Question

In Problems 13 and 14 solve the given initial-value problem.

${\mathit{X}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\frac{\mathbf{1}}{\mathbf{2}}& \mathbf{0}\\ \mathbf{1}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\end{array}\mathbf{\right)}\mathit{X}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{3}\\ \mathbf{5}\end{array}\mathbf{\right)}$

Short answer

$$\begin{gathered} K_1=\left(\begin{array}{c}1\\ 1\end{array}\right),X_1=\left(\begin{array}{c}1\\ 1\end{array}\right)e^{\frac{1}{2}t} \\ {K}_2=\left(\begin{array}{c}0\\ 1\end{array}\right),X_2=\left(\begin{array}{c}0\\ 1\end{array}\right)e^{-\frac{1}{2}t} \\ X\left(t\right)=3\left(\begin{array}{c}1\\ 1\end{array}\right)e^{\frac{1}{2}t}+2\left(\begin{array}{c}0\\ 1\end{array}\right)e^{-\frac{1}{2}t} \end{gathered}$$

Step-by-step solution

Distinct Real Eigenvalues

When the $\mathit{n}\mathbf{\times }\mathit{n}$matrix A possesses n distinct real eigenvalues $\mathbf{ }{\mathit{\lambda }}_{\mathbf{1}}\mathbf{,}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{,}\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{.}\mathbf{,}\mathbf{ }{\mathit{\lambda }}_{\mathbf{n}}$, then a set of n linearly independent eigenvectors ${\mathit{K}}_{\mathbf{1}}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }{\mathit{K}}_{\mathbf{2}}\mathbf{,}\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{ }\mathbf{.}\mathbf{,}\mathbf{ }{\mathit{K}}_{\mathbf{n}}$can always be found, and${\mathit{X}}_{\mathbf{1}}\mathbf{=}\mathbf{ }\mathbf{ }{\mathit{K}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{A}\mathbf{2}\mathbf{t}}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{X}}_{\mathbf{2}}\mathbf{=}\mathbf{ }\mathbf{ }{\mathit{K}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{A}\mathbf{2}\mathbf{t}}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{X}}_{\mathbf{n}}\mathbf{=}\mathbf{ }\mathbf{ }{\mathit{K}}_{\mathbf{n}}{\mathit{e}}^{\mathbf{A}\mathbf{2}\mathbf{t}}\mathbf{,}$ is a fundamental set of solutions of (2) on the interval $(-\infty , \infty )$.

$\mathit{X}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{ }{\mathit{K}}_{\mathbf{1}}\mathbf{ }{\mathit{e}}^{\mathbf{\lambda }{}_{\mathbf{1}}\mathbf{t}}\mathbf{+}\mathbf{ }{\mathit{c}}_{\mathbf{2}}\mathbf{ }{\mathit{K}}_{\mathbf{2}}\mathbf{ }{\mathit{e}}^{\mathbf{\lambda }{}_{\mathbf{2}}\mathbf{t}}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{.}\mathbf{ }\mathbf{ }\mathbf{+}\mathbf{ }{\mathit{c}}_{\mathbf{3}}\mathbf{ }{\mathit{K}}_{\mathbf{3}}\mathbf{ }{\mathit{e}}^{\mathbf{\lambda }{}_{\mathbf{n}}\mathbf{t}}$

Solving the problem

We have,

$$\begin{gathered} X^{\text{'}}=\left(\frac{1}{2} 0 \\ 1 -\frac{1}{2}\right)X \end{gathered}$$

Where,

$$\begin{gathered} A=\left( \frac{1}{2} 0 \\ 1 -\frac{1}{2} \right) \end{gathered}$$

Now,

$$\begin{gathered} \det \left(A-\lambda I\right)=0 \to \left|\frac{1}{2}-\lambda 0 \\ 1 -\frac{1}{2} -\lambda \right|=0 \\ \left( \frac{1}{2}-\lambda \right)\left(- \frac{1}{2}-\lambda \right) =0 \end{gathered}$$

So, our eigenvalues are${\lambda }_1= \frac{1}{2}=0.5$ and${\lambda }_2=- \frac{1}{2}=-0.5$

Substitute for λ1=0.5

For${\lambda }_1=0.5$:

localid="1663963094284" $$\begin{gathered} \left(A-0.5I | 0\right)=\left( 0.5-0.5 \\ 1\right. \left.0 \\ -0.5-0.5\right| \left. 0 \\ 0\right) \\ =\left(0 \\ 1\right. \left. 0 \\ -1 \right|\left.0 \\ 0\right) \end{gathered}$$

$K_1 -K_2=\cup \to K_1 =K_2$

Letting $k_2=1$yields $k_1=1$. This gives an eigenvector and corresponding solution vector:

$$\begin{gathered} K_1=\left(1 \\ 1\right) , X_1=\left( 1 \\ 1\right) e^{\frac{1}{2}t} \end{gathered}$$

Substitute for λ2=-0.5

For${\lambda }_2=-0.5$:

$$\begin{gathered} \left(A + 0.5I | 0\right)= \left( 0.5-\left(-0.5\right) \\ 1\right. \left.0 \\ -0.5-\left(-0.5\right)\right| \left. 0 \\ 0\right) \\ =\left(1 \\ 1\right. \left. 0 \\ 0 \right|\left.0 \\ 0\right) \end{gathered}$$

we see that $k_1=0$, as for$k{ }_2$ we will choose an arbitrary value, say $k_2=1$.This gives an eigenvector and a corresponding solution vector

$$\begin{gathered} K_2=\left(0 \\ 1\right) , X_2=\left( 0 \\ 1\right)e^{-\frac{1}{2}t} \end{gathered}$$

Finally, the general solution is given by

$$\begin{gathered} X=c_1 X_1 + c_2 X_2 = c_1\left(1 \\ 1\right) e^{\frac{1}{2}t}+c_2\left(0 \\ 1\right)e^{-\frac{1}{2}t} \end{gathered}$$

Applying the initial value condition

$$\begin{gathered} X\left(0\right)=\left(3 \\ 5\right) \end{gathered}$$

$$\begin{gathered} \left(3 \\ 5\right)=c_1\left(1 \\ 1\right)+c_2\left( 0 \\ 1\right) \end{gathered}$$

Here we have,

$$\begin{gathered} c_1 =3 \\ {c}_1 +c_2 =5 \end{gathered}$$

Substituting(1) into (2) we can solve for,

$3+c_2=5 \to c_2=2$

Here the solution that satisfies the initial conditionis,

$$\begin{gathered} X\left(t\right) =3\left(1 \\ 1\right) e^{\frac{1}{2}t}+2\left( 0 \\ 1\right)e^{-\frac{1}{2}t} \end{gathered}$$