4.7: Variation of Parameters for Nonhomogeneous Linear Systems

Variation of Parameters for Nonhomogeneous Linear Systems

We now consider the nonhomogeneous linear system

$$\begin{eqnarray*} {\bf y}' = A(t) {\bf y} + {\bf f}(t), \end{eqnarray*}$$

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

The next theorem is analogous to Theorems $(2.3.2)$ and $(3.1.5)$. 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 $(4.7E.21)$).

Theorem $\PageIndex{1}$

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

$$\begin{eqnarray*} {\bf y} = {\bf y}_p + c_1 {\bf y}_1 + c_2 {\bf y}_2 + \cdots + c_n {\bf y}_n, \end{eqnarray*}$$

where $c_1,$ $c_2,$ $\dots$, $c_n$ are constants.

Proof

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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,

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

where

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

is a fundamental set of solutions of the complementary system. In Section 4.3 we saw that $Y'=A(t)Y$. We seek a particular solution of

$$\begin{equation} \label{eq:4.7.1} {\bf y}'=A(t){\bf y}+{\bf f}(t) \end{equation}$$

of the form

$$\begin{equation} \label{eq:4.7.2} {\bf y}_p=Y{\bf u}, \end{equation}$$

where ${\bf u}$ is to be determined. Differentiating $\eqref{eq:4.7.2}$ 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 $\eqref{eq:4.7.1}$ shows that ${\bf y}_p=Y{\bf u}$ is a solution of $\eqref{eq:4.7.1}$ if and only if

$$\begin{eqnarray*} Y {\bf u}' = {\bf f}. \end{eqnarray*}$$

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 $(4.7E.22)$ sketches a proof that this method is analogous to the method of variation of parameters discussed in Sections $3.4$ for scalar linear equations.

Example $\PageIndex{1}$

(a) Find a particular solution of the system

$$\begin{equation} \label{eq:4.7.3} {\bf y}' = \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] {\bf y} + \left[ \begin{array}{c}2e^{4t}\\e^{4t} \end{array}\right], \end{equation}$$

which we considered in Example $(4.2.1)$.

(b)
Find the general solution of $\eqref{eq:4.7.3}$.

Answer

(a) The complementary system is

$$\begin{equation} \label{eq:4.7.4} {\bf y}' = \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] {\bf y}. \end{equation}$$

The characteristic polynomial of the coefficient matrix is

$$\begin{eqnarray*} \left| \begin{array} \\ 1-\lambda & 2 \\ 2 & a-\lambda \end{array} \right| = (\lambda + 1) (\lambda - 3). \end{eqnarray*}$$

Using the method of Section $4.4$, we find that

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

are linearly independent solutions of $\eqref{eq:4.7.4}$. Therefore

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

is a fundamental matrix for $\eqref{eq:4.7.4}$. We seek a particular solution ${\bf y}_p=Y{\bf u}$ of $\eqref{eq:4.7.3}$, where $Y{\bf u}'={\bf f}$; that is,

$$\begin{eqnarray*} \left[ \begin{array} \\ e^{3t} & e^{-t} \\ e^{3t} & -e^{-t} \end{array} \right] \left[ \begin{array} \\ {u_1'} \\ {u_2'} \end{array} \right] = \left[ \begin{array} \\ 2e^{4t} \\ e^{4t} \end{array} \right]. \end{eqnarray*}$$

The determinant of $Y$ is the Wronskian

$$\begin{eqnarray*} \left| \begin{array} \\ e^{3t} & e^{-t} \\ e^{3t} & -e^{-t} \end{array} \right| = -2e^{2t}. \end{eqnarray*}$$

By Cramer's rule,

$$\begin{eqnarray*} \begin{array} \\ u_1' &=& -\displaystyle{1 \over 2e^{2t}} \left| \begin{array} \\ 2e^{4t} & e^{-t} \\ e^{4t} & -e^{-t} \end{array} \right| &=& \displaystyle{3e^{3t} \over 2e^{2t}} &=& \displaystyle {3 \over 2} e^t, \\ u_2' &=& -\displaystyle {1 \over 2e^{2t}} \left| \begin{array} \\ e^{3t} & 2e^{4t} \\ e^{3t} & e^{4t} \end{array} \right| &=& \displaystyle {e^{7t} \over 2e^{2t}} &=& \displaystyle {1 \over 2} e^{5t}. \end{array} \end{eqnarray*}$$

Therefore

$$\begin{eqnarray*} {\bf u}' = {1 \over 2} \left[ \begin{array} \\ 3e^t \\ e^{5t} \end{array} \right]. \end{eqnarray*}$$

Integrating and taking the constants of integration to be zero yields

$$\begin{eqnarray*} {\bf u} = {1 \over 10} \left[ \begin{array} \\ 15e^t \\ e^{5t} \end{array} \right], \end{eqnarray*}$$

so

$$\begin{eqnarray*} {\bf y}_p = Y {\bf u} = \displaystyle{1 \over 10} \left[ \begin{array} \\ e^{3t} & e^{-t} \\ e^{3t} & -e^{-t} \end{array} \right] \left[ \begin{array} \\ 15e^t \\ e^{5t} \end{array} \right] = \phantom{0} \displaystyle{1 \over 5} \left[ \begin{array} \\ 8e^{4t} \\ 7e^{4t} \end{array} \right] \end{eqnarray*}$$

is a particular solution of $\eqref{eq:4.7.3}$.

(b) From Theorem $(4.7.1)$, the general solution of $\eqref{eq:4.7.3}$ is

$$\begin{equation} \label{eq:4.7.5} {\bf y}={\bf y}_p+c_1{\bf y}_1+c_2{\bf y}_2= {1\over5}\left[\begin{array}{c}8e^{4t}\\7e^{4t}\end{array}\right] +c_1\left[\begin{array}{r}e^{3t}\\e^{3t}\end{array}\right] +c_2\left[\begin{array}{r}e^{-t}\\-e^{-t}\end{array}\right], \end{equation}$$

which can also be written as

$$\begin{eqnarray*} {\bf y} = {1 \over 5} \left[ \begin{array} \\ 8e^{4t} \\ 7e^{4t} \end{array} \right] + \left[ \begin{array} \\ e^{3t} & e^{-t} \\ e^{3t} & -e^{-t} \end{array} \right] {\bf c}, \end{eqnarray*}$$

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

Writing $\eqref{eq:4.7.5}$ in terms of coordinates yields

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

so our result is consistent with Example $(4.2.1)$.

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 $\PageIndex{2}$

Find a particular solution of

$$\begin{equation} \label{eq:4.7.6} {\bf y}'=\left[\begin{array}{cc}2&2e^{-2t}\\2e^{2t}&4\end{array}\right]{\bf y} + \left[ \begin{array} \\ 1 \\ 1 \end{array} \right], \end{equation}$$

given that

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

is a fundamental matrix for the complementary system.

Answer

We seek a particular solution ${\bf y}_p=Y{\bf u}$ of $\eqref{eq:4.7.6}$ where $Y{\bf u}'={\bf f}$; that is,

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

The determinant of $Y$ is the Wronskian

$$\begin{eqnarray*} \left| \begin{array} \\ e^{4t} & -1 \\ e^{6t} & e^{2t} \end{array} \right| = 2e^{6t}. \end{eqnarray*}$$

By Cramer's rule,

$$\begin{eqnarray*} \begin{array} \\ u_1' &=& \displaystyle {1 \over 2e^{6t}} \left| \begin{array} \\ 1 & -1 \\ 1 & e^{2t} \end{array} \right| &=& \displaystyle{e^{2t} + 1 \over 2e^{6t}} &=& \displaystyle{e^{4t} + e^{-6t} \over 2} \\ u_2' &=& \displaystyle {1 \over 2e^{6t}} \left| \begin{array} \\ e^{4t} & 1 \\ e^{6t} & 1 \end{array} \right| &=& \displaystyle{e^{4t} - e^{6t} \over 2 e^{6t}} &=& \displaystyle{e^{-2t} - 1 \over 2}. \end{array} \end{eqnarray*}$$

Therefore

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

Integrating and taking the constants of integration to be zero yields

$$\begin{eqnarray*} {\bf u} = - {1 \over {24}} \left[ \begin{array} \\ 3e^{-4t} + 2e^{-6t} \\ 6e^{-2t} + 12t \end{array} \right], \end{eqnarray*}$$

so

$$\begin{eqnarray*} {\bf y}_p = Y {\bf u} = -\displaystyle{1 \over 24} \left[ \begin{array} \\ e^{4t} & -1 \\ e^{6t} & e^{2t} \end{array} \right] \left[ \begin{array} \\ 3e^{-4t} + 2e^{-6t} \\ 6e^{-2t} + 12t \end{array} \right] = \displaystyle {1 \over 24} \left[ \begin{array} \\ 4e^{-2t} + 12t - 3 \\ -3e^{2t}(4t+1)-8 \end{array} \right] \end{eqnarray*}$$

is a particular solution of $\eqref{eq:4.7.6}$.

Example $\PageIndex{3}$

Find a particular solution of

$$\begin{equation} \label{eq:4.7.7} {\bf y}'=-{2\over t^2 }\left[\begin{array}{cc}t&-3t^2\\1&-2t\end{array}\right]{\bf y} +t^2 \left[ \begin{array} \\ 1 \\ 1 \end{array} \right], \end{equation}$$

given that

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

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

Answer

We seek a particular solution ${\bf y}_p=Y{\bf u}$ of $\eqref{eq:4.7.7}$ where $Y{\bf u}'={\bf f}$; that is,

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

The determinant of $Y$ is the Wronskian

$$\begin{eqnarray*} \left| \begin{array} \\ 2t & 3t^2 \ 1 & 2t \end{array} \right| = t^2. \end{eqnarray*}$$

By Cramer's rule,

$$\begin{eqnarray*} \begin{array} \\ u_1' &=& \displaystyle {1 \over t^2} \left| \begin{array} \\ t^2 & 3t^2 \\ t^2 & 2t \end{array} \right| &=& \displaystyle{2t^3 - 3t^4 \over t^2} &=& 2t - 3t^2,\\ u_2' &=& \displaystyle {1 \over t^2} \left| \begin{array} \\ 2t & t^2 \\ 1 & t^2 \end{array} \right| &=& \displaystyle {2t^3 - t^2 \over t^2} &=& 2t-1. \end{array} \end{eqnarray*}$$

Therefore

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

Integrating and taking the constants of integration to be zero yields

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

so

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

is a particular solution of $\eqref{eq:4.7.7}$.

Example $\PageIndex{4}$

(a) Find a particular solution of

$$\begin{equation} \label{eq:4.7.8} {\bf y}'= \left[ \begin{array} \\ 2 & {-1} & {-1} \\ 1 & 0 & {-1} \\ 1 & {-1} & 0 \end{array} \right]{\bf y}+\left[\begin{array}{c}e^{t}\\0\\e^{-t} \end{array}\right]. \end{equation}$$

(b) Find the general solution of $\eqref{eq:4.7.8}$.

Answer

(a) The complementary system for $\eqref{eq:4.7.8}$ is

$$\begin{equation} \label{eq:4.7.9} {\bf y}' = \left[ \begin{array} \\ 2 & {-1} & {-1} \\ 1 & 0 & {-1} \\ 1 & {-1} & 0 \end{array} \right] {\bf y}. \end{equation}$$

The characteristic polynomial of the coefficient matrix is

$$\begin{eqnarray*} \left| \begin{array} \\ 2 - \lambda & -1 & -1 \\ 1 & -\lambda & -1 \\ 1 & -1 & -\lambda \end{array} \right| = -\lambda (\lambda-1)^2. \end{eqnarray*}$$

Using the method of Section $4.4$, we find that

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

are linearly independent solutions of $\eqref{eq:4.7.9}$. Therefore

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

is a fundamental matrix for $\eqref{eq:4.7.9}$. We seek a particular solution ${\bf y}_p=Y{\bf u}$ of $\eqref{eq:4.7.8}$, where
$Y{\bf u}'={\bf f}$; that is,

$$\begin{eqnarray*} \left[ \begin{array} \\ 1 & e^t & e^t \\ 1 & e^t & 0 \\ e & 0 & e^t \end{array} \right] \left[ \begin{array} \\ {u_1'} \\ {u_2'} \\ {u_3'} \end{array} \right] = \left[ \begin{array} \\ e^t \\ 0 \\ e^{-t} \end{array} \right]. \end{eqnarray*}$$

The determinant of $Y$ is the Wronskian

$$\begin{eqnarray*} \left| \begin{array} \\ 1 & e^t & e^t \\ 1 & e^t & 0 \\ 1 & 0 & e^t \end{array} \right| = -e^{2t}. \end{eqnarray*}$$

Thus, by Cramer's rule,

$$\begin{eqnarray*} \begin{array} \\ u_1' &=& - \displaystyle {1 \over e^{2t}} \left| \begin{array} \\ e^t & e^t & e^t \\ 0 & e^t & 0 \\ e^{-t} & 0 & e^t \end{array} \right| &=& -\displaystyle {e^{3t} - e^t \over e^{2t}} &=& e^{-t} - e^t \\ u_2' &=& -\displaystyle{1 \over e^{2t}} \left| \begin{array} \\ 1 & e^t & e^t \\ 1 & 0 & 0 \\ 1 & e^{-t} & e^t \end{array} \right| &=& -\displaystyle{1 - e^{2t} \over e^{2t}} &=& 1 - e^{-2t} \\ u_3' &=& -\displaystyle{1 \over e^{2t}} \left| \begin{array} \\ 1 & e^t & e^t \\ 1 & e^t & 0 \\ 1 & 0 & e^{-t} \end{array} \right| &=& \displaystyle{e^{2t} \over e^{2t}} &=& 1. \end{array} \end{eqnarray*}$$

Therefore

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

Integrating and taking the constants of integration to be zero yields

$$\begin{eqnarray*} {\bf u} = \left[ \begin{array} \\ -e^t - e^{-t} \\ \displaystyle{e^{-2t} \over 2} + t \\ t \end{array} \right], \end{eqnarray*}$$

so

$$\begin{eqnarray*} \begin{array} \\ {\bf y}_p = Y {\bf u} &=& \left[ \begin{array} \\ 1 & e^t & e^t \\ 1 & e^t & 0 \\ 1 & 0 & e^t \end{array} \right] \left[ \begin{array} \\ -e^t - e^{-t} \\ \displaystyle{e^{-2t} \over 2} + t \\ t \end{array} \right] &=& \left[ \begin{array} \\ e^t(2t-1) -\displaystyle{e^{-t} \over 2} \\ e^t(t-1) - \displaystyle{e^{-t} \over 2} \\ e^t(t-1) - \displaystyle{e^{-t}} \end{array} \right] \end{array} \end{eqnarray*}$$

is a particular solution of $\eqref{eq:4.7.8}$.

(b) From Theorem $(4.7.1)$ the general solution of $\eqref{eq:4.7.8}$ is

$$\begin{eqnarray*} {\bf y} = {\bf y}_p + c_1 {\bf y}_1 + c_2{\bf y}_2 + c_3{\bf y}_3 = \left[ \begin{array} \\ e^t(2t-1) - \displaystyle{e^{-t} \over 2} \\ e^t(t-1) - \displaystyle{e^{-t} \over 2} \\ e^t(t-1) - \displaystyle{e^{-t}} \end{array} \right] + c_1 \left[ \begin{array} \\ 1 \\ 1 \\ 1 \end{array} \right] + c_2 \left[ \begin{array} \\ e^t \\ e^t \\ 0 \end{array} \right] + c_3 \left[ \begin{array} \\ e^t \\ 0 \\ e^t \end{array} \right], \end{eqnarray*}$$

which can be written as

$$\begin{eqnarray*} {\bf y} = {\bf y}_p + Y{\bf c} = \left[ \begin{array} \\ e^t(2t-1) - \displaystyle{e^{-t} \over 2} \\ e^t(t-1) - \displaystyle{e^{-t} \over 2} \\ e^t(t-1) - \displaystyle{e^{-t}} \end{array} \right] + \left[ \begin{array} \\ 1 & e^t & e^t \\ 1 & e^t & 0 \\ 1 & 0 & e^t \end{array} \right] {\bf c} \end{eqnarray*}$$

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

Example $\PageIndex{5}$

Find a particular solution of

$$\begin{equation} \label{eq:4.7.10} {\bf y}'={1\over2} \left[\begin{array}{ccc}3&e^{-t}&-e^{2t}\\0&6&0\\-e^{-2t}&e^{-3t}&-1\end{array}\right] {\bf y}+\left[\begin{array}{c}1\\e^t\\e^{-t}\end{array}\right], \end{equation}$$

given that

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

is a fundamental matrix for the complementary system.

Answer

We seek a particular solution of $\eqref{eq:4.7.10}$ in the form ${\bf y}_p=Y{\bf u}$, where $Y{\bf u}'={\bf f}$; that is,

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

The determinant of $Y$ is the Wronskian

$$\begin{eqnarray*} \left| \begin{array} \\ e^t & 0 & e^{2t} \\ 0 & e^{3t} & e^{3t} \\ e^{-t} & 1 & 0 \end{array} \right| = -2e^{4t}. \end{eqnarray*}$$

By Cramer's rule,

$$\begin{eqnarray*} \begin{array} \\ u_1' &=& -\displaystyle{1 \over 2e^{4t}} \left| \begin{array} \\ 1 & 0 & e^{2t} \\ e^t & e^{3t} & e^{3t} \\ e^{-t} & 1 & 0 \end{array} \right| &=& \displaystyle{e^{4t} \over 2 e^{4t}} &=& \displaystyle{1 \over 2} \\ u_2' &=& -\displaystyle{1 \over 2 e^{4t}} \left| \begin{array} \\ e^t & 1 & e^{2t} \\ 0 & e^t & e^{3t} \\ e^{-t} & e^{-t} & 0 \end{array} \right| &=& \displaystyle{e^{3t} \over 2e^{4t}} &=& \displaystyle{1 \over 2} e^{-t} \\ u_3' &=& -\displaystyle{1 \over 2e^{4t}} \left| \begin{array} \\ e^t & 0 & 1 \\ 0 & e^{3t} & e^t \\ e^{-t} & 1 & e^{-t} \end{array} \right| &=& -\displaystyle{e^{3t} - 2e^{2t} \over 2e^{4t}} &=& \displaystyle{2e^{-2t} - e^{-t} \over 2} \end{array} \end{eqnarray*}$$

Therefore

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

Integrating and taking the constants of integration to be zero yields

$$\begin{eqnarray*} {\bf u} = {1 \over 2} \left[ \begin{array} \\ t \\ -e^{-t} \\ e^{-t} - e^{-2t} \end{array} \right], \end{eqnarray*}$$

so

$$\begin{eqnarray*} {\bf y}_p = Y {\bf u} = {1 \over 2} \left[ \begin{array} \\ e^t & 0 & e^{2t} \\ 0 & e^{3t} & e^{3t} \\ e^{-t} & 1 & 0 \end{array} \right] \left[ \begin{array} \\ t \\ -e^{-t} \\ e^{-t} - e^{-2t} \end{array} \right] = \displaystyle{1 \over 2} \left[ \begin{array} \\ e^t(t+1) - 1 \\ -e^t \\ e^{-t}(t-1) \end{array} \right] \end{eqnarray*}$$

is a particular solution of $\eqref{eq:4.7.10}$.