91Ó°ÊÓ

Let$b\left(t\right)$is the balance in the account at time$t$and$b\left(0\right)=1000$.

The bank is paying interest at a continuous rate of$5\%$per year.

Ngozi makes deposits in to the account at a continuous rate of$s\left(t\right)$.

And$s\left(t\right)$is increasing at a continuous rate of$7\%$per year also that$s\left(0\right)=1000$.

The setup of a linear system as follows.

$\left|\begin{array}{c}\frac{db}{dt}=?b+?s\\ \frac{ds}{dt}=?b+?s\end{array}\right|$

Since, the bank is already paying interest at a continuous rate of$5\%$per year and Ngozi makes deposits into the account at a continuous rate of$s\left(t\right)$per year.

The first equation can written as follows.

$\left|\begin{array}{c}\frac{db}{dt}=0.05b+1s\\ \frac{ds}{dt}=?b+?s\end{array}\right|$

Since,$s\left(t\right)$is increasing at a continuous rate of$7\%$per year.

Therefore, the equation can be written as follows.

$\left|\begin{array}{c}\frac{db}{dt}=0.05b+1s\\ \frac{ds}{dt}=0b+0.07s\end{array}\right|$

Hence, the setup of the linear system is$\left|\begin{array}{c}\frac{db}{dt}=0.05b+1s\\ \frac{ds}{dt}=0b+0.07s\end{array}\right|$

Consider the linear system of equation as follows.

$\left|\begin{array}{c}\frac{db}{dt}=0.05b+1s\\ \frac{ds}{dt}=0b+0.07s\end{array}\right|$

Rewrite the system in the form of$\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$as follows.

$\left[\begin{array}{c}\frac{db}{dt}\\ \frac{ds}{dt}\end{array}\right]=\left[\begin{array}{cc}0.05& 1\\ 0& 0.07\end{array}\right]\left[\begin{array}{c}b\\ s\end{array}\right]$

Since the coefficient matrix$A=\left[\begin{array}{cc}0.05& 1\\ 0& 0.07\end{array}\right]$is a diagonal matrix, the eigenvalues are the diagonal entries.

The eigenvalues are as follows.

$\begin{array}{l}{\mathrm{λ}}_1=0.05\\ {\mathrm{λ}}_2=0.07\end{array}$

Now, find the corresponding eigenvectors for eigenvalue${\mathrm{λ}}_1=0.05$ as follows.

$\begin{array}{c}(A-\mathrm{λ}I)=0\\ \left[\begin{array}{cc}0.05-\left(0.05\right)& 1\\ 0& 0.07-\left(0.05\right)\end{array}\right]\left[\begin{array}{c}{u}_1\\ u_2\end{array}\right]=\left[\begin{array}{c}0\\ 00\end{array}\right]\\ \left[\begin{array}{cc}0& 1\\ 0& 0.02\end{array}\right]\left[\begin{array}{c}{u}_1\\ u_2\end{array}\right]=\left[\begin{array}{c}0\\ 00\end{array}\right]\\ \left[\begin{array}{c}0u_1+1u_2\\ 0u_1+0.02u_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

The corresponding equations are as follows.

$\begin{array}{c}0u_1+1u_2=0\\ 0u_1+0.2u_2=0\end{array}$

Solving the equations as follows.

$\begin{array}{c}0u_1+1u_2=0\\ u_2=-0u_1\\ u_2=0\end{array}$

Therefore, the eigenvectors as follows.

$\begin{array}{c}\stackrel{→}{u}=\left[\begin{array}{c}{u}_1\\ u_2\end{array}\right]\\ =\left[\begin{array}{c}{u}_1\\ 0\end{array}\right]\\ \stackrel{→}{u}=u_1\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Similarly, find the corresponding eigenvectors for eigenvalue${\mathrm{λ}}_1=0.07$as follows.

$\begin{array}{c}(A-\mathrm{λ}I)=0\\ \left[\begin{array}{cc}0.05-\left(0.07\right)& 1\\ 0& 0.07-\left(0.07\right)\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}0.02& 1\\ 0& 0\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{c}0.02v_1+1v_2\\ 0v_1+0v_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

The corresponding equations are as follows.

$\begin{array}{c}0.02v_1+1v_2=0\\ 0v_1+0v_2=0\end{array}$

Solving the equations as follows.

$\begin{array}{c}0.02v_1+1v_2=0\\ 0.02v_1=-v_2\\ v_1=\frac{-1v_2}{0.02}\\ v_1=50v_2\end{array}$

Therefore, the eigenvectors as follows.

$\begin{array}{c}\stackrel{→}{v}=\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]\\ =\left[\begin{array}{c}50v_2\\ v_2\end{array}\right]\\ \stackrel{→}{v}=v_2\left[\begin{array}{c}50\\ 1\end{array}\right]\end{array}$

Thus, the eigenvectors are as follows.

$\left[\begin{array}{c}1\\ 0\end{array}\right]\text{ }$and$\text{ â¶Ä‰}\left[\begin{array}{c}50\\ 1\end{array}\right]$

The general solution to the system can be written as follows.

$\begin{array}{c}\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=c_1{e}^{{\mathrm{λ}}_{1}t}\stackrel{→}{u}+c_2{e}^{{\mathrm{λ}}_{2}t}\stackrel{→}{v}\\ =c_1{e}^{0.05t}\left[\begin{array}{c}1\\ 0\end{array}\right]+c_2{e}^{0.07t}\left[\begin{array}{c}50\\ 1\end{array}\right]\\ =\left[\begin{array}{c}{c}_1{e}^{0.05t}\\ 0\end{array}\right]+\left[\begin{array}{c}50c_2{e}^{0.07t}\\ c_2{e}^{0.07t}\end{array}\right]\\ \left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_1{e}^{0.05t}+50c_2{e}^{0.07t}\\ c_2{e}^{0.07t}\end{array}\right]\end{array}$

Now substitute the value 0 forin$\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_1{e}^{0.05t}+50c_2{e}^{0.07t}\\ c_2{e}^{0.07t}\end{array}\right]$as follows.

$\begin{array}{l}\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_1{e}^{0.05t}+50c_2{e}^{0.07t}\\ c_2{e}^{0.07t}\end{array}\right]\\ \left[\begin{array}{c}b\left(0\right)\\ s\left(0\right)\end{array}\right]=\left[\begin{array}{c}{c}_1{e}^{0.05\left(0\right)}+50c_2{e}^{0.07\left(0\right)}\\ c_2{e}^{0.07\left(0\right)}\end{array}\right]\\ \left[\begin{array}{c}1000\\ 1000\end{array}\right]=\left[\begin{array}{c}{c}_1+50c_2\\ c_2\end{array}\right]\end{array}$

Now, equating the corresponding entries of the matrix on sides of the equation as follows.

$\begin{array}{c}{c}_1+50c_2=1000\\ c_2=1000\end{array}$

Substitute the value$1000$for$c_2$in$c_1+50c_2=1000$as follows.

$\begin{array}{c}{c}_1+50c_2=1000\\ c_1+50\left(1000\right)=1000\\ c_1+50000=1000\\ c_1=1000-50000\end{array}$

Simplify further as follows.

$c_1=−49,000$

Substitute the value$−49,000$ for$c_1$ and$1000$ for$c_2$ in$\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_1{e}^{0.05t}+50c_2{e}^{0.07t}\\ c_2{e}^{0.07t}\end{array}\right]$ as follows.

$\begin{array}{l}\left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}{c}_1{e}^{0.05t}+50c_2{e}^{0.07t}\\ c_2{e}^{0.07t}\end{array}\right]\\ \left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}-49000e^{0.05t}+50\left(1000\right)c_2{e}^{0.07t}\\ 1000e^{0.07t}\end{array}\right]\\ \left[\begin{array}{c}b\left(t\right)\\ s\left(t\right)\end{array}\right]=\left[\begin{array}{c}-49000e^{0.05t}+50000c_2{e}^{0.07t}\\ 1000e^{0.07t}\end{array}\right]\end{array}$

Equating the corresponding entries on both sides of the equation as follows.

$\begin{array}{l}b\left(t\right)=-49000e^{0.05t}+50000c_2{e}^{0.07t}\\ s\left(t\right)=1000e^{0.07t}\end{array}$

Thus, the equations are $b\left(t\right)=-49000e^{0.05t}+50000c_2{e}^{0.07t},\text{ â¶Ä‰}s\left(t\right)=1000e^{0.07t}$.