Chapter 8, Problem 1RE

fill in the blanks.

1. The vector $X=k\left(\begin{array}{c}4\\ 5\end{array}\right)$ is a solution of

$X^{\prime }=\left(\begin{array}{cc}1& 4\\ 2& -1\end{array}\right)X-\left(\begin{array}{c}8\\ 1\end{array}\right)$

for k = _____________.

To determine

The blank in the statement: “The vector $X=k\left(\begin{array}{c}4\\ 5\end{array}\right)$ is a solution of $X^{\prime }=\left(\begin{array}{cc}1& 4\\ 2& -1\end{array}\right)X-\left(\begin{array}{c}8\\ 1\end{array}\right)$ for $k=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$.”

Answer to Problem 1RE

The vector $X=k\left(\begin{array}{c}4\\ 5\end{array}\right)$ is a solution of $X^{\prime }=\left(\begin{array}{cc}1& 4\\ 2& -1\end{array}\right)X-\left(\begin{array}{c}8\\ 1\end{array}\right)$ for $\underset{\_}{k=\frac{1}{3}}$.

Explanation of Solution

Given:

The systems of differential equation is $X^{\prime }=\left(\begin{array}{cc}1& 4\\ 2& -1\end{array}\right)X-\left(\begin{array}{c}8\\ 1\end{array}\right)$.

Calculation:

The vector is $X=k\left(\begin{array}{c}4\\ 5\end{array}\right)$ and the differentiation is $X^{\prime }=\left(\begin{array}{c}0\\ 0\end{array}\right)$ as the given vector is constant.

Substitute $k\left(\begin{array}{c}4\\ 5\end{array}\right)$ for $X$ and $\left(\begin{array}{c}0\\ 0\end{array}\right)$ for $X^{\prime }$ for in equation $X^{\prime }=\left(\begin{array}{cc}1& 4\\ 2& -1\end{array}\right)X-\left(\begin{array}{c}8\\ 1\end{array}\right)$.

$\begin{array}{c}{X}^{\prime }=\left(\begin{array}{cc}1& 4\\ 2& -1\end{array}\right)k\left(\begin{array}{c}4\\ 5\end{array}\right)-\left(\begin{array}{c}8\\ 1\end{array}\right)\\ \left(\begin{array}{c}0\\ 0\end{array}\right)=\left(\begin{array}{cc}1& 4\\ 2& -1\end{array}\right)\left(\begin{array}{c}4k\\ 5k\end{array}\right)-\left(\begin{array}{c}8\\ 1\end{array}\right)\\ \left(\begin{array}{c}0\\ 0\end{array}\right)=\left(\begin{array}{c}1\left(4k\right)+4\left(5k\right)\\ 2\left(4k\right)-1\left(5k\right)\end{array}\right)-\left(\begin{array}{c}8\\ 1\end{array}\right)\\ \left(\begin{array}{c}0\\ 0\end{array}\right)=\left(\begin{array}{c}4k+20k\\ 8k-5k\end{array}\right)-\left(\begin{array}{c}8\\ 1\end{array}\right)\end{array}$

Further simplify the above equation.

$\begin{array}{l}\left(\begin{array}{c}0\\ 0\end{array}\right)=\left(\begin{array}{c}24k\\ 3k\end{array}\right)-\left(\begin{array}{c}8\\ 1\end{array}\right)\\ \left(\begin{array}{c}0\\ 0\end{array}\right)=\left(\begin{array}{c}24k-8\\ 3k-1\end{array}\right)\end{array}$

From above equation, $3k-1=0$ so the value of $k$ is calculated as follows:

$\begin{array}{c}3k-1=0\\ 3k=1\\ k=\frac{1}{3}\end{array}$

Therefore, the value of $k$ is $\frac{1}{3}$.

Thus, the vector $X=k\left(\begin{array}{c}4\\ 5\end{array}\right)$ is a solution of $X^{\prime }=\left(\begin{array}{cc}1& 4\\ 2& -1\end{array}\right)X-\left(\begin{array}{c}8\\ 1\end{array}\right)$ for $\underset{\_}{k=\frac{1}{3}}$.