Question

Find the general solution of the given system

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}4& -5\\ 5& -4\end{array}\right)\mathit{X}$

Short answer

The general solution of the given derivative system is

$X=c_1\left(\begin{array}{c}5\cos 3t\\ 4\cos 3t+3\sin 3t\end{array}\right)+c_2\left(\begin{array}{c}5\sin 3t\\ 4\sin 3t-3\cos 3t\end{array}\right)$

Step-by-step solution

Finding of the eigenvalues of matrix of coefficient from the characteristics equation

The theorem of General Solution-Homogeneous System states that

Let ${\lambda }_1,{\lambda }_2,....,{\lambda }_n$be n distinct real eigenvalues of the coefficient matrix of A the homogeneous system and let $K_1,K_2,...,K_n$be the corresponding eigenvectors. Then the general solution of the homogeneous system on the interval $\left(\infty ,-\infty \right)$is given by,

role="math" localid="1667976643450" $X=c_1{K}_1{e}^{{\lambda }_{1}t}+c_2{K}_2{e}^{{\lambda }_{2}t}+...+c_n{K}_n{e}^{{\lambda }_{n}t}$

The given matrix form is,

$X\text{'}=\left(\begin{array}{cc}4& -5\\ 5& -4\end{array}\right)X$

Where

$A=\left(\begin{array}{cc}4& -5\\ 5& -4\end{array}\right)$

We need to find the eigenvalues and eigenvectors of the matrix of the coefficient.

Hence,$det\left|\begin{array}{c}A-\lambda I\end{array}\right|=0\to \left|\begin{array}{cc}4-\lambda & 5\\ -5& -4-\lambda \end{array}\right|=0$

$$\begin{gathered} \left(4-\lambda \right)\left(-4-\lambda \right)-\left(-25\right)=0 \\ -16-4\lambda +4\lambda +{\lambda }^2+25=0 \\ {\lambda }^2+9=0 \\ \left(\lambda +3i\right)\left(\lambda -3i\right)=0 \end{gathered}$$

So our eigenvalues are ${\lambda }_1=3i,{\lambda }_2=\overline{{\lambda }_1}=-3i$.

Finding of the corresponding eigenvectors and the general solution

For ${\lambda }_1=3i$, we have

$$\begin{gathered} (A-3I|0)=\left(\begin{array}{ccc}4-3i& 5& 0\\ -5& -4-3i& 0\end{array}\right) \\ =\left(\begin{array}{ccc}4-3i& -5& 0\\ -2& 1-3i& 0\end{array}\right) \end{gathered}$$

Apply row operation,$\left(4-3i\right)R_2-5R_1\to R_2$$\left(4-3i\right)R_2-5R_1\to R_2$

$\left(\begin{array}{ccc}4-3i& -5& 0\\ 0& 0& 0\end{array}\right)$

So now we have,

$\left(4-3i\right)k_1-5k_2=0\to k_1=\frac{5}{4-3i}{k}_2$

Choosing $k_2=4-3i$ yields k₁ =5. This gives an eigenvector,

$K_1=\left(\begin{array}{c}5\\ 4-3i\end{array}\right)=\left(\begin{array}{c}5\\ 4\end{array}\right)+i\left(\begin{array}{c}0\\ -3\end{array}\right)$

So,

$B_1=\left(\begin{array}{c}5\\ 4\end{array}\right),B_2=\left(\begin{array}{c}0\\ -3\end{array}\right)$

Also, ${\lambda }_1=\alpha +\beta i=0+3i$.

Where$\alpha =0,\beta =3$

Therefore,

$$\begin{gathered} X_1=\left[B_1\cos \beta t-B_2\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}5\cos 3t\\ 4\cos 3t+3\sin 3t\end{array}\right) \end{gathered}$$

And

$$\begin{gathered} X_2=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}5\sin 3t\\ 4\sin 3t-3c0os3t\end{array}\right) \end{gathered}$$

Thus the general solution that, we could finally conclude is,

$$\begin{gathered} X=c_1{X}_1+c_2{X}_2 \\ =c_1\left(\begin{array}{c}5\cos 3t\\ 4\cos 3t+3\sin 3t\end{array}\right)+c_2\left(\begin{array}{c}5\sin 3t\\ 4\sin 3t-3\cos 3t\end{array}\right) \end{gathered}$$