4.2E: Exercises

Exercise \(\PageIndex{1}\)

Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants \(c_1\) and \(c_2\).

1. \(\begin{array}{ccl}y'_1&=&2y_1 + 4y_2\\[4pt] y_2'&=&4y_1+2y_2;\end{array} \quad {\bf y}=c_1 \left[\begin{array}{c} 1\\1\end{array}\right] e^{6t}+c_2 \left[\begin{array}{c} 1\\-1\end{array}\right]e^{-2t}\)
2. \(\begin{array}{ccl}y'_1&=&-2y_1 - 2y_2\\[4pt] y_2'&=&-5y_1 + \phantom{2}y_2;\end{array} \quad {\bf y}=c_1 \left[\begin{array}{c} 1\\1\end{array}\right] e^{-4t}+c_2 \left[\begin{array}{c} -2\\5\end{array}\right] e^{3t}\)
3. \(\begin{array}{ccr}y'_1&=&-4y_1 -10y_2\\[4pt] y_2'&=&3y_1 + \phantom{1}7y_2;\end{array} \quad {\bf y}=c_1 \left[\begin{array}{c} -5\\3\end{array}\right] e^{2t}+c_2 \left[\begin{array}{c} 2\\-1\end{array}\right] e^t\)
4. \(\begin{array}{ccl}y'_1&=&2y_1 +\phantom{2}y_2 \\[4pt] y_2'&=&\phantom{2}y_1 + 2y_2;\end{array} \quad {\bf y}=c_1\left[\begin{array}{c} 1\\1\end{array}\right] e^{3t}+c_2 \left[\begin{array}{c} 1\\-1\end{array}\right] e^t\)

Exercise \(\PageIndex{2}\)

Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants \(c_1\), \(c_2\), and \(c_3\).

(a)\(\begin{array}{ccr}y'_1&=&- y_1+2y_2 + 3y_3 \\
y_2'&=&y_2 + 6y_3\\y_3'&=&- 2y_3;\end{array}\)

\(\mathbf{y} = c_1 \left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right] e^t + c_2 \left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] e^{-t} + c_3 \left[\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right] e^{-2t}.\)
 

(b) \(\begin{array}{ccc}y'_1&=&\phantom{2y_1+}2y_2 + 2y_3 \\
y_2'&=&2y_1\phantom{+2y_2} + 2y_3\\y_3'&=&2y_1 +
2y_2;\phantom{+2y_3}\end{array}\)

\(\mathbf{y} = c_1 \left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right] e^{-2t }+ c_2 \left[\begin{array}{c} 0 \\ 1 \\ -1 \end{array}\right] e^{-2t} + c_3 \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] e^{4t}.\)

 

(c) \(\begin{array}{ccr}y'_1&=&-y_1 +2y_2 + 2y_3\\
y_2'&=&2y_1 -\phantom{2}y_2 +2y_3\\y_3'&=&2y_1 + 2y_2
-\phantom{2}y_3;\end{array}\)

\(\mathbf{y} = c_1 \left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right] e^{-3t }+ c_2 \left[\begin{array}{c} 0 \\ -1 \\  1 \end{array}\right] e^{-3t} + c_3 \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] e^{3t}.\)

 

(d) \(\begin{array}{ccr}y'_1&=&3y_1 - \phantom{2}y_2 -\phantom{2}y_3
\\ y_2'&=&-2y_1 + 3y_2 + 2y_3\\y_3'&=&\phantom{-}4y_1 -\phantom{3}y_2 -
2y_3;\end{array}\)

\(\mathbf{y} = c_1 \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] e^{2t }+ c_2 \left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right] e^{3t} + c_3 \left[\begin{array}{c} 1 \\ -3 \\ 7 \end{array}\right] e^{-t}.\)

 

Exercise \(\PageIndex{3}\)

Rewrite the initial value problem in matrix form and verify that the given vector function is a solution.

(a) \begin{eqnarray*}y'_1 &=&\phantom{-2}y_1+\phantom{4}y_2\\
y_2'&=&-2y_1 + 4y_2,\end{eqnarray*}

\begin{eqnarray*}y_1(0)&=&1\\y_2(0)&=&0;\end{eqnarray*}

\({\bf y}=2 \left[\begin{array}{c} 1\\1\end{array}\right] e^{2t}- \left[\begin{array}{c} 1\\2 \end{array}\right] e^{3t}\)

 

(b) \begin{eqnarray*}y'_1 &=&5y_1 + 3y_2 \\
y_2'&=&- y_1 + y_2,\end{eqnarray*}

\begin{eqnarray*}y_1(0)&=&12\\y_2(0)&=&-6;\end{eqnarray*}

\({\bf y}=3 \left[\begin{array}{c} 1\\-1\end{array}\right] e^{2t}+3 \left[\begin{array}{c} 3\\-1 \end{array}\right] e^{4t}\)

 

Exercise \(\PageIndex{4}\)

Rewrite the initial value problem in matrix form and verify that the given vector function is a solution.

(a) \begin{eqnarray*}y'_1&=&6y_1 + 4y_2 + 4y_3 \\
y_2'&=&-7y_1 -2y_2 - y_3,\\y_3'&=&7y_1 + 4y_2 + 3y_3,\end{eqnarray*}

\begin{eqnarray*}y_1(0)&=&3\\ y_2(0)&=&-6\\ y_3(0)&=&4\end{eqnarray*}

\(\mathbf{y} =  \left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right] e^{6t }+ 2 \left[\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right] e^{2t} +  \left[\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right] e^{-t}.\)

 

(b) \begin{eqnarray*}y'_1&=& \phantom{-}8y_1 + 7y_2 +\phantom{1}7y_3 \\
y_2'&=&-5y_1 -6y_2 -\phantom{1}9y_3,\\y_3'&=& \phantom{-}5y_1 + 7y_2 +10y_3,\end{eqnarray*}

\begin{eqnarray*}y_1(0)&=&2\\ y_2(0)&=&-4\\ y_3(0)&=&3\end{eqnarray*}

\(\mathbf{y} = \left[\begin{array}{c} 1 \\ -1\\ 1 \end{array}\right] e^{8t }+ \left[\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right] e^{3t} +  \left[\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right] e^{t}.\)

 

Exercise \(\PageIndex{5}\)

Rewrite the system in matrix form and verify that the given vector function satisfies the ystem for any choice of the constants \(c_1\) and \(c_2\).

(a) \begin{eqnarray*}y'_1&=&-3y_1+2y_2+3-2t \\ y_2'&=&-5y_1+3y_2+6-3t\end{eqnarray*}

\(\mathbf{y} = c_1 \begin{bmatrix} 2\cos t \\ 3\cos t - \sin t \end{bmatrix} + c_2 \begin{bmatrix} 2\sin t \\ 3\sin t + \cos t \end{bmatrix} + \begin{bmatrix} 1 \\ t \end{bmatrix}\)

 

(b) \begin{eqnarray*}y'_1&=&3y_1+y_2-5e^t \\ y_2'&=&-y_1+y_2+e^t\end{eqnarray*}

\(\mathbf{y} = c_1 \begin{bmatrix} -1 \\ 1 \end{bmatrix} e^{2t} + c_2 \begin{bmatrix} 1+t \\ -t \end{bmatrix} e^{2t} + \begin{bmatrix} 1 \\ 3 \end{bmatrix} e^t \)

 

(c) \begin{eqnarray*}y'_1&=&-y_1-4y_2+4e^t+8te^t \\ y_2'&=&-y_1-\phantom{4}y_2+e^{3t}+(4t+2)e^t\end{eqnarray*}

\(\mathbf{y} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{-3t} + c_2 \begin{bmatrix} -2 \\ 1 \end{bmatrix} e^t + \begin{bmatrix} e^{3t} \\ 2te^t \end{bmatrix}\)

(d) \begin{eqnarray*}y'_1&=&-6y_1-3y_2+14e^{2t}+12e^t \\ y_2'&=&\phantom{6}y_1-2y_2+7e^{2t}-12e^t\end{eqnarray*}

\(\mathbf{y} = c_1 \begin{bmatrix} -3\\1\end{bmatrix} e^{-5t}+c_2  \begin{bmatrix} -1\\1\end{bmatrix} e^{-3t}+ \begin{bmatrix}{c}e^{2t}+3e^t\\2e^{2t}-3e^t\end{bmatrix}\)

Exercise \(\PageIndex{6}\)

Convert the linear scalar equation

\begin{equation} \label{eq:4.2E.1}
P_0 (t) y^{(n)} + P_1 (t) y^{(n-1)} + \cdots + P_n (t) y(t) = F(t)
\end{equation}

into an equivalent \(n\times n\) system

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

and show that \(A\) and \({\bf f}\) are continuous on an interval \((a,b)\) if and only if \eqref{eq:4.2E.1} is normal on \((a,b)\).

Answer

Let  \(y_i=y^{(i-1)}\), \(i=1,2,\dots,n\); then \(y_i'=y_{i+1}\),
\(i=1,2,\dots,n-1\) and \(P_0(t)y_n'+P_1(t)y_n+\cdots+P_n(t)y_1=F(t)\), so
\[A=-{1\over P_0}
\left[\begin{array}{ccccc}
0&1&0&\cdots&0\\
0&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&1\\
P_n&P_{n-1}&P_{n-2}&\cdots&P_1
\end{array}\right]
\quad \mbox{and} \quad
{\bf f}={1\over P_0}
\left[\begin{array}{c}
0\\ 0\\\vdots\\0\\F
\end{array}\right].
\]
If \(P_0,P_1,\dots,P_n\) and \(F\) are continuous and \(P_0\)
has no zeros on \((a,b)\), then \(P_1/P_0,\dots,P_n/P_0\)  and
\(F/P_0\) are continuous on \((a,b)\).

Exercise \(\PageIndex{7}\)

A matrix function

\begin{eqnarray*}
Q(t) = q_{rs}
\end{eqnarray*}

is said to be \( \textcolor{blue}{\mbox{differentiable}} \) if its entries \(\{q_{ij}\}\) are differentiable. Then the \( \textcolor{blue}{\mbox{derivative}} \) \(Q'\) is defined by

\begin{eqnarray*}
Q'(t) = q'_{rs}.
\end{eqnarray*}

(a) Prove: If \(P\) and \(Q\) are differentiable matrices such that \(P+Q\) is defined and if \(c_1\) and \(c_2\) are constants, then

\begin{eqnarray*}
(c_1 P + c_2 Q)' = c_1 P' + c_2 Q'.
\end{eqnarray*}

(b) Prove: If \(P\) and \(Q\) are differentiable matrices such that \(PQ\) is defined, then

\begin{eqnarray*}
(PQ)' = P'Q + PQ'.
\end{eqnarray*}

Answer

(a)
\((c_1P+c_2Q)'_{ij}=
(c_1p_{ij}+c_2q_{ij})'=c_1p_{ij}'+c_2q_{ij}'=(c_1P'+c_2Q')_{ij}\) ;
hence
\((c_1P+c_2Q)'=c_1P'+c_2Q'\) .

(b) Let \(P\) be \(k\times r\) and \(Q\) be \( r\times s\) ; then
\(PQ\) is \(k\times s\) and \((PQ)_{ij}=\displaystyle\sum_{l=1}^rp_{i,l}q_{\l j}\) .
Therefore,
\((PQ)_{ij}'=\displaystyle\sum_{\l=1}^rp_{i\l}'q_{\l j}
+\sum_{\l=1}^rp_{i\l}q_{\l j}'=(P'Q)_{ij}+(PQ')_{ij}\) . Therefore,
\((PQ)'=P'Q+PQ'\) .

Exercise \(\PageIndex{8}\)

Verify that \(Y' = AY\).

(a) \(Y = \left[ \begin{array} \\ {e^{6t}} & {e^{-2t}}\\ {e^{6t}} & {-e^{-2t}} \end{array} \right], \quad A = \left[ \begin{array} \\ 2 & 4 \\ 4 & 2 \end{array} \right]\)

(b) \( Y = \left[\begin{array} \\ {e^{-4t}} & {-2e^{3t}} \\ {e^{-4t}} & {5e^{3t}} \end{array} \right], \quad A = \left[ \begin{array} \\ {-2} & {-2} \\ {-5} & {1} \end{array} \right] \)

(c) \( Y = \left[ \begin{array} \\ {-5e^{2t}} & {2e^t} \\ {3e^{2t}} & {-e^t} \end{array} \right], \quad A = \left[ \begin{array} \\ {-4} & {-10} \\ 3 & 7 \end{array} \right] \)

(d) \( Y = \left[ \begin{array} \\ {e^{3t}} & {e^t} \\ {e^{3t}} & {-e^t} \end{array} \right], \quad A = \left[ \begin{array} \\ 2 & 1 \\ 1 & 2 \end{array} \right] \)

(e) \(Y = \left[ \begin{array} \\ {e^t} & {e^{-t}} & {e^{-2t}} \\ {e^t} & 0 & {-2e^{-2t}} \\ 0 & 0 & {e^{-2t}} \end{array} \right], \quad A = \left[ \begin{array} \\ {-1} & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & {-2} \end{array} \right] \)

(f) \( Y = \left[ \begin{array} \\ {-e^{-2t}} & {-e^{-2t}} & {e^{4t}} \\ 0 & {\phantom{-} e^{-2t}} & {e^{4t}} \\ {e^{-2t}} & 0 & { e^{4t}} \end{array} \right], \quad A = \left[ \begin{array} \\ 0 & 2 & 2 \\ 2 & 0 & 2 \\ 2 & 2 & 0 \end{array} \right] \)

(g) \( Y = \left[ \begin{array} \\ {e^{3t}} & {e^{-3t}} & 0 \\ {e^{3t}} & 0 & {-e{-3t}} \\ {e^{3t}} & {e^{-3t}} & {\phantom{-}e^{-3t}} \end{array} \right], \quad A = \left[ \begin{array} \\ {-9} & 6 & 6 \\ {-6} & 3 & 6 \\ {-6} & 6 & 3 \end{array} \right] \)

(h) \( Y = \left[ \begin{array} \\ {e^{2t}} & {e^{3t}} & {e^{-t}} \\ 0 & {-e^{-3t}} & {-3e^{-t}} \\ {e^{2t}} & {e^{3t}} & {7e^{-t}} \end{array} \right], \quad A = \left[ \begin{array} \\ 3 & {-1} & {-1} \\ {-2} & 3 & 2 \\ 4 & {-1} & {-2} \end{array} \right] \)

Exercise \(\PageIndex{9}\)

Suppose \( {\bf y}_1 = \left[ \begin{array} \\ y_{11} \\ y_{21} \end{array} \right] \quad \mbox{and} \quad {\bf y}_2 = \left[ \begin{array} \\ y_{12} \\ y_{22} \end{array} \right] \) are solutions of the homogeneous system

\begin{equation} \label{eq:4.2E.2}
{\bf y}' = A(t) {\bf y},
\end{equation}

and define \( Y = \left[ \begin{array} \\ \; y_{11} \; y_{12} \\ \; y_{21} \; y_{22} \end{array} \right] \).

(a) Show that \(Y'=AY\).

(b) Show that if \({\bf c}\) is a constant vector then \({\bf y}= Y{\bf c}\) is a solution of \eqref{eq:4.2E.2}.

(c) State generalizations of part (a) and part (b) for \(n\times n\) systems.

Exercise \(\PageIndex{10}\)

Suppose \(Y\) is a differentiable square matrix.

(a) Find a formula for the derivative of \(Y^2\).

(b) Find a formula for the derivative of \(Y^n\), where \(n\) is any positive integer.

(c) State how the results obtained in part (a) and part (b) are analogous to results from calculus concerning scalar functions.

Answer

(a)
From Exercise~4.2.7(b) with \(P=Q=X\) ,
\((X^2)'=(XX)'=X'X+XX'\) .

(b) By starting from
 Exercise~4.2.7(b) and using induction it can be shown
if \(P_1,P_2,\dots,P_n\) are square matrices of the same order, then
 \((P_1P_2\cdots P_n)'=P_1'P_2\cdots P_n+P_1P_2'\cdots P_n+\cdots+
P_1P_2\cdots P_n'\) . Taking \(P_1=P_2=\cdots=P_n=X\)  yields
(A) \((Y^n)'=Y'Y^{n-1}+YY'Y^{n-2}+Y^2Y'Y^{n-3}+\cdots+Y^{n-1}Y'=\displaystyle{
\sum_{r=0}^{n-1}}Y^rY'Y^{n-r-1}\) .

(c)
If \(Y\) is a scalar function, then (A) reduces to
the familiar result \((Y^n)'=nY^{n-1}Y'\) .
 

Exercise \(\PageIndex{11}\)

It can be shown that if \(Y\) is a differentiable and invertible square matrix function, then \(Y^{-1}\) is differentiable.

(a) Show that \((Y^{-1})' = -Y^{-1}Y'Y^{-1}\).
Hint: Differentiate the identity \(Y^{-1}Y=I\).

(b) Find the derivative of \(Y^{-n}=\left[Y^{-1}\right]^n\), where \(n\) is a positive integer.

(c) State how the results obtained in part (a) and part (b) are analogous to results from calculus concerning scalar functions.

Exercise \(\PageIndex{12}\)

Show that Theorem \((4.2.1)\) implies Theorem \((3.1.1)\).

Hint: Write the scalar equation

\begin{eqnarray*}
P_0(x)y^{(n)} + P_1(x)y^{(n-1)} + \cdots + P_n(x)y = F(x)
\end{eqnarray*}

as an \(n\times n\) system of linear equations.

Answer

From Exercise~4.2.6, the initial value problem
(A) \(P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x)\) ,
\(Y(x_0)=k_0,\ y'(x_0)=k_1,\dots,\ y^{(n-1)}(x_0)=k_{n-1}\) is
equivalent to the initial value problem (B) \({\bf y'}=A(t){\bf y}+{\bf
f}(t)\) , with
$$A=-{1\over P_0}
\left[\begin{array}{ccccc}
0&1&0&\cdots&0\\
0&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&1\\
P_n&P_{n-1}&P_{n-2}&\cdots&P_1
\end{array}\right],
{\bf f}={1\over P_0}
\left[\begin{array}{c}
0\\ 0\\\vdots\\0\\F
\end{array}\right],
\mbox{\ and \ }
{\bf k}=
\left[\begin{array}{c}
k_0\\k_1\\\vdots\\k_{n-1}
\end{array}\right].
$$
Since Theorem~4.2.1 implies that (B) has a unique solution on
\((a,b)\) , it follows that (A) does also.

Exercise \(\PageIndex{13}\)

Suppose \({\bf y}\) is a solution of the \(n\times n\) system \({\bf y}'=A(t){\bf y}\) on \((a,b)\), and that the \(n\times n\) matrix \(P\) is invertible and differentiable on \((a,b)\). Find a matrix \(B\) such that the function \({\bf x}=P{\bf y}\) is a solution of \({\bf x}'=B{\bf x}\) on \((a,b)\).