Question

Question: In Problems 5-14solve the given linear system.

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

Short answer

The solution for the linear system $X\text{'}=\left(\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right)X+\left(\begin{array}{c}1\\ cott\end{array}\right)$is.

$X=c_1\left(\begin{array}{c}\cos t\\ \cos t-\sin t\end{array}\right)+c_2\left(\begin{array}{c}\sin t\\ \cos t+\sin t\end{array}\right)+\begin{array}{rcl}& & \left(\begin{array}{c}-1\\ -1\end{array}\right)\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \left(\begin{array}{c}-\mathrm{sint}\\ -\mathrm{cost}-\mathrm{sint}\end{array}\right)\end{array}\begin{array}{rcl}& & \ln \end{array}\begin{array}{rcl}& & \left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\end{array}$

Step-by-step solution

Define matrix exponential

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

$A^0=l,A^1=A,A^2=A·A,A^3=A^2·A,.....,A^k=A^{k-1}·A$

where is the unit matrix of order.

Therefore the infinite matrix power series is

$l+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\dots +\frac{t^k}{k!}{A}^k+..$

Now, the matrix exponential is defined as the sum of the infinite matrix power series. It is denoted by the expression. It is given by the formula,

$e^{tA}=\sum _{k\to 0}^A\frac{t^k}{k!}A$

Find the eigenvalue of the given linear system.

We are given that,

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

where

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

Now we find the characteristic equation of the coefficient matrix,

$$\begin{gathered} det(A-\lambda l)=0\to \left|\begin{array}{cc}-1-\lambda & 1\\ -2& 1-\lambda \end{array}\right| \\ \left(-1-\lambda \right)\left(1-\lambda \right)+2=0 \\ -1-\lambda +\lambda +{\lambda }^2+2=0 \\ {\lambda }^2+1=0 \\ {\lambda }^2=-1 \\ \lambda =\pm i \end{gathered}$$

So we have eigenvalues of ${\mathrm{\lambda }}_1=\mathrm{i},{\mathrm{\lambda }}_2=\overline{{\mathrm{\lambda }}_1}=-\mathrm{i}$.

Find the eigenvector and solution vector of the given linear system.

Solving,

$\begin{array}{rcl}(A-\left(i\right)\overline{)I}0)& =& \left(\begin{array}{ccc}-1-i& 1& 0\\ -2& 1-i& 0\end{array}\right)\end{array}$

Apply row operation,

$$\begin{gathered} \left(\frac{1}{2}+\frac{1}{2}i\right)R_2-R_1\to R_2 \\ =\left(\begin{array}{ccc}-1-\mathrm{i}& 1& 0\\ 0& 0& 0\end{array}\right) \end{gathered}$$

We have single equation as,

$-\left(1+\mathrm{i}\right){\mathrm{k}}_1+{\mathrm{k}}_2=0\to {\mathrm{k}}_1=\frac{1}{\left(1+\mathrm{i}\right)}{\mathrm{k}}_2$

Choosing $k_2=1+\mathrm{i}$we have$k_1=1$.

This gives an eigenvector $K=\left(\left.\begin{array}{c}1\\ 1+i\end{array}\right)=\left(\begin{array}{c}1\\ 1\end{array}\right)+i\left(\begin{array}{c}0\\ 1\end{array}\right.\right)$.

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

where,

$$\begin{gathered} \lambda =\alpha +\beta i \\ \lambda =0+i \end{gathered}$$

Hence the eigenvector is $K=\left(\left.\begin{array}{c}1\\ 1+i\end{array}\right)=\left(\begin{array}{c}1\\ 1\end{array}\right)+i\left(\begin{array}{c}0\\ 1\end{array}\right.\right)$.

Find the complementary function of the given linear system

Therefore,

$$\begin{gathered} \begin{array}{cc}{X}_1=\left[B_1,\cos \beta t-B_2\sin \beta t\right]e^{\alpha t}& \end{array} \\ \left(\begin{array}{c}1\\ 1\end{array}\right)\cos t-\left(\begin{array}{c}0\\ 1\end{array}\right)\sin t=\left(\begin{array}{c}\cos t\\ \cos t-\sin t\end{array}\right) \end{gathered}$$

and

$$\begin{gathered} \begin{array}{cc}{X}_1=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t}& \end{array} \\ \left(\begin{array}{c}0\\ 1\end{array}\right)\cos t+\left(\begin{array}{c}1\\ 1\end{array}\right)\sin t=\left(\begin{array}{c}sint\\ \cos t+\sin t\end{array}\right) \end{gathered}$$

Hence the complementary function is $X_c=c_1\left(\begin{array}{c}\cos t\\ \cos t-\sin t\end{array}\right)+c_2\left(\begin{array}{c}\sin t\\ \cos t+\sin t\end{array}\right)$.

Find the inverse matrix of ϕ(t)

Using the eigenvectors construct the matrix of $\mathrm{\varphi }\left(\mathrm{t}\right)$.

$\mathrm{\varphi }\left(\mathrm{t}\right)=\left(\begin{array}{cc}\mathrm{cost}& \mathrm{sint}\\ \mathrm{cost}-\mathrm{sint}& \mathrm{cost}+\mathrm{sint}\end{array}\right)$

We need to make sure that $\varphi \left(t\right)$is an invertible matrix by checking the determinant.

$\begin{array}{rcl}\left|\mathrm{\varphi }\left(\mathrm{t}\right)\right|& =& \left|\begin{array}{cc}\mathrm{cost}& \mathrm{sint}\\ \mathrm{cost}-\mathrm{sint}& \mathrm{cost}+\mathrm{sint}\end{array}\right|\\ & =& \left(\mathrm{cost}\right)\left(\mathrm{cost}+\mathrm{sint}\right)-\left(\mathrm{sint}\right)\left(\mathrm{cost}-\mathrm{sint}\right)\\ & =& {\cos }^2\mathrm{t}+\mathrm{costsint}-\mathrm{costsint}+{\sin }^2\mathrm{t}\\ & =& 1\ne 0\end{array}$

Since the value of the determinant is not equal to zero, the matrix is invertible.

So now,

role="math" localid="1668066627009" ${\mathrm{\varphi }}^{-1}\left(\mathrm{t}\right)=\left(\begin{array}{cc}\mathrm{cost}+\mathrm{sint}& -\mathrm{sint}\\ \mathrm{sint}-\mathrm{cost}& \mathrm{cost}\end{array}\right)$

Hence the inverse matrix of is role="math" localid="1668066648903" ${\mathrm{\varphi }}^{-1}\left(\mathrm{t}\right)=\left(\begin{array}{cc}\mathrm{cost}+\mathrm{sint}& -\mathrm{sint}\\ \mathrm{sint}-\mathrm{cost}& \mathrm{cost}\end{array}\right)$.

Find the particular solution of the given linear system by integration.

${\int }_{}^{}\cos \mathrm{t}\cot \mathrm{dt}={\int }_{}^{}\frac{{\cos }^2\mathrm{t}}{\sin \mathrm{t}}\mathrm{dt}$Obtaining the particular solution,

role="math" localid="1668066995253" $$\begin{gathered} X_p=\varphi \left(t\right){\int }_{}^{}{\varphi }^{-1}\left(t\right)F\left(t\right)dt \\ =\left(\begin{array}{cc}\mathrm{cost}& \mathrm{sint}\\ \mathrm{cost}-\mathrm{sint}& \mathrm{cost}+\mathrm{sint}\end{array}\right){\int }_{}^0\left(\begin{array}{cc}\mathrm{cost}+\mathrm{sint}& -\mathrm{sint}\\ \mathrm{sint}-\mathrm{cost}& \mathrm{cost}\end{array}\right)\left(\begin{array}{c}1\\ \cot \mathrm{t}\end{array}\right)\mathrm{dt} \\ =\left(\begin{array}{cc}\mathrm{cost}& \mathrm{sint}\\ \mathrm{cost}-\mathrm{sint}& \mathrm{cost}+\mathrm{sint}\end{array}\right){\int }_{}^0\left(\begin{array}{c}\mathrm{cost}+\mathrm{sint}-\mathrm{sintcot}\mathrm{t}\\ \mathrm{sint}-\mathrm{cost}+\cos \mathrm{t}\cot \mathrm{t}\end{array}\right)\mathrm{dt} \end{gathered}$$

Apply trigonometric identity$\cos \mathrm{t}=\mathrm{sint}\cot \mathrm{t}$

$$\begin{gathered} =\left(\begin{array}{cc}\mathrm{cost}& \mathrm{sint}\\ \mathrm{cost}-\mathrm{sint}& \mathrm{cost}+\mathrm{sint}\end{array}\right){\int }_{}^0\left(\begin{array}{c}\mathrm{cost}+\mathrm{sint}-\cos \mathrm{t}\\ \mathrm{sint}-\mathrm{cost}+\cos \mathrm{t}\cot \mathrm{t}\end{array}\right)\mathrm{dt} \\ =\left(\begin{array}{cc}\mathrm{cost}& \mathrm{sint}\\ \mathrm{cost}-\mathrm{sint}& \mathrm{cost}+\mathrm{sint}\end{array}\right){\int }_{}^0\left(\begin{array}{c}\mathrm{sint}\\ \mathrm{sint}-\mathrm{cost}+\cos \mathrm{t}\cot \mathrm{t}\end{array}\right)\mathrm{dt} \end{gathered}$$

Integrate,

${\int }_{}^{}\cos \mathrm{t}\cot \mathrm{t}\mathrm{dt}$

Simplify using trigonometric identity$cot\mathrm{t}=\frac{\cos \mathrm{t}}{\sin \mathrm{t}}$

${\int }_{}^{}\cos \mathrm{t}\cot \mathrm{t}\mathrm{dt}={\int }_{}^{}\frac{{\cos }^2\mathrm{t}}{\sin \mathrm{t}}\mathrm{dt}$

Use trigonometric identity ${\cos }^2\mathrm{t}=1-{\sin }^2\mathrm{t}$,

$$\begin{gathered} ={\int }_{}^{}\frac{1-si{n}^2\mathrm{t}}{\sin \mathrm{t}}\mathrm{dt} \\ ={\int }_{}^{}\frac{1}{\sin t}dt-{\int }_{}^{}\sin tdt \end{gathered}$$

Use trigonometric identity,$\mathrm{cosec}\mathrm{t}=\frac{1}{\sin \mathrm{t}}$

$$\begin{gathered} ={\int }_{}^{}\mathrm{cosect}\mathrm{dt}-{\int }_{}^{}\mathrm{sint}\mathrm{dt} \\ =\mathrm{cost}-\ln \left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right| \end{gathered}$$

Hence the value after integration is $\mathrm{cost}-\ln \left|\mathrm{cosect}+\mathrm{cott}\right|$.

Find the general solution of the given linear system by integration

To get the general solution, solve,

$\begin{array}{rcl}{\mathrm{X}}_{\mathrm{p}}& =& \left(\begin{array}{cc}\mathrm{cost}& \mathrm{sint}\\ \mathrm{cost}-\mathrm{sint}& \mathrm{cost}+\mathrm{sint}\end{array}\right){\int }_{}^0\left(\begin{array}{c}\mathrm{sint}\\ \mathrm{sint}-\mathrm{cost}+\cos \mathrm{t}\cot \mathrm{t}\end{array}\right)\mathrm{dt}\\ & =& \left(\begin{array}{cc}\mathrm{cost}& \mathrm{sint}\\ \mathrm{cost}-\mathrm{sint}& \mathrm{cost}+\mathrm{sint}\end{array}\right)\left(\begin{array}{c}-\mathrm{cost}\\ \mathrm{sint}-\ln \left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\end{array}\right)\\ & =& \left(\begin{array}{c}-{\cos }^2\mathrm{t}-{\sin }^2\mathrm{t}-\mathrm{sintln}\left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\\ -{\cos }^2\mathrm{t}+\mathrm{costsint}-\mathrm{costsint}-\mathrm{costln}\left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|++{\sin }^2\mathrm{t}-\mathrm{sintln}\left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\end{array}\right)\\ & =& \left(\begin{array}{c}-\left({\cos }^2\mathrm{t}+{\sin }^2\mathrm{t}\right)-\mathrm{sintln}\left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\\ -\left({\cos }^2\mathrm{t}+{\sin }^2\mathrm{t}\right)-\left(\mathrm{cost}+\mathrm{sint}\right)\ln \left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\end{array}\right)\\ & =& \left(\begin{array}{c}-1-\mathrm{sintln}\left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\\ -1-\left(\mathrm{cost}+\mathrm{sint}\right)\ln \left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\end{array}\right)\\ & =& \left(\begin{array}{c}-1\\ -1\end{array}\right)+\left(\begin{array}{c}-\mathrm{sint}\\ -\mathrm{cost}-\mathrm{sint}\end{array}\right)\ln \left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\end{array}$

Use the formula,

role="math" localid="1668068436075" $$\begin{gathered} X=X_c+X_p \\ X=c_1\left(\begin{array}{c}\cos t\\ \cos t-\sin t\end{array}\right)+c_2\left(\begin{array}{c}\sin t\\ \cos t+\sin t\end{array}\right)+\begin{array}{rcl}& & \left(\begin{array}{c}-1\\ -1\end{array}\right)\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \left(\begin{array}{c}-\mathrm{sint}\\ -\mathrm{cost}-\mathrm{sint}\end{array}\right)\end{array}\begin{array}{rcl}& & \ln \end{array}\begin{array}{rcl}& & \left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\end{array} \end{gathered}$$

Therefore the general solution of the system is$X=c_1\left(\begin{array}{c}\cos t\\ \cos t-\sin t\end{array}\right)+c_2\left(\begin{array}{c}\sin t\\ \cos t+\sin t\end{array}\right)+\begin{array}{rcl}& & \left(\begin{array}{c}-1\\ -1\end{array}\right)\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \left(\begin{array}{c}-\mathrm{sint}\\ -\mathrm{cost}-\mathrm{sint}\end{array}\right)\end{array}\begin{array}{rcl}& & \ln \end{array}\begin{array}{rcl}& & \left|\mathrm{cosec}\mathrm{t}+\cot \mathrm{t}\right|\end{array}$