10.7 Variation of Parameters for Nonhomogeneous Linear Systems

We now consider the nonhomogeneous linear system

\[ {\bf y}'= A(t){\bf y}+{\bf f}(t), \]

where $A$ is an $n\times n$ matrix function and ${\bf f}$ is an $n$-vector forcing function. Associated with this system is the complementary system ${\bf y}'=A(t){\bf y}$.

The next theorem is analogous to Theorems 5.3.2 and 9.1.5. It shows how to find the general solution of ${\bf y}'=A(t){\bf y}+{\bf f}(t)$ if we know a particular solution of ${\bf y}'=A(t){\bf y}+{\bf f}(t)$ and a fundamental set of solutions of the complementary system. We leave the proof as an exercise (Exercise 21).

Theorem 10.7.1

Suppose the $n\times n$ matrix function $A$ and the $n$-vector function ${\bf f}$ are continuous on $(a,b).$ Let ${\bf y}_p$ be a particular solution of ${\bf y}'=A(t){\bf y}+{\bf f}(t)$ on $(a,b)$, and let $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ be a fundamental set of solutions of the complementary equation ${\bf y}'=A(t){\bf y}$ on $(a,b)$. Then ${\bf y}$ is a solution of ${\bf y}'=A(t){\bf y}+{\bf f}(t)$ on $(a,b)$ if and only if

\[ {\bf y}={\bf y}_p+c_1{\bf y}_1+c_2{\bf y}_2+\cdots+c_n{\bf y}_n, \]

where $c_1,$ $c_2,$ …, $c_n$ are constants.

Finding a Particular Solution of a Nonhomogeneous System

We now discuss an extension of the method of variation of parameters to linear nonhomogeneous systems. This method will produce a particular solution of a nonhomogenous system ${\bf y}'=A(t){\bf y}+{\bf f}(t)$ provided that we know a fundamental matrix for the complementary system. To derive the method, suppose $Y$ is a fundamental matrix for the complementary system; that is,

\[ Y=\left[\begin{array}{cccc} y_{11}&y_{12}&\cdots&y_{1n} \\ y_{21}&y_{22}&\cdots&y_{2n}\\ \vdots&\vdots&\ddots&\vdots \\ y_{n1}&y_{n2}&\cdots&y_{nn} \\ \end{array} \right], \]

where

\[ {\bf y}_1=\left[\begin{array}{c}y_{11}\\y_{21}\\ \vdots\\ y_{n1}\end{array}\right],\quad {\bf y}_2=\left[\begin{array}{c}y_{12}\\y_{22}\\ \vdots\\ y_{n2}\end{array}\right],\quad \cdots,\quad {\bf y}_n=\left[\begin{array}{c}y_{1n}\\y_{2n}\\ \vdots\\ y_{nn}\end{array}\right] \]

is a fundamental set of solutions of the complementary system. In Section 10.3 we saw that $Y'=A(t)Y$. We seek a particular solution of

\begin{equation} {\bf y}'=A(t){\bf y}+{\bf f}(t) \tag{10.7.1}\end{equation}

of the form

\begin{equation} {\bf y}_p=Y{\bf u}, \tag{10.7.2}\end{equation}

where ${\bf u}$ is to be determined. Differentiating (10.7.2) yields

\begin{eqnarray*} {\bf y}_p'&=&Y' {\bf u}+Y {\bf u}' \\ &=&A Y {\bf u}+Y {\bf u}'\mbox{ (since $Y'=AY$)} \\ &=& A{\bf y}_p+Y {\bf u}'\mbox{ (since $Y{\bf u}={\bf y}_p$)}. \end{eqnarray*}

Comparing this with (10.7.1) shows that ${\bf y}_p=Y{\bf u}$ is a solution of (10.7.1) if and only if

\[ Y{\bf u}'={\bf f}. \]

Thus, we can find a particular solution ${\bf y}_p$ by solving this equation for ${\bf u}'$, integrating to obtain ${\bf u}$, and computing $Y{\bf u}$. We can take all constants of integration to be zero, since any particular solution will suffice.

Exercise 22 sketches a proof that this method is analogous to the method of variation of parameters discussed in Sections 5.7 and 9.4 for scalar linear equations.

Example 10.7.1

Find a particular solution of the system

\begin{equation} {\bf y}'=\twobytwo1221{\bf y}+\left[\begin{array}{c}2e^{4t}\\e^{4t} \end{array}\right], \tag{10.7.3}\end{equation}

which we considered in Example 10.2.1.
Find the general solution of (10.7.3).

Solution (a) The complementary system is

\begin{equation} {\bf y}'=\twobytwo1221{\bf y}. \tag{10.7.4}\end{equation}

The characteristic polynomial of the coefficient matrix is

\[ \left|\begin{array}{cc}1-\lambda&2\\2&1-\lambda\end{array}\right|= (\lambda+1)(\lambda-3). \]

Using the method of Section 10.4, we find that

\[ {\bf y}_1=\left[\begin{array}{r}e^{3t}\\e^{3t}\end{array}\right] \mbox{\quad and \quad} {\bf y}_2=\left[\begin{array}{r}e^{-t}\\-e^{-t}\end{array}\right] \]

are linearly independent solutions of (10.7.4). Therefore

\[ Y=\left[\begin{array}{rr}e^{3t}&e^{-t}\\e^{3t}&-e^{-t}\end{array}\right] \]

is a fundamental matrix for (10.7.4). We seek a particular solution ${\bf y}_p=Y{\bf u}$ of (10.7.3), where $Y{\bf u}'={\bf f}$; that is,

\[ \left[\begin{array}{rr}e^{3t}&e^{-t}\\e^{3t}&-e^{-t}\end{array}\right] \twocol{u_1'}{u_2'} =\left[\begin{array}{c}2e^{4t}\\e^{4t}\end{array}\right]. \]

The determinant of $Y$ is the Wronskian

\[ \left|\begin{array}{rr}e^{3t}&e^{-t}\\e^{3t}&-e^{-t}\end{array}\right| =-2e^{2t}. \]

By Cramer’s rule,

\[ \begin{array}{ccccccl} u_1'&=&-\dst{1\over2e^{2t}} \left|\begin{array}{cr}2e^{4t}&e^{-t}\\e^{4t}&-e^{-t} \end{array}\right|&=&\dst{3e^{3t}\over2e^{2t}}&= &\dst{3\over2}e^t,\\[12pt] u_2'&=&-\dst{1\over2e^{2t}} \left|\begin{array}{rr}e^{3t}&2e^{4t}\\e^{3t}&e^{4t} \end{array}\right|&=&\dst{e^{7t}\over2e^{2t}}&=&\dst{1\over2}e^{5t}. \end{array} \]

Therefore

\[ {\bf u}'={1\over2}\left[\begin{array}{c}3e^t\\e^{5t}\end{array}\right]. \]

Integrating and taking the constants of integration to be zero yields

\[ {\bf u}={1\over10}\left[\begin{array}{c}15e^t\\e^{5t}\end{array}\right], \]

\[ {\bf y}_p=Y{\bf u}= \dst{1\over10}\left[\begin{array}{rr}e^{3t}&e^{-t}\\[6pt] e^{3t}&-e^{-t}\end{array}\right] \left[\begin{array}{c}15e^t\\e^{5t}\end{array}\right] =\phantom{0} \dst{1\over5}\left[\begin{array}{c}8e^{4t}\\7e^{4t}\end{array}\right] \]

is a particular solution of (10.7.3).

Solution (b) From Theorem 10.7.1, the general solution of (10.7.3) is

\begin{equation} {\bf y}={\bf y}_p+c_1{\bf y}_1+c_2{\bf y}_2= {1\over5}\left[\begin{array}{c}8e^{4t}\\7e^{4t}\end{array}\right] +c_1\left[\begin{array}{r}e^{3t}\\e^{3t}\end{array}\right] +c_2\left[\begin{array}{r}e^{-t}\\-e^{-t}\end{array}\right], \tag{10.7.5}\end{equation}

which can also be written as

\[ {\bf y}= {1\over5}\left[\begin{array}{c}8e^{4t}\\7e^{4t}\end{array}\right] +\left[\begin{array}{rr}e^{3t}&e^{-t}\\e^{3t}&-e^{-t}\end{array}\right] {\bf c}, \]

where ${\bf c}$ is an arbitrary constant vector.

Writing (10.7.5) in terms of coordinates yields

\begin{eqnarray*} y_1&=&{8\over5}e^{4t}+c_1e^{3t}+c_2e^{-t} \\[6pt] y_2&=&{7\over5}e^{4t}+c_1e^{3t}-c_2e^{-t}, \end{eqnarray*}

so our result is consistent with Example 10.2.1. ∎.

If $A$ isn’t a constant matrix, it’s usually difficult to find a fundamental set of solutions for the system ${\bf y}'=A(t){\bf y}$. It is beyond the scope of this text to discuss methods for doing this. Therefore, in the following examples and in the exercises involving systems with variable coefficient matrices we’ll provide fundamental matrices for the complementary systems without explaining how they were obtained.

Example 10.7.2

Find a particular solution of

\begin{equation} {\bf y}'=\left[\begin{array}{cc}2&2e^{-2t}\\2e^{2t}&4\end{array}\right]{\bf y}+\twocol11, \tag{10.7.6}\end{equation}

given that

\[ Y=\left[\begin{array}{cc} e^{4t}&-1\\e^{6t}&e^{2t}\end{array}\right] \]

is a fundamental matrix for the complementary system.

Solution We seek a particular solution ${\bf y}_p=Y{\bf u}$ of (10.7.6) where $Y{\bf u}'={\bf f}$; that is,

\[ \left[\begin{array}{cc} e^{4t}&-1\\e^{6t}&e^{2t}\end{array}\right] \twocol{u_1'}{u_2'}=\twocol11. \]

The determinant of $Y$ is the Wronskian

\[ \left|\begin{array}{cc} e^{4t}&-1\\e^{6t}&e^{2t}\end{array}\right|=2e^{6t}. \]

By Cramer’s rule,

\[ \begin{array}{cccccll} u_1'&=&\dst{1\over2e^{6t}}\left|\begin{array}{cc}1&-1\\1&e^{2t} \end{array}\right|&=&\dst{e^{2t}+1\over2e^{6t}}&=&\dst{e^{-4t}+e^{-6t}\over2} \\[12pt] u_2'&=&\dst{1\over2e^{6t}}\left|\begin{array}{cc}e^{4t}&1\\e^{6t}&1 \end{array}\right|&=&\dst{e^{4t}-e^{6t}\over2e^{6t}}&=&\dst{e^{-2t}-1\over2}. \end{array} \]

Therefore

\[ {\bf u}'={1\over2}\left[\begin{array}{c}e^{-4t}+e^{-6t}\\e^{-2t}-1\end{array}\right]. \]

Integrating and taking the constants of integration to be zero yields

\[ {\bf u}=-{1\over24}\left[\begin{array}{c}3e^{-4t}+2e^{-6t}\\6e^{-2t}+12t \end{array}\right], \]

\[ {\bf y}_p=Y{\bf u}= -\displaystyle{1\over24}\left[\begin{array}{cc}e^{4t}&-1\\e^{6t}&e^{2t}\end{array}\right] \left[\begin{array}{c}3e^{-4t}+2e^{-6t}\\6e^{-2t}+12t\end{array}\right] =\displaystyle{1\over24}\left[\begin{array}{c}4e^{-2t}+12t-3\\-3e^{2t}(4t+1)-8\end{array}\right] \]

is a particular solution of (10.7.6).

Example 10.7.3

Find a particular solution of

\begin{equation} {\bf y}'=-{2\over t^2 }\left[\begin{array}{cc}t&-3t^2\\1&-2t\end{array}\right]{\bf y} +t^2\twocol11, \tag{10.7.7}\end{equation}

given that

\[ Y=\left[\begin{array}{cc}2t&3t^2\\1&2t\end{array}\right] \]

is a fundamental matrix for the complementary system on $(-\infty,0)$ and $(0,\infty)$.

Solution We seek a particular solution ${\bf y}_p=Y{\bf u}$ of (10.7.7) where $Y{\bf u}'={\bf f}$; that is,

\[ \left[\begin{array}{cc}2t&3t^2\\1&2t\end{array}\right]\twocol{u_1'}{u_2'} =\twocol{t^2}{t^2}. \]

The determinant of $Y$ is the Wronskian

\[ \left|\begin{array}{cc}2t&3t^2\\1&2t\end{array}\right|=t^2. \]

By Cramer’s rule,

\[ \begin{array}{ccccccl} u_1'&=&\dst{1\over t^2}\left|\begin{array}{cc}t^2&3t^2\\t^2&2t \end{array}\right|&=&\dst{2t^3-3t^4\over t^2}&=&2t-3t^2, \\[12pt] u_2'&=&\dst{1\over t^2}\left|\begin{array}{cc}2t&t^2\\1&t^2 \end{array}\right|&=&\dst{2t^3-t^2\over t^2}&=&2t-1. \end{array} \]

Therefore

\[ {\bf u}'=\left[\begin{array}{c}2t-3t^2\\2t-1\end{array}\right]. \]

Integrating and taking the constants of integration to be zero yields

\[ {\bf u}=\left[\begin{array}{c}t^2-t^3\\t^2-t \end{array}\right], \]

\[ {\bf y}_p=Y{\bf u}= \left[\begin{array}{cc}2t&3t^2\\1&2t\end{array}\right] \left[\begin{array}{c}t^2-t^3\\t^2-t\end{array}\right] =\left[\begin{array}{c}t^3(t-1)\\t^2(t-1)\end{array}\right] \]

is a particular solution of (10.7.7).

Example 10.7.4

Find a particular solution of

\begin{equation} {\bf y}'=\threebythree2{-1}{-1}10{-1}1{-1}0{\bf y}+\left[\begin{array}{c}e^{t}\\0\\e^{-t} \end{array}\right]. \tag{10.7.8}\end{equation}
Find the general solution of (10.7.8).

Solution (a) The complementary system for (10.7.8) is

\begin{equation} {\bf y}'=\threebythree2{-1}{-1}10{-1}1{-1}0{\bf y}. \tag{10.7.9}\end{equation}

The characteristic polynomial of the coefficient matrix is

\[ \left|\begin{array}{ccc}2-\lambda&-1&-1\\1&-\lambda&-1\\1&-1&-\lambda \end{array}\right|= -\lambda(\lambda-1)^2. \]

Using the method of Section 10.4, we find that

\[ {\bf y}_1=\left[\begin{array}{c}1\\1\\1\end{array}\right],\quad {\bf y}_2=\left[\begin{array}{c}e^t\\e^t\\0\end{array}\right], \mbox{\quad and \quad} {\bf y}_3=\left[\begin{array}{c}e^t\\0\\e^t\end{array}\right] \]

are linearly independent solutions of (10.7.9). Therefore

\[ Y=\left[\begin{array}{ccc}1&e^t&e^t\\1&e^t&0\\1&0&e^t\end{array}\right] \]

is a fundamental matrix for (10.7.9). We seek a particular solution ${\bf y}_p=Y{\bf u}$ of (10.7.8), where $Y{\bf u}'={\bf f}$; that is,

\[ \left[\begin{array}{ccc}1&e^t&e^t\\1&e^t&0\\1&0&e^t\end{array}\right] \threecol{u_1'}{u_2'}{u_3'}= \left[\begin{array}{c}e^t\\0\\e^{-t}\end{array}\right]. \]

The determinant of $Y$ is the Wronskian

\[ \left|\begin{array}{ccc}1&e^t&e^t\\1&e^t&0\\1&0&e^t\end{array}\right| =-e^{2t}. \]

Thus, by Cramer’s rule,

\[ \begin{array}{cccccll} u_1'&=&-\dst{1\over e^{2t}}\left|\begin{array}{ccc}e^t&e^t&e^t\\0&e^t&0\\e^{-t}&0&e^t \end{array}\right|&=&-\dst{e^{3t}-e^t\over e^{2t}}&=&e^{-t}-e^t\\[18pt] u_2'&=&-\dst{1\over e^{2t}}\left|\begin{array}{ccc}1&e^t&e^t\\1&0&0\\1&e^{-t}&e^t \end{array}\right|&=&-\dst{1-e^{2t}\over e^{2t}}&=&1-e^{-2t}\\[18pt] u_3'&=&-\dst{1\over e^{2t}}\left|\begin{array}{ccc}1&e^t&e^t\\1&e^t&0\\1&0&e^{-t} \end{array}\right|&=&\dst{e^{2t}\over e^{2t}}&=&1. \end{array} \]

Therefore

\[ {\bf u}'=\left[\begin{array}{c}e^{-t}-e^t\\1-e^{-2t}\\1\end{array}\right]. \]

Integrating and taking the constants of integration to be zero yields

\[ {\bf u}=\left[\begin{array}{c}-e^t-e^{-t}\\\dst{e^{-2t}\over2}+t \\t\end{array}\right], \]

\[ \begin{array}{ccccl} {\bf y}_p=Y{\bf u}&=& \left[\begin{array}{ccc}1&e^t&e^t\\1&e^t&0\\1&0&e^t\end{array}\right] \left[\begin{array}{c}-e^t-e^{-t}\\\dst{e^{-2t}\over2}+t\\t\end{array}\right] &=&\left[\begin{array}{c}e^t(2t-1)-\dst{e^{-t}\over2}\\[5pt]e^t(t-1)-\dst{e^{-t} \over2}\\[5pt]e^t(t-1)-\dst{e^{-t}} \end{array}\right] \end{array} \]

is a particular solution of (10.7.8).

Solution (a) From Theorem 10.7.1 the general solution of (10.7.8) is

\[ {\bf y}={\bf y}_p+c_1{\bf y}_1+c_2{\bf y}_2+c_3{\bf y}_3= \left[\begin{array}{c}e^t(2t-1)-\dst{e^{-t}\over2}\\[5pt]e^t(t-1)-\dst{e^{-t} \over2}\\[5pt]e^t(t-1)-\dst{e^{-t}} \end{array}\right]+ c_1\left[\begin{array}{c}1\\1\\1\end{array}\right]+ c_2\left[\begin{array}{c}e^t\\e^t\\0\end{array}\right] +c_3\left[\begin{array}{c}e^t\\0\\e^t\end{array}\right], \]

which can be written as

\[ {\bf y}={\bf y}_p+Y{\bf c}= \left[\begin{array}{c}e^t(2t-1)-\dst{e^{-t}\over2}\\[5pt]e^t(t-1)-\dst{e^{-t} \over2}\\[5pt]e^t(t-1)-\dst{e^{-t}} \end{array}\right]+ \left[\begin{array}{ccc}1&e^t&e^t\\1&e^t&0\\1&0&e^t\end{array}\right]{\bf c} \]

where ${\bf c}$ is an arbitrary constant vector.

Example 10.7.5

Find a particular solution of

\begin{equation} {\bf y}'={1\over2} \left[\begin{array}{ccc}3&e^{-t}&-e^{2t}\\0&6&0\\-e^{-2t}&e^{-3t}&-1\end{array}\right] {\bf y}+\left[\begin{array}{c}1\\e^t\\e^{-t}\end{array}\right], \tag{10.7.10}\end{equation}

given that

\[ Y=\left[\begin{array}{ccc}e^t&0&e^{2t}\\0&e^{3t}&e^{3t}\\e^{-t}&1&0 \end{array}\right] \]

is a fundamental matrix for the complementary system.

Solution We seek a particular solution of (10.7.10) in the form ${\bf y}_p=Y{\bf u}$, where $Y{\bf u}'={\bf f}$; that is,

\[ \left[\begin{array}{ccc}e^t&0&e^{2t}\\0&e^{3t}&e^{3t}\\e^{-t}&1&0 \end{array}\right]\threecol{u_1'}{u_2'}{u_3'}= \left[\begin{array}{c}1\\e^t\\e^{-t}\end{array}\right]. \]

The determinant of $Y$ is the Wronskian

\[ \left|\begin{array}{ccc}e^t&0&e^{2t}\\0&e^{3t}&e^{3t}\\e^{-t}&1&0 \end{array}\right|=-2e^{4t}. \]

By Cramer’s rule,

\[ \begin{array}{ccccccl} u_1'&=&-\dst{1\over 2e^{4t}}\left|\begin{array}{ccc}1&0&e^{2t}\\e^t&e^{3t}&e^{3t} \\e^{-t}&1&0 \end{array}\right|&=&\dst{e^{4t}\over2e^{4t}}&=&\dst{1\over2} \\[18pt] u_2'&=&-\dst{1\over 2e^{4t}}\left|\begin{array}{ccc}e^t&1&e^{2t}\\0&e^t&e^{3t} \\e^{-t}&e^{-t}&0 \end{array}\right|&=&\dst{e^{3t}\over2e^{4t}}&=&\dst{1\over2}e^{-t}\\[18pt] u_3'&=&-\dst{1\over 2e^{4t}}\left|\begin{array}{ccc}e^t&0&1\\0&e^{3t}&e^t \\e^{-t}&1&e^{-t} \end{array}\right|&=&-\dst{e^{3t}-2e^{2t}\over2e^{4t}}&=&\dst{2e^{-2t}-e^{-t} \over2}. \end{array} \]

Therefore

\[ {\bf u}'={1\over2}\left[\begin{array}{c}1\\e^{-t}\\2e^{-2t}-e^{-t}\end{array}\right]. \]

Integrating and taking the constants of integration to be zero yields

\[ {\bf u}={1\over2}\left[\begin{array}{c}t\\-e^{-t}\\e^{-t}-e^{-2t} \end{array}\right], \]

\[ {\bf y}_p=Y{\bf u}= {1\over2}\left[\begin{array}{ccc}e^t&0&e^{2t}\\0&e^{3t}&e^{3t}\\e^{-t}&1&0 \end{array}\right] \left[\begin{array}{c}t\\-e^{-t}\\e^{-t}-e^{-2t} \end{array}\right]=\dst{1\over2}\left[\begin{array}{c}e^t(t+1)-1\\ -e^t\\e^{-t}(t-1)\end{array}\right] \]

is a particular solution of (10.7.10).

10.7 Exercises

In Exercises 1–10 find a particular solution.

$\dst{{\bf y}'=\twobytwo{-1}{-4}{-1}{-1}{\bf y}+\left[\begin{array}{cc}21e^{4t}\\8e^{-3t}\end{array}\right]}$

Show answer

$\dst{\left[\begin{array}{c}5e^{4t}+e^{-3t}(2+8t)\\ -e^{4t}-e^{-3t}(1-4t)\end{array}\right]}$
$\dst{{\bf y}'={1\over5}\twobytwo{-4}3{-2}{-11}{\bf y}+\left[\begin{array}{cc}50e^{3t}\\10e^{-3t}\end{array}\right]}$

Show answer

$\dst{\left[\begin{array}{c}13e^{3t}+3e^{-3t}\\ -e^{3t}-11e^{-3t}\end{array}\right]}$
$\dst{{\bf y}'=\twobytwo1221{\bf y}+\left[\begin{array}{cc}1\\t\end{array}\right]}$

Show answer

$\dst{{1\over9}\left[\begin{array}{c}7-6t\\ -11+3t\end{array}\right]}$
$\dst{{\bf y}'=\twobytwo{-4}{-3}65{\bf y}+\left[\begin{array}{cc}2\\-2e^t\end{array}\right]}$

Show answer

$\dst{\left[\begin{array}{c}5-3e^t\\ -6+5e^t\end{array}\right]}$
$\dst{{\bf y}'=\twobytwo{-6}{-3}1{-2}{\bf y}+\left[\begin{array}{cc}4e^{-3t}\\4e^{-5t}\end{array}\right]}$

Show answer

$\dst{\left[\begin{array}{c}e^{-5t}(3+6t)+e^{-3t}(3-2t)\\ -e^{-5t}(3+2t)-e^{-3t}(1-2t)\end{array}\right]}$
$\dst{{\bf y}'=\twobytwo01{-1}0{\bf y}+\left[\begin{array}{cc}1\\t\end{array}\right]}$

Show answer

$\dst{\left[\begin{array}{c}t\\0 \end{array}\right]}$
$\dst{{\bf y}'=\threebythree31{-1}351{-6}24{\bf y}+\left[\begin{array}{cc}3\\6\\3\end{array}\right]}$

Show answer

$\dst{-{1\over6}\left[\begin{array}{c}2-6t\\7+6t\\1-12t \end{array}\right]}$
$\dst{{\bf y}'=\threebythree3{-1}{-1}{-2}324{-1}{-2}{\bf y}+\left[\begin{array}{cc}1\\e^t\\e^t\end{array}\right]}$

Show answer

$\dst{-{1\over6}\left[\begin{array}{c}3e^t+4\\6e^t-4\\10 \end{array}\right]}$
$\dst{{\bf y}'=\threebythree{-3}222{-3}222{-3}{\bf y}+\left[\begin{array}{cc}e^t\\e^{-5t}\\e^t\end{array}\right]}$

Show answer

$\dst{1\over18}{\left[\begin{array}{c}e^t(1+12t)-e^{-5t}(1+6t)\\-2e^t(1-6t)- e^{-5t}(1-12t)\\e^t(1+12t)-e^{-5t}(1+6t) \end{array}\right]}$
$\dst{{\bf y}'={1\over3}\threebythree11{-3}{-4}{-4}3{-2}10{\bf y}+\left[\begin{array}{cc}e^t\\e^t\\e^t\end{array}\right]}$

Show answer

$\dst{{1\over3}\left[\begin{array}{r}2e^t\\e^t\\2e^t \end{array}\right]}$

In Exercises 11–20 find a particular solution, given that $Y$ is a fundamental matrix for the complementary system.

$\dst{{\bf y}'={1\over t}\left[\begin{array}{rc}1&t\\-t&1\end{array}\right]{\bf y} +t\left[\begin{array}{cc} \cos t\\\sin t\end{array}\right]; \quad Y=t\left[\begin{array}{rc}\cos t&\sin t\\-\sin t&\cos t \end{array}\right]}$

Show answer

$\dst{\left[\begin{array}{c}t\sin t\\0 \end{array}\right]}$
$\dst{{\bf y}'={1\over t}\left[\begin{array}{cc}1&t\\t&1\end{array}\right]{\bf y} +\left[\begin{array}{c} t\\t^2\end{array}\right]; \quad Y=t\left[\begin{array}{cr} e^t&e^{-t}\\e^t&-e^{-t} \end{array}\right]}$

Show answer

$\dst{-\left[\begin{array}{c}t^2\\2t \end{array}\right]}$
$\dst{{\bf y}'={1\over t^2-1}\left[\begin{array}{rr}t&-1\\-1&t\end{array}\right]{\bf y}+ t\left[\begin{array}{r} 1\\-1\end{array}\right]; \quad Y=\left[\begin{array}{cc} t&1\\1&t \end{array}\right]}$

Show answer

$\dst{(t-1)\left(\ln|t-1|+t\right)\left[\begin{array}{r}1\\-1 \end{array}\right]}$
$\dst{{\bf y}'={1\over 3}\left[\begin{array}{cc}1&-2e^{-t}\\2e^t&-1\end{array}\right]{\bf y}+ \left[\begin{array}{c}e^{2t} \\e^{-2t}\end{array}\right]; \quad Y=\left[\begin{array}{cc} 2&e^{-t}\\e^t&2 \end{array}\right]}$

Show answer

$\dst{{1\over9}\left[\begin{array}{c}5e^{2t}-e^{-3t}\\e^{3t}-5e^{-2t} \end{array}\right]}$
$\dst{{\bf y}'={1\over 2t^4}\left[\begin{array}{cc}3t^3&t^6\\1&-3t^3\end{array}\right]{\bf y}+ {1\over t}\left[\begin{array}{c} t^2\\1\end{array}\right]; \quad Y={1\over t^2}\left[\begin{array}{rc} t^3&t^4\\-1&t \end{array}\right]}$

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$\dst{{1\over4t}\left[\begin{array}{c}2t^3\ln|t|+t^3(t+2)\\ 2\ln|t|+3t-2 \end{array}\right]}$
$\dst{{\bf y}'= \left[\begin{array}{cc}\dst{1\over t-1} &-\dst{e^{-t}\over t-1}\\[6pt]\dst{e^t\over t+1}&\dst{1\over t+1}\end{array}\right]{\bf y}+\left[\begin{array}{c} t^2-1\\t^2-1\end{array}\right]; \quad Y=\left[\begin{array}{cc} t&e^{-t}\\e^t&t \end{array}\right]}$

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$\dst{{1\over2}\left[\begin{array}{c}te^{-t}(t+2)+(t^3-2)\\ te^t(t-2)+(t^3+2) \end{array}\right]}$
$\dst{{\bf y}'={1\over t}\threebythree110021{-2}22{\bf y} +\threecol121 \quad Y=\left[\begin{array}{ccr} t^2&t^3&1\\t^2&2t^3&-1\\0&2t^3&2 \end{array}\right]}$

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$\dst{-\left[\begin{array}{c}t\\t\\t \end{array}\right]}$
$\dst{{\bf y}'= \left[\begin{array}{ccc}3&e^t&e^{2t}\\e^{-t}&2&e^t\\e^{-2t}&e^{-t}&1\end{array}\right]{\bf y} +\left[\begin{array}{c} e^{3t}\\0 \\ 0\end{array}\right]; \quad Y=\left[\begin{array}{crr} e^{5t}&e^{2t}&0\\e^{4t}&0&e^t\\e^{3t}&-1&-1 \end{array}\right]}$

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$\dst{1\over4}\left[\begin{array}{cccc}-3e^t\\ 1\\e^{-t}\end{array}\right]$
$\dst{{\bf y}'={1\over t}\left[\begin{array}{crc}1&t&0\\0&1&t\\0&-t&1\end{array}\right]{\bf y} +\left[\begin{array}{c} t\\t \\t \end{array}\right]; \quad Y=t\left[\begin{array}{crr} 1&\cos t&\sin t \\0&-\sin t&\cos t\\0&-\cos t&-\sin t \end{array}\right]}$

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$\left[\begin{array}{cccc}2t^2+t\\t\\-t\end{array}\right]$
$\dst{{\bf y}'=-{1\over t}\left[\begin{array}{crr}e^{-t}&-t&1-e^{-t}\\e^{-t}&1&-t-e^{-t}\\e^{-t}&-t &1-e^{-t}\end{array}\right]{\bf y} +{1\over t}\left[\begin{array}{c} e^t\\0 \\e^t \end{array}\right]; \quad Y={1\over t}\left[\begin{array}{crc} e^t&e^{-t}&t\\e^t&-e^{-t}&e^{-t}\\e^t&e^{-t}&0 \end{array}\right]}$

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$\dst{{e^t\over4t}\left[\begin{array}{c}2t+1\\2t-1\\2t+1 \end{array}\right]}$
Prove Theorem 10.7.1.
1. Convert the scalar equation
  
  \[ P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=F(t) \tag*{\rm (A)} \]
  
  into an equivalent $n\times n$ system
  
  \[ {\bf y}'=A(t){\bf y}+{\bf f}(t). \tag*{\rm (B)} \]
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
  
  \[ P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=0 \tag*{\rm (C)} \]
  
  on $(a,b)$. Find a corresponding fundamental matrix $Y$ for
  
  \[ {\bf y}'=A(t){\bf y} \tag*{\rm (D)} \]
  
  on $(a,b)$ such that
  
  \[ y=c_1y_1+c_2y_2+\cdots+c_ny_n \]
  
  is a solution of (C) if and only if ${\bf y}=Y{\bf c}$ with
  
  \[ {\bf c}=\left[\begin{array}{c}c_1\\c_2\\\vdots\\c_n\end{array}\right] \]
  
  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
  
  \[ {\bf u}=\left[\begin{array}{c}u_1\\u_2\\\vdots\\u_n\end{array}\right]. \]
  
  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).
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(a) $\dst{{\bf y}'= \left[\begin{array}{cccc} 0&1&\cdots&0\\0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&1\\ -P_n(t)/P_0(t)&-P_{n-1}/P_0(t)& \cdots&-P_1(t)/P_0(t)\end{array} \right]{\bf y}+ \left[\begin{array}{c} 0\\0\\\vdots\\F(t)/P_0(t)\end{array}\right].}$

(b) ${\dst{\left[\begin{array}{cccc} y_1&y_2&\cdots&y_n\\y_1'&y_2'&\cdots&y_n'\\ \vdots&\vdots&\ddots&\vdots\\y_1^{(n-1)}&y_2^{(n-1)}&\cdots&y_n^{(n-1)} \end{array}\right]}}$
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

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

is

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