Question

To find the general solution of the given system.

6.${\mathit{X}}^{\mathbf{\text{'}}}\mathbf{=}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\mathit{X}$

Short answer

The general solution of the given matrix is$X=C_1\left(\begin{array}{c}\cos ht\\ \sin ht\end{array}\right)+C_2\left(\begin{array}{c}\sin ht\\ \cos ht\end{array}\right)$

Step-by-step solution

Definition of Matrix Exponential.

• Exponential matrix is a square matrix function similar to a regular exponential function. Used to solve a system of linear differential equations.
• In Lie group theory, the matrix exponential function provides the connection between the matrix Lie algebra and the corresponding Lie group.
• For any $n\times n$matrix A,
• $X=e^{At}C$
  

Calculate the Matrix exponential eAt.

Given that,

$X^{\text{'}}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)X$

By using

$X=e^{At}C$ (1)

To find the general solution of the given system. Now, let

role="math" localid="1664270101523" $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$

By using this equation

$e^{At}=I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots \left(2\right)$

Compute e^At . So,

A² = A.A

$$\begin{gathered} =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =I \end{gathered}$$

A³=A².A

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =A \end{gathered}$$

A⁴ =A². A²

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =I \end{gathered}$$

A⁵ = A⁴.A

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =A \end{gathered}$$

Thus,

$A^k=\left\{\begin{array}{l}A,fork=1,3,5,....\\ I,fork=2,4,6....\end{array}\right.$

Therefore, using (2), we have

$e^{At}=I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$

$=I\left(\begin{array}{c}1+\frac{t^2}{2!}+\frac{t^3}{3!}+......\end{array}\right)+A\left(t+\frac{t^3}{3!}+\frac{t^5}{5!}+.....\right)$

Notice that these are Taylor series expansions of the hyperbolic functions cosht and sinht,s0,

$$\begin{gathered} =\mathbf{I}\left(\cosh t\right)+\mathbf{A}\left(\sinh t\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\cos ht+\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\sin ht \\ =\left(\begin{array}{cc}\cos ht& 0\\ 0& \cos ht\end{array}\right)\left(\begin{array}{cc}0& \sin ht\\ \sin ht& 0\end{array}\right) \\ =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right) \end{gathered}$$

Find the General solution of given matrix.

Finally, using (1) to find the general solution of the system,

$$\begin{gathered} =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right) \\ =\left(\begin{array}{cc}{C}_1\cos ht& C_2\sin ht\\ C_1\sin ht& C_2\cos ht\end{array}\right) \\ =C_1\left(\begin{array}{c}\cos ht\\ \sin ht\end{array}\right)+C_2\left(\begin{array}{c}\sin ht\\ \cos ht\end{array}\right) \end{gathered}$$

Therefore, the general solution of the given matrix is,

$X=C_1\left(\begin{array}{c}\cos ht\\ \sin ht\end{array}\right)+C_2\left(\begin{array}{c}\sin ht\\ \cos ht\end{array}\right)$