Question

To find the general solution of the given system.

9.$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\mathit{X}\mathbf{+}\left(\begin{array}{c}3\\ -1\end{array}\right)$

Short answer

The general solution for the given matrix is $X=C_3\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_4\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+\left(\begin{array}{c}-3\\ \frac{1}{2}\end{array}\right)$

Step-by-step solution

Definition for non-homogeneous system.

• For a nonhomogeneous system of linear first-order differential equations it can be shown that the general solution of X'=AX + F(t), where A is an n*n matrix of constants, is
• $\mathbf{X}=e^{\mathbf{A}t}\mathbf{C}+e^{\mathbf{A}t}{\int }_{t_0}^t{e}^{-\mathbf{A}s}\mathbf{F}\left(s\right)ds.$
  
• Since the matrix exponential e^At is a fundamental matrix, it is always nonsingular and e^-As = (e^As)^-1.

Calculate the Matrix exponential eAt.

Given that,

$X\text{'}=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)X+\left(\begin{array}{c}3\\ -1\end{array}\right)$

By using

$\mathbf{X}=e^{\mathbf{A}t}\mathbf{C}+e^{\mathbf{A}t}{\int }_{t_0}^t{e}^{-\mathbf{A}s}\mathbf{F}\left(s\right)ds$

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

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

By using

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

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

^{$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 16\end{array}\right) \end{gathered}$$}

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

Therefore, using (2), we have

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

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

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.

Now, using (1), we have

$$\begin{gathered} X=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right){\int }_{t_0}^t\left(\begin{array}{cc}{e}^{-s}& 0\\ 0& e^{-2s}\end{array}\right)\left(\begin{array}{c}3\\ -1\end{array}\right)ds \\ =C_1\left(\begin{array}{c}{e}^t\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ e^{2t}\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right){\int }_0^t\left(\begin{array}{c}3e^{-s}\\ -e^{-2s}\end{array}\right)ds \\ =C_1\left(\begin{array}{c}{e}^t\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ e^{2t}\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right){{\left(\begin{array}{c}3e^{-s}\\ \frac{1}{2}{e}^{-2s}\end{array}\right)}^t}_0 \\ =C_1\left(\begin{array}{c}{e}^t\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ e^{2t}\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{c}3e^{-s}+3\\ \frac{1}{2}{e}^{-2s}-\frac{1}{2}\end{array}\right) \end{gathered}$$$$\begin{gathered} =C_1\left(\begin{array}{c}{e}^t\\ 0\end{array}\right)+C_2\left(\begin{array}{c}0\\ e^{2t}\end{array}\right)+\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{c}-3+3e^t\\ \frac{1}{2}-\frac{1}{2}{e}^{2t}\end{array}\right) \\ =C_1\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_2\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+3\left(\begin{array}{c}1\\ 0\end{array}\right)e^t-\frac{1}{2}\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+\left(\begin{array}{c}-3\\ \frac{1}{2}\end{array}\right) \\ =\left(\begin{array}{c}1\\ 0\end{array}\right)e^t[C_1+3]+\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}[C_2-\frac{1}{2}]+\left(\begin{array}{c}-3\\ \frac{1}{2}\end{array}\right) \end{gathered}$$

Let C1 +3 =C3 and C4=C2-1/2,

Therefore, the general solution for the given matrix is

$X=C_3\left(\begin{array}{c}1\\ 0\end{array}\right)e^t+C_4\left(\begin{array}{c}0\\ 1\end{array}\right)e^{2t}+\left(\begin{array}{c}-3\\ \frac{1}{2}\end{array}\right)$