n$$ is a solution of (A).

23. Suppose the $$n\times n$$ matrix function $$A$$ and the $$n$$–vector function $${\bf f}$$ are continuous on $$(a,b)$$. Let $$t_0$$ be in $$(a,b)$$, let $${\bf k}$$ be an arbitrary constant vector, and let $$Y$$ be a fundamental matrix for the homogeneous system $${\bf y}'=A(t){\bf y}$$. Use variation of parameters to show that the solution of the initial value problem

${\bf y}'=A(t){\bf y}+{\bf f}(t),\quad {\bf y}(t_0)={\bf k}\nonumber$

is

${\bf y}(t)=Y(t)\left( Y^{-1}(t_0){\bf k}+\int_{t_0}^t Y^{-1}(s){\bf f}(s)\, ds\right).\nonumber$

This page titled 10.7.1: Variation of Parameters for Nonhomogeneous Linear Systems (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.