Question

In Problems 1and 2 fill in the blanks.

The vector$\mathit{X}\mathbf{}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{\left(}\begin{array}{c}\mathbf{-}\mathbf{1}\\ \mathbf{1}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{9}\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathbf{\left(}\begin{array}{c}\mathbf{5}\\ \mathbf{2}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{7}\mathbf{t}}$is the solution of the initial-value problem $\mathit{X}\mathbf{\text{'}}\mathbf{}\mathbf{=}\left[\begin{array}{cc}1& 10\\ 6& -3\end{array}\right]\mathit{X}\mathbf{,}\mathbf{}\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\left(\begin{array}{c}2\\ 0\end{array}\right)$for c₁= and c₂= .

Short answer

The vector $\mathit{X}\mathbf{}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{\left(}\begin{array}{c}\mathbf{-}\mathbf{1}\\ \mathbf{1}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{9}\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathbf{\left(}\begin{array}{c}\mathbf{5}\\ \mathbf{2}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{7}\mathbf{t}}$is the solution of the initial-value problem $\mathit{X}\mathbf{\text{'}}\mathbf{}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{1}& \mathbf{10}\\ \mathbf{6}& \mathbf{-}\mathbf{3}\end{array}\mathbf{\right]}\mathit{X}\mathbf{,}\mathbf{}\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\left(}\begin{array}{c}\mathbf{2}\\ \mathbf{0}\end{array}\mathbf{\right)}$ for C₁ =$-\frac{3}{4}and{C}_2=\frac{1}{4}$.

Step-by-step solution

Define matrix exponential.

Consider a square matrix A of size $n\times n$. This matrix can contain either complex numbers or real numbers. The matrix can be calculated as

$A^0=I,A^1=A,A^2=A.A,A^3=A^2.A,.....,A^k=A.\underset{ktimes}{A}.....$

where I is the unit matrix of order n.

Therefore, the infinite matrix power series is $I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$.

Now, the matrix exponential is defined as the sum of the infinite matrix power series. It is denoted by the expression e^At. It is given by the formula, $e^{tA}=\sum _{k=0}^{\infty }\frac{t^k}{k!}{A}^k$.

Find the value of c1 and c2.

It is given that, $X=c_1\left(\begin{array}{c}-1\\ 1\end{array}\right)e^{-9\mathrm{t}}+c_2\left(\begin{array}{c}5\\ 2\end{array}\right)e^{7\mathrm{t}}$.

Applying the initial condition $X\left(0\right)=\left(\begin{array}{c}2\\ 0\end{array}\right)$.

$$\begin{gathered} \left(\begin{array}{c}2\\ 0\end{array}\right)=C_1\left(\begin{array}{c}-1\\ 1\end{array}\right)+C_2\left(\begin{array}{c}5\\ 3\end{array}\right) \\ \left(\begin{array}{c}2\\ 0\end{array}\right)=\left(\begin{array}{cc}-C_1& 5C_2\\ C_1& 3C_2\end{array}\right) \end{gathered}$$

Now it has two equations,

$-C_1+5C_2=2$ ............(1)

$C_1+3C_2=0$ …… (2)

From the second equation, C₁ = -3C₂.

Substitute the value C₁ = -3C₂into $-C_1+5C_2=2$.

$$\begin{gathered} 3C_2+5C_2=2 \\ 8C_2=2 \\ {C}_2=\frac{1}{4} \\ {C}_1=-3\left(\begin{array}{c}1\\ 4\end{array}\right)=-\left(\begin{array}{c}3\\ 4\end{array}\right) \end{gathered}$$

Hence, the values of C₁ = - (3/4) and C₂ = (1/4).