Question

In problems 31 and 32 solve the given initial-value problem.

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{2}& \mathbf{4}\\ \mathbf{-}\mathbf{1}& \mathbf{6}\end{array}\mathbf{\right)}\mathit{X}\mathbf{,}\mathbf{}\mathbf{}\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{-}\mathbf{1}\\ \mathbf{6}\end{array}\mathbf{\right)}$

Short answer

The general solution of the given system using the given initial-value$X\left(0\right)=\left(\begin{array}{c}-1\\ 6\end{array}\right)$is found to be $X=-7\left(\begin{array}{c}2\\ 1\end{array}\right)e^{4t}+13\left[\left(\left(\begin{array}{c}2\\ 1\end{array}\right)t+\left(\begin{array}{c}1\\ 1\end{array}\right)\right)e^{4t}\right]$.

Step-by-step solution

Definition of general solution-Homogeneous Systems

Let ${\mathit{\lambda }}_{\mathbf{1}}\mathbf{,}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{\lambda }}_{\mathbf{n}}$be ndistinct real eigen values of the coefficient matrix Aof the homogeneous system and let ${\mathit{K}}_{\mathbf{1}}\mathbf{,}{\mathit{K}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{K}}_{\mathbf{n}}$be the correspondingeigen vectors.

Then the general solution on the interval $(-\infty ,\infty )$is given by

$\mathit{X}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{K}}_{\mathbf{1}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{1}}\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{K}}_{\mathbf{2}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{2}}\mathbf{t}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{n}}{\mathit{K}}_{\mathbf{n}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{n}}\mathbf{t}}$

We are given,

$X\text{'}=\left(\begin{array}{cc}2& 4\\ -1& 6\end{array}\right)X$

Where, $A=\left(\begin{array}{cc}2& 4\\ -1& 6\end{array}\right)$

Eigen values of the coefficient matrix 

Obtaining the eigen values of the coefficient matrix,

$$\begin{gathered} det\left(\mathrm{A}-\lambda \mathrm{I}\right)=0 \\ \left|\begin{array}{cc}2-\lambda & 4\\ -1& 6-\lambda \end{array}\right|=0 \\ \left(2-\lambda \right)\left(6-\lambda \right)-\left(-4\right)=0 \\ 12-2\lambda -6\lambda +{\lambda }^2+4=0 \\ {\lambda }^2-8\lambda +16=0 \\ \left(\lambda -4\right)\left(\lambda -4\right)=0 \end{gathered}$$

So our eigen values are $\lambda =4$, which is a multiplicity of two.

Eigen vector and its corresponding solution vector

For $\lambda =4$:

localid="1664244600795" $$\begin{gathered} \left(\mathrm{A}-4\mathrm{I}\right)\mathrm{K}=0 \\ \left(\begin{array}{cc}2-4& 4\\ -1& 6-4\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}-2& 4\\ -1& 2\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

Apply row operation, $2R_2-R_1\to R_2$:

$\left(\begin{array}{cc}-2& 4\\ 0& 0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

So we have $-2k_1+4k_2=0\to k_1=2k_2$

Choosing $k_2=1$yields $k_1=2$.

This gives an eigen vector and a corresponding solution vector

$K=\left(\begin{array}{c}2\\ 1\end{array}\right),X_1=\left(\begin{array}{c}2\\ 1\end{array}\right)e^{4t}$

Next,

localid="1664244729125" $$\begin{gathered} \left(A-4I\right)P=K \\ \left(\begin{array}{cc}2-4& 4\\ -1& 6-4\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right) \\ \left(\begin{array}{cc}-2& 4\\ -1& 2\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right) \end{gathered}$$

Apply row operation, $2R_2-R_1\to R_2$:

$\left(\begin{array}{cc}-2& 4\\ 0& 0\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)$

So we have $-2p_1+4p_2=2\to 4p_2=2+2p_1$

Choosing $p_1=1$yields $p_2=1$.

This gives an eigen vector and a corresponding solution vector

localid="1664245083074" $P=\left(\begin{array}{c}1\\ 1\end{array}\right),X_2=\left(\begin{array}{c}2\\ 1\end{array}\right)t{e}^{4t}+\left(\begin{array}{c}1\\ 1\end{array}\right)e^{4t}$

General solution of given system

Finally, the general solution is given by

$X=c_1{X}_1+c_2{X}_2$

Substituting the values, we get,

$X=c_1\left(\begin{array}{c}2\\ 1\end{array}\right)t{e}^{4t}+c_2\left[\left(\left(\begin{array}{c}2\\ 1\end{array}\right)t+\left(\begin{array}{c}1\\ 1\end{array}\right)\right)e^{4t}\right]...\left(A\right)$

Applying initial-value condition

Now, Apply the given initial-value condition $X\left(0\right)=\left(\begin{array}{c}-1\\ 6\end{array}\right)$in (A),

$\left(\begin{array}{c}-1\\ 6\end{array}\right)=c_1\left(\begin{array}{c}2\\ 1\end{array}\right)+c_2\left(\begin{array}{c}2\\ 1\end{array}\right)$

So we have,

$$\begin{gathered} -1=2c_1+c_2...\left(1\right) \\ 6=c_1+c_2...\left(2\right) \end{gathered}$$

Subtracting (2) from (1), we obtain $-1-6=2c_1-c_1 \to c_1=-7$

Now, substitute c₁ in (1), $-1=2(-7)+c_2 \to c_2=13$

Thus, the solution is given by

localid="1664245740424" $X=-7\left(\begin{array}{c}2\\ 1\end{array}\right)e^{4t}+13\left[\left(\left(\begin{array}{c}2\\ 1\end{array}\right)t+\left(\begin{array}{c}1\\ 1\end{array}\right)\right)e^{4t}\right]$

Therefore, the general solution of the given system using the given initial-value $X\left(0\right)=\left(\begin{array}{c}-1\\ 6\end{array}\right)$ is found to be $X=-7\left(\begin{array}{c}2\\ 1\end{array}\right)e^{4t}+13\left[\right(\left(\begin{array}{c}2\\ 1\end{array}\right)t+\left(\begin{array}{c}1\\ 1\end{array}\right)\left)e^{4t}\right]$