Question

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

$\mathbf{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}3& 1\\ -1& 1\end{array}\right)\mathbf{X}\mathbf{+}\left(\begin{array}{c}-2\\ 1\end{array}\right){\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

Short answer

The solution for the linear system $X\text{'}=\left(\begin{array}{cc}3& 1\\ -1& 1\end{array}\right)X+\left(\begin{array}{c}-2\\ 1\end{array}\right)e^{2t}$is

$\begin{array}{rcl}X& =& c_1\left(\begin{array}{c}1\\ -1\end{array}\right)e^{2t}+c_2\left[\left(\begin{array}{c}1\\ -1\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}1\\ 0\end{array}\right)e^{2t}\right]+\left(\begin{array}{c}-\frac{1}{2}{t}^2{e}^{2t}-2t{e}^{2t}\\ \frac{1}{2}{t}^2{e}^{2t}+t{e}^{2t}\end{array}\right)\end{array}$

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,

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

Find the eigenvalue of the given linear system.

We are given that,

$X\text{'}=\left(\begin{array}{cc}3& 1\\ -1& 1\end{array}\right)X+\left(\begin{array}{c}-2\\ 1\end{array}\right)e^{2t}$

where,

$A=\left(\begin{array}{cc}3& 1\\ -1& 1\end{array}\right)$

Now we find the characteristic equation of the coefficient matrix,

$\begin{array}{rcl}det(A-\lambda l)& =& 0\to \left|\begin{array}{cc}3-\lambda & 1\\ -1& 1-\lambda \end{array}\right|\\ \left(3-\lambda \right)\left(1-\lambda \right)+1& =& 0\\ 3-3\lambda -\lambda +{\lambda }^2+1& =& 0\\ 4-2\lambda +{\lambda }^2& =& 0\\ {\left(\lambda -2\right)}^2& =& 0\\ \lambda & =& 2\end{array}$

So we have eigenvalues of $\lambda =2$with algebraic multiplicity 2.

Find the complementary function of the given linear system.

Solving,

$\begin{array}{rcl}(A-2I)K& =& 0\to \left(\begin{array}{cc}3-2& 1\\ -1& 1-2\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$

Apply row operation,

Choosing $k_2=-1$we have$k_1=1$.

$$\begin{gathered} K=\left(\begin{array}{c}1\\ -1\end{array}\right),X_1=\left(\begin{array}{c}1\\ -1\end{array}\right)e^{2t} \\ \left(A-2I\right)P=K\to \left(\begin{array}{cc}3-2& 1\\ -1& 1-2\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}1\\ -1\end{array}\right) \\ \left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}1\\ -1\end{array}\right) \end{gathered}$$

Apply row operation,

$$\begin{gathered} R_2+R_1\to R_1 \\ \left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right) \\ {p}_1+p_2=1 \end{gathered}$$

Choosing $p_2=0$we have $p_1=1$.

This gives an eigenvector $P=\left(\begin{array}{c}1\\ 0\end{array}\right),X_2=\left(\begin{array}{c}1\\ -1\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}1\\ 0\end{array}\right)e^{2t}$.

Hence the complementary function is $X_c=c_1\left(\begin{array}{c}1\\ -1\end{array}\right)e^{2t}+c_2\left[\left(\begin{array}{c}1\\ -1\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}1\\ 0\end{array}\right)e^{2t}\right]$.

Find the inverse matrix of ϕ(t).

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

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

We need to make sure that is an invertible matrix by checking the determinant.

$\begin{array}{rcl}\left|\varphi \left(t\right)\right|& =& \left|\begin{array}{cc}{e}^{2t}& t{e}^{2t}+e^{2t}\\ -e^{2t}& -t{e}^{2t}\end{array}\right|\\ & =& \left(e^{2t}\right)\left(-t{e}^{2t}\right)+\left(t{e}^{2t}+e^{2t}\right)\left(e^{2t}\right)\\ & =& t{e}^{4t}-t{e}^{4t}+e^{4t}\\ & =& e^{4t}\ne 0\end{array}$

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

So now,

${\varphi }^{-1}\left(t\right)=\frac{1}{e^{4t}}\left(\begin{array}{cc}-t{e}^{2t}& -t{e}^{2t}-e^{2t}\\ e^{2t}& e^{2t}\end{array}\right)=\left(\begin{array}{cc}-t{e}^{-2t}& -t{e}^{-2t}-e^{-2t}\\ e^{-2t}& e^{-2t}\end{array}\right)$

Hence the inverse matrix of is ${\varphi }^{-1}\left(t\right)=\left(\begin{array}{cc}-t{e}^{-2t}& -t{e}^{-2t}-e^{-2t}\\ e^{-2t}& e^{-2t}\end{array}\right)$.

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

To get the general solution, solve,

.$\begin{array}{rcl}{X}_p& =& \varphi \left(t\right){\int }_{}^{}{\varphi }^{-1}\left(t\right)F\left(t\right)dt\\ & =& \left(\begin{array}{cc}{e}^{2t}& t{e}^{2t}+e^{2t}\\ -e^{2t}& -t{e}^{2t}\end{array}\right){\int }_{}^{}\left(\begin{array}{cc}-t{e}^{-2t}& -t{e}^{-2t}-e^{-2t}\\ e^{-2t}& e^{-2t}\end{array}\right)\left(\begin{array}{c}-2e^{2t}\\ e^{2t}\end{array}\right)dt\\ & =& \left(\begin{array}{cc}{e}^{2t}& t{e}^{2t}+e^{2t}\\ -e^{2t}& -t{e}^{2t}\end{array}\right){\int }_{}^{}\left(\begin{array}{c}2t-t-1\\ -2+1\end{array}\right)dt\\ & =& \left(\begin{array}{cc}{e}^{2t}& t{e}^{2t}+e^{2t}\\ -e^{2t}& -t{e}^{2t}\end{array}\right){\int }_{}^{}\left(\begin{array}{c}t-1\\ -1\end{array}\right)dt\\ & =& \left(\begin{array}{cc}{e}^{2t}& t{e}^{2t}+e^{2t}\\ -e^{2t}& -t{e}^{2t}\end{array}\right)\left(\begin{array}{c}\frac{t^2}{2}-t\\ -t\end{array}\right)\\ & =& \left(\begin{array}{c}\frac{1}{2}{t}^2{e}^{2t}-t{e}^{2t}-t^2{e}^{2t}-t{e}^{2t}\\ -\frac{1}{2}{t}^2{e}^{2t}+t{e}^{2t}+t^2{e}^{2t}\end{array}\right)\\ & =& \left(\begin{array}{c}-\frac{1}{2}{t}^2{e}^{2t}-2t{e}^{2t}\\ \frac{1}{2}{t}^2{e}^{2t}+t{e}^{2t}\end{array}\right)\end{array}$

Use the formula,

$\begin{array}{rcl}X& =& X_c+X_p\\ X& =& c_1\left(\begin{array}{c}1\\ -1\end{array}\right)e^{2t}+c_2\left[\left(\begin{array}{c}1\\ -1\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}1\\ 0\end{array}\right)e^{2t}\right]+\left(\begin{array}{c}-\frac{1}{2}{t}^2{e}^{2t}-2t{e}^{2t}\\ \frac{1}{2}{t}^2{e}^{2t}+t{e}^{2t}\end{array}\right)\end{array}$

Therefore the general solution of the system is $\begin{array}{rcl}X& =& c_1\left(\begin{array}{c}1\\ -1\end{array}\right)e^{2t}+c_2\left[\left(\begin{array}{c}1\\ -1\end{array}\right)t{e}^{2t}+\left(\begin{array}{c}1\\ 0\end{array}\right)e^{2t}\right]+\left(\begin{array}{c}-\frac{1}{2}{t}^2{e}^{2t}-2t{e}^{2t}\\ \frac{1}{2}{t}^2{e}^{2t}+t{e}^{2t}\end{array}\right)\end{array}$.