Question

Find the general solution of the given system

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

Short answer

The general solution of the given derivative system is

$X=c_1\left(\begin{array}{c}-\cos t+\sin t\\ 2\cos t\end{array}\right)+c_2\left(\begin{array}{c}-\cos t-\sin t\\ 2\sin t\end{array}\right)$

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${\mathit{\lambda }}_{\mathbf{1}}\mathbf{,}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{\lambda }}_{\mathbf{n}}$be n distinct real eigenvalues of the coefficient matrix A 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$(\infty ,-\infty )$is given by,

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

Rewrite the given system in matrix form

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

Where

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

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

Hence

$$\begin{gathered} \det \left|A-\lambda I\right|=0 \\ \left|\begin{array}{cc}1-\lambda & 1\\ -2& -1-\lambda \end{array}\right|=0 \\ \left(1-\lambda \right)\left(-1-\lambda \right)-\left(-2\right)=0 \\ -1-\lambda +\lambda +{\lambda }^2+2=0 \\ {\lambda }^2+1=0 \\ \left(\lambda +i\right)\left(\lambda -i\right)=0 \\ {\lambda }_1=i,\lambda {}_2=\overline{{\lambda }_1}=-i \end{gathered}$$

Finding of the corresponding eigenvectors and the general solution

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

$\left(A-iI|0\right)=\left(\begin{array}{cc}1-i& 1\\ -2& -1-i\end{array}\overline{)\begin{array}{c}\\ \end{array}}\right)$

Apply row operation$\left(\frac{1}{1-i}\right)R_1\to R_1$

localid="1664299414962" $$\begin{gathered} =\left(\begin{array}{cc}1& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \left(1-i\right)}$}\right.\\ -2& -1-i\end{array}\overline{)\begin{array}{c}0\\ 0\end{array}}\right) \\ =\left(\begin{array}{cc}1& \raisebox{1ex}{${\displaystyle \left(1+i\right)}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 2}$}\right.\\ -2& -\left(1+i\right)\end{array}\overline{)\begin{array}{c}0\\ 0\end{array}}\right) \end{gathered}$$

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

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

So here we have $k_1+\left(\frac{1+i}{2}\right)k_2=0\to k_1=-\left(\frac{1+i}{2}\right)k_2$

Choosing $k_2=2$yields $k_1=-\left(1+i\right)$. This gives an eigenvector,

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

As for ${\lambda }_2=-i$, notice that this eigenvalue is the conjugate of ${\lambda }_1$, which means that our eigenvector$K_2$will be the conjugate of our first eigenvector $K_1$. Now, here we have

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

Also, localid="1664300386168" $\beta =1,\alpha =0$. Therefore,

localid="1664300081291" $$\begin{gathered} X_1=\left[B_1\cos \beta t-B_2\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}-1\\ 0\end{array}\right)\cos t+\left(\begin{array}{c}-1\\ 2\end{array}\right)\sin t \\ =\left(\begin{array}{c}-\cos t-\sin t\\ 2\sin t\end{array}\right) \end{gathered}$$

Thus the general solution is,

localid="1664300147692" $$\begin{gathered} X=c_1{X}_1+c_2{X}_2 \\ =c_1\left(\begin{array}{c}-\cos t+\sin t\\ 2\cos t\end{array}\right)+c_2\left(\begin{array}{c}-\cos t-\sin t\\ 2\sin t\end{array}\right) \end{gathered}$$