Question

To find the general solution of the given system.

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

Short answer

The general solution for the given matrix is$X=C_3\left(\begin{array}{c}\cos ht\\ \sin ht\end{array}\right)+C_4\left(\begin{array}{c}\sin ht\\ \cos ht\end{array}\right)-\left(\begin{array}{c}1\\ 1\end{array}\right)$

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,

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

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

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

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

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

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

^{localid="1664351930975" $$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =I \end{gathered}$$}

A⁵ = A⁴.A

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

Therefore, using (2), we have

$$\begin{gathered} e^{At}=I+At+I\frac{t^2}{2!}+A\frac{t^3}{3!}+I\frac{t^4}{4!}+..... \\ =I\left(1+\frac{t^2}{2!}+\frac{t^4}{4!}.....\right)+A\left(t+\frac{t^3}{3!}+\frac{t^5}{5!}....\right) \end{gathered}$$

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

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

Find the General solution of given matrix.

Finally, using (1), we have

$$\begin{gathered} =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right){\int }_{t_0}^t\left(\begin{array}{cc}\cos hs& -\sin hs\\ -\sin hs& \cos hs\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)ds \\ =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right){\int }_0^t\left(\begin{array}{cc}\cos hs& -\sin hs\\ \cos hs& -\sin hs\end{array}\right)ds \\ =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right){\left(\begin{array}{cc}\sin hs& -\cos hs\\ sinhs& -\cos hs\end{array}\right)}_0^t \\ =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right){\left(\begin{array}{cc}\sin ht& -\cos ht+1\\ sinht& -\cos ht+1\end{array}\right)}_0^t \end{gathered}$$

$$\begin{gathered} =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{cc}-\cos h^{2}t+\cos ht+& \sin h^{2}t+\sin ht\\ \sin h^{2}t+\sin ht-& \cos h^{2}t+\cos ht\end{array}\right) \\ =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{cc}-(\cos h^{2}t-\sin h^{2}t)+& \cos ht+\sin ht\\ -(\cos h^{2}t-\sin h^{2}t)+& \sin ht+\cos ht\end{array}\right) \end{gathered}$$

Apply hyperbolic identity ,${\cosh }^{2}t-{\sinh }^{2}t=1$

$$\begin{gathered} =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)+\left(\begin{array}{cc}-1+& \cos ht+\sin ht\\ -1+& \sin ht+\cos ht\end{array}\right) \\ =C_1\left(\begin{array}{c}\cos ht\\ \sin ht\end{array}\right)+C_2\left(\begin{array}{c}\sin ht\\ \cos ht\end{array}\right)-\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}\cos ht\\ \sin ht\end{array}\right)+\left(\begin{array}{c}\sin ht\\ \cos ht\end{array}\right) \\ =\left(\begin{array}{c}\cos ht\\ \sin ht\end{array}\right)(C_1+1)+\left(\begin{array}{c}\sin ht\\ \cos ht\end{array}\right)[C_2+1]-\left(\begin{array}{c}1\\ 1\end{array}\right) \end{gathered}$$

C₃=C₁+1 and C₄=C₂=1

$=C_3\left(\begin{array}{c}\cos ht\\ \sin ht\end{array}\right)+C_4\left(\begin{array}{c}\sin ht\\ \cos ht\end{array}\right)-\left(\begin{array}{c}1\\ 1\end{array}\right)$

Therefore, the general solution for the given matrix,

$X=C_3\left(\begin{array}{c}\cos ht\\ \sin ht\end{array}\right)+C_4\left(\begin{array}{c}\sin ht\\ \cos ht\end{array}\right)-\left(\begin{array}{c}1\\ 1\end{array}\right)$