Question

Find the general solution of the given system

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{4}\mathit{x}\mathbf{+}\mathbf{5}\mathit{y} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{6}\mathit{y} \end{gathered}$$

Short answer

The general solution of the given derivative system is

$X=c_1\left[\left(\begin{array}{c}\cos 3t+3\sin 3t\\ 2\cos 3t\end{array}\right)e^{5t}\right]+c_2\left[\left(\begin{array}{c}\sin 3t-3\cos 3t\\ 2\sin 3t\end{array}\right)e^{5t}\right]$

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 ${\mathit{\lambda }}_{\mathbf{1}}\mathbf{,}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{\lambda }}_{\mathbf{n}}$be distinct real eigenvalues of the coefficient matrix of the homogeneous system and let ${\mathit{K}}_{\mathbf{1}}\mathbf{,}{\mathit{K}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{K}}_{\mathbf{n}}$be the corresponding eigenvectors. Then the general solution of the homogeneous system on the interval $\left(\infty ,-\infty \right)$is given by,$\mathit{X}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{K}}_{\mathbf{1}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{1}}\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{K}}_{\mathbf{2}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{2}}\mathbf{t}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{n}}{\mathit{K}}_{\mathbf{n}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{n}}\mathbf{t}}$.

Rewrite the given system in matrix form

$X\text{'}=\left(\begin{array}{cc}4& 5\\ -2& 6\end{array}\right)\left(X\right)$

Where

$A=\left(\begin{array}{cc}4& 5\\ -2& 6\end{array}\right)$

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

Hence

$$\begin{gathered} det\left|A-\lambda |\right|=0\to \left|\begin{array}{cc}4-\lambda & 5\\ -2& 6-\lambda \end{array}\right| \\ \left(4-\lambda \right)\left(6-\lambda \right)-\left(-10\right)=0 \\ 24-4\lambda -6\lambda +{\lambda }^2+10=0 \\ {\lambda }^2-10\lambda +34=0 \end{gathered}$$

Where

$$\begin{gathered} \lambda =\frac{10\pm \sqrt{100-4\left(34\right)}}{2} \\ =\frac{10\pm \sqrt{-36}}{2} \\ =\frac{10\pm 6i}{2} \\ =5\pm 3 \end{gathered}$$

So our eigenvalues are ${\lambda }_1=5+3i,{\lambda }_2=\overline{{\lambda }_1}=5-3i$.

Finding of the corresponding eigenvectors and the general solution

For ${\lambda }_1=5+3i$, we have

$$\begin{gathered} (A+(5+3i\left)\mathrm{l}\right|0)=\left(\begin{array}{cc}4-(5+3i)& 5\\ -2& 6-(5+3i)\end{array}|\begin{array}{c}0\\ 0\end{array}\right) \\ =\left(\begin{array}{c}-1-3i\\ -2\end{array}\begin{array}{c}5\\ 1-3i\end{array}|\left|\begin{array}{c}0\\ 0\end{array}\right|\right) \end{gathered}$$

Apply row operation,$\frac{-1}{1+3i}{R}_1\to R_1$

$=\left(\begin{array}{cc}1& \frac{-5}{\left(1+3i\right)}\\ -2& 1-3i\end{array}|\begin{array}{c}0\\ 0\end{array}\right)$

Apply row operation,$R_2+2R_1\to R_2$

$\left(\begin{array}{cc}1& \frac{-5}{\left(1+3i\right)}\\ 0& 1-3i\end{array}|\begin{array}{c}0\\ 0\end{array}\right)$

Multiply$-\frac{5}{1+3i}$ by the conjugate, which will yield $-\frac{1-3i}{2}$,

$=\left(\begin{array}{cc}1& \raisebox{1ex}{$-1(1-3i)$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& 0\end{array}|\begin{array}{c}0\\ 0\end{array}\right)$

Apply row operation$-2R_1\to R_1$

$=\left(\begin{array}{cc}-2& 1-3i\\ 0& 0\end{array}|\frac{0}{0}\right)$

So now we have,

$-2k+\left(1-3\right)k_2=0\to k_1=\frac{1-3i}{2}{k}_2$

Choosing $k_2-2$yields $k1-1-3i$. This gives an eigenvector,

$K_1=\left(\begin{array}{c}1-3i\\ 2\end{array}\right)=\left(\begin{array}{c}1\\ 2\end{array}\right)+i\left(\begin{array}{c}-3\\ 0\end{array}\right)$

So,

$B_1=\left(\begin{array}{c}1\\ 2\end{array}\right),B_2=\left(\begin{array}{c}-3\\ 0\end{array}\right)$

Also,${\lambda }_1=\alpha +\beta i=5+3i$ .

Where$\alpha =5,\beta =3$

Therefore,

$$\begin{gathered} X_1=\left[B_1\cos \beta t-B_2\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}1\\ 2\end{array}\cos 3t-\begin{array}{c}-3\\ 0\end{array}\sin 3t\right)e^{5t}=\left(\begin{array}{c}\cos 3t-3\sin 3t\\ 2\cos 3t\end{array}\right)e^{5t} \end{gathered}$$

And

$$\begin{gathered} X_2=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}-3\\ 0\end{array}\cos 3t+\begin{array}{c}1\\ 2\end{array}\sin 3t\right)e^{5t}=\left(\begin{array}{c}\sin 3t-3\cos 3t\\ 2\sin 3t\end{array}\right)e^{5t} \end{gathered}$$

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

$$\begin{gathered} X=c_1{X}_1+c_2{X}_2 \\ =c_1\left(\begin{array}{c}\cos 3t+3\sin 3t\\ 2\cos 3t\end{array}\right)e^{5t}+c_2\left(\begin{array}{c}\sin 3t+3\cos 3t\\ 2\sin 3t\end{array}\right)e^{5t} \end{gathered}$$