Question

To find the general solution of the given system.

5.$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\mathit{X}$

Short answer

The general solution of given matrix is$X=C_1\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_2\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}$

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,

$X\text{'}=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)X$

By using

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

To find the general solution of the given system.

Now, let

localid="1664342224258" $A=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$

By using this equation,

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

Compute e^At. So,

A² = A.A

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right) \end{gathered}$$

A³ = A².A

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 8\end{array}\right) \end{gathered}$$

A⁴ = A² . A²

^{$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 8\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 8\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 16\end{array}\right) \end{gathered}$$}

$A^k=\left(\begin{array}{cc}1& 0\\ 0& 2^k\end{array}\right),fork=1,2,3,4...$

therefore, using (2), we have

$e^{At}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)t+\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)\frac{t^2}{2!}+\left(\begin{array}{cc}1& 0\\ 0& 8\end{array}\right)\frac{t^3}{3!}$

$=\left(\begin{array}{cc}1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}& 0\\ 0& 1+2t+\frac{2^2{t}^2}{2!}+\frac{2^3{t}^3}{3!}+\frac{2^4{t}^4}{4!}\end{array}\right)$

The power series in the first and second rows of the matrix represent, respectively, e^t and e^2t, therefore,

$=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)$

Find the General solution of given matrix.

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

$$\begin{gathered} X=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right) \\ =\left(\begin{array}{c}{C}_1{e}^t\\ C_2{e}^{2t}\end{array}\right) \\ =C_1\left(\begin{array}{c}{e}^t\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ e^{2t}\end{array}\right) \\ =C_1\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_2\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t} \end{gathered}$$

Therefore, the general solution of given matrix is$X=C_1\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_2\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}$