Question

Consider the linear system X' = AXof two differential equations, where Ais a real coefficient matrix. What is the general solution of the system if it is known that ${\mathit{\lambda }}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{2}\mathit{i}$is an eigenvalue and K=$\left(\begin{array}{c}1\\ i\end{array}\right)$is a corresponding eigenvector?

Short answer

The general solution of the system is $X=c_1\left(\begin{array}{c}\cos 2t\\ -\sin 2t\end{array}\right)e^t+c_2\left(\begin{array}{c}\sin 2t\\ \cos 2t\end{array}\right)e^t$.

Step-by-step solution

 Step 1: Define matrix exponential.

Consider a square matrix A of size n*n. This matrix can contain either complex numbers or real numbers. The matrix can be calculated as:

$A^0=I,A^1=A,A^2=A.A,A^3=A^2.A,.....,A^k=A.\underset{ktimes}{A}.....$

where I is the unit matrix of order n.

Therefore, the infinite matrix power series is $I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$.

Now, the matrix exponential is defined as the sum of the infinite matrix power series.

It is denoted by the expression e^At. It is given by the formula,$e^{tA}=\sum _{k=0}^{\infty }\frac{t^k}{k!}{A}^k$.

Find the general solution of the system.

It is given that, $K=\left(\begin{array}{c}1\\ i\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)+i\left(\begin{array}{c}0\\ 1\end{array}\right)$.

The column vectors are therefore,$B_1=\left(\begin{array}{c}1\\ 0\end{array}\right),B_2=\left(\begin{array}{c}0\\ 1\end{array}\right)$.

Also, $\lambda =\alpha +\beta i\to {\lambda }_2=1+2i$where $\alpha =1,\beta =2$.

Now can obtain the corresponding solution vectors,

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

And the other vector is obtained as:

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

Finally, it concludes the general solution of the system is;

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

Hence$c_1\left(\begin{array}{c}\cos 2t\\ -\sin 2t\end{array}\right)e^t+c_2\left(\begin{array}{c}\sin 2t\\ \cos 2t\end{array}\right)e^t$ is the general solution of the system.