Question

In Problems33and 34 use (14) to solve the given initial-value problem.

34.$\mathbf{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)\mathbf{X}\mathbf{+}\left(\begin{array}{c}1/t\\ 1/t\end{array}\right)\mathbf{,}\mathbf{X}\left(1\right)\mathbf{=}\left(\begin{array}{c}2\\ -1\end{array}\right)$

Short answer

The general solution of the system for$\mathrm{X}\text{'}=\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)\mathrm{X}+\left(\begin{array}{c}1/\mathrm{t}\\ 1/\mathrm{t}\end{array}\right)$is$\mathrm{X}=\left(\begin{array}{c}3\\ 3\end{array}\right)\mathrm{t}-\left(\begin{array}{c}1\\ 4\end{array}\right)+\left(\begin{array}{c}1\\ 1\end{array}\right)\mathrm{in}\left(\mathrm{t}\right)$

Step-by-step solution

Determine the complex Eigen values

The polynomial of degree $\mathrm{n}$of the variable $\mathrm{\lambda }$ is the characteristic polynomial of the

$\mathrm{n}\times \mathrm{n}$matrix A .

$\mathrm{p}\left(\mathrm{\lambda }\right)=\det (\mathrm{\lambda I}-\mathrm{A})$

Its root is the eigenvalue of A.

Determine the Eigen values

Given

$\mathrm{X}\text{'}=\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)\mathrm{X}+\left(\begin{array}{c}\frac{1}{\mathrm{t}}\\ \frac{1}{\mathrm{t}}\end{array}\right)$

Where

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

$\mathrm{F}\left(\mathrm{t}\right)=\left(\begin{array}{c}\frac{1}{\mathrm{t}}\\ \frac{1}{\mathrm{t}}\end{array}\right)$

Obtaining the eigenvalues of the coefficient matrix,

$\det (\mathrm{A}-\mathrm{\lambda I})=0\to \left|\begin{array}{cc}1-\mathrm{\lambda }& -1\\ 1& -1-\mathrm{\lambda }\end{array}\right|=0$

$(1-\mathrm{\lambda })(-1-\mathrm{\lambda })+1=0$

$-1-\mathrm{\lambda }+\mathrm{\lambda }+{\mathrm{\lambda }}^2+1=0$

${\mathrm{\lambda }}^2=0$

We have $\mathrm{\lambda }=0$, which is an eigenvalue of multiplicity two.

Determine the eigenvector and a corresponding solution vector

Solving,

$(\mathrm{A}-\mathrm{\lambda I})\mathrm{K}=0\to \left(\begin{array}{cc}1-0& -1\\ 1& -1-0\end{array}\right)\left(\begin{array}{c}{\mathrm{k}}_1\\ {\mathrm{k}}_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

$\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)\left(\begin{array}{c}{\mathrm{k}}_1\\ {\mathrm{k}}_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Apply row operation ${\mathrm{R}}_2-{\mathrm{R}}_1\to {\mathrm{R}}_2$:

$\left(\begin{array}{cc}1& -1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{\mathrm{k}}_1\\ {\mathrm{k}}_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Here we have

role="math" localid="1668091804247" ${\mathrm{k}}_1-{\mathrm{k}}_2=0\to {\mathrm{k}}_1={\mathrm{k}}_2$

Choosing v yields ${\mathrm{k}}_1=1$.

This gives an eigenvector and a corresponding solution vector

$\mathrm{K}=\left(\begin{array}{c}1\\ 1\end{array}\right),{\mathrm{X}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$

$(\mathrm{A}-\mathrm{\lambda I})\mathrm{P}=\mathrm{K}\to \left(\begin{array}{cc}1-0& -1\\ 1& -1-0\end{array}\right)\left(\begin{array}{c}{\mathrm{p}}_1\\ {\mathrm{p}}_2\end{array}\right)=\left(\begin{array}{c}1\\ 1\end{array}\right)$

$\left(\begin{array}{cc}1& -1\\ 0& -1\end{array}\right)\left(\begin{array}{c}{\mathrm{p}}_1\\ {\mathrm{p}}_2\end{array}\right)=\left(\begin{array}{c}1\\ 1\end{array}\right)$

Apply row operation ${\mathrm{R}}_2-{\mathrm{R}}_1\to {\mathrm{R}}_2$:

$\left(\begin{array}{cc}1& -1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{\mathrm{p}}_1\\ {\mathrm{p}}_2\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)$

Here we have

${\mathrm{k}}_1-{\mathrm{k}}_2=1$

Choosing ${\mathrm{k}}_2=0$ yields ${\mathrm{k}}_1=1$.

This gives an eigenvector and a corresponding solution vector

$\mathrm{P}=\left(\begin{array}{c}1\\ 0\end{array}\right),{\mathrm{X}}_2=\left(\begin{array}{c}1\\ 1\end{array}\right)\mathrm{t}+\left(\begin{array}{c}1\\ 0\end{array}\right)$

Determine the general solution of the system

The entries in ${\mathrm{X}}_1$form the first column of $\mathrm{\varphi }\left(\mathrm{t}\right)$, and the entries in ${\mathrm{X}}_2$form the second

column of $\mathrm{\varphi }\left(\mathrm{t}\right)$.

Therefore,

$\mathrm{\varphi }\left(\mathrm{t}\right)=\left(\begin{array}{cc}1& 1+\mathrm{t}\\ 1& \mathrm{t}\end{array}\right)$

We want to make sure that $\mathrm{\varphi }\left(\mathrm{t}\right)$is an invertible matrix by checking the determinant,

where

$\left|\begin{array}{c}\mathrm{\varphi }\left(\mathrm{t}\right)\end{array}\right|=\left|\begin{array}{cc}1& 1+t\\ 1& t\end{array}\right|$

role="math" localid="1668092965069" $=t-(1+t)$

$=-1\ne 0$

Since the determinant does not equal zero, the matrix is in fact, an invertible matrix.

So now,

$\begin{array}{c}{\mathrm{\varphi }}^{-1}\left(\mathrm{t}\right)\end{array}=-\left(\begin{array}{cc}\mathrm{t}& -1-\mathrm{t}\\ -1& 1\end{array}\right)$

$=\left(\begin{array}{cc}-\mathrm{t}& 1+\mathrm{t}\\ 1& -1\end{array}\right)$

Using

$\mathrm{X}=\mathrm{\varphi }\left(\mathrm{t}\right){\mathrm{\varphi }}^{-1}\left(1\right)\times \left(1\right)+\mathrm{\varphi }\left(\mathrm{t}\right){\int }_0^{\mathrm{t}}{\mathrm{\varphi }}^{-1}\left(\mathrm{s}\right)\mathrm{F}\left(\mathrm{s}\right)\mathrm{ds}$

Where

$\mathrm{X}\left(1\right)=\left(\begin{array}{c}2\\ -1\end{array}\right)$

First, we have

$\mathrm{\varphi }\left(\mathrm{t}\right){\mathrm{\varphi }}^{-1}\left(1\right)\times \left(1\right)=\left(\begin{array}{cc}1& 1+\mathrm{t}\\ 1& \mathrm{t}\end{array}\right)\left(\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right)\left(\begin{array}{c}2\\ -1\end{array}\right)$

$=\left(\begin{array}{cc}1& 1+\mathrm{t}\\ 1& \mathrm{t}\end{array}\right)\left(\begin{array}{c}-4\\ 3\end{array}\right)$

$=\left(\begin{array}{cc}-1& +3\mathrm{t}\\ -4& +3\mathrm{t}\end{array}\right)$

Substituting into (1),

$\mathrm{X}=\left(\begin{array}{cc}-1& +3\mathrm{t}\\ -4& +3\mathrm{t}\end{array}\right)+\left(\begin{array}{cc}1& 1+\mathrm{t}\\ 1& \mathrm{t}\end{array}\right){\int }_0^{\mathrm{t}}\left(\begin{array}{cc}-\mathrm{s}& 1-\mathrm{s}\\ 1& -1\end{array}\right)\left(\begin{array}{c}\frac{1}{\mathrm{s}}\\ \frac{1}{\mathrm{s}}\end{array}\right)\mathrm{ds}$

$=\left(\begin{array}{cc}-1& +3\mathrm{t}\\ -4& +3\mathrm{t}\end{array}\right)+\left(\begin{array}{cc}1& 1+\mathrm{t}\\ 1& \mathrm{t}\end{array}\right){\int }_0^{\mathrm{t}}\left(\begin{array}{c}\frac{1}{\mathrm{s}}\\ 0\end{array}\right)\mathrm{ds}$

$=\left(\begin{array}{cc}-1& +3\mathrm{t}\\ -4& +3\mathrm{t}\end{array}\right)+\left(\begin{array}{cc}1& 1+\mathrm{t}\\ 1& \mathrm{t}\end{array}\right)\left(\begin{array}{c}\mathrm{In}\left(\mathrm{s}\right)\\ 0\end{array}\right){|}_0^{\mathrm{t}}$

role="math" localid="1668094727447" $=\left(\begin{array}{cc}-1& +3\mathrm{t}\\ -4& +3\mathrm{t}\end{array}\right)+\left(\begin{array}{cc}1& 1+\mathrm{t}\\ 1& \mathrm{t}\end{array}\right)\left(\begin{array}{c}\mathrm{In}\left(t\right)\\ 0\end{array}\right)$

$=\left(\begin{array}{cc}-1& +3\mathrm{t}\\ -4& +3\mathrm{t}\end{array}\right)+\left(\begin{array}{c}\mathrm{In}\left(\mathrm{t}\right)\\ \mathrm{In}\left(\mathrm{t}\right)\end{array}\right)$

$=\left(\begin{array}{c}3\\ 3\end{array}\right)\mathrm{t}-\left(\begin{array}{c}1\\ 4\end{array}\right)+\left(\begin{array}{c}1\\ 1\end{array}\right)\mathrm{In}\left(\mathrm{t}\right)$

Therefore, the general solution of the system for $\mathrm{X}\text{'}=\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)\mathrm{X}+\left(\begin{array}{c}1/\mathrm{t}\\ 1/\mathrm{t}\end{array}\right)\mathrm{is}$

$\mathrm{X}=\left(\begin{array}{c}3\\ 3\end{array}\right)\mathrm{t}-\left(\begin{array}{c}1\\ 4\end{array}\right)+\left(\begin{array}{c}1\\ 1\end{array}\right)\mathrm{In}\left(\mathrm{t}\right)$