Question

In Problems$\mathbf{5}\mathbf{-}\mathbf{14}$solve the given linear system.

$\mathbf{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}2& 8\\ 0& 4\end{array}\right)\mathbf{X}\mathbf{+}\left(\begin{array}{c}2\\ 16t\end{array}\right)$

Short answer

The solution for the linear system $\mathrm{X}\text{'}=\left(\begin{array}{cc}2& 8\\ 0& 4\end{array}\right)\mathrm{X}+\left(\begin{array}{c}2\\ 16\mathrm{t}\end{array}\right)$

$X=c_1\left(\left.\begin{array}{c}1\\ 0\end{array}\right)e^{-2t}+c_2\left(\begin{array}{c}4\\ 1\end{array}\right)e^{4t}+\left(\begin{array}{c}-11+16|\\ -1-4t\end{array}\right.\right)$

Step-by-step solution

Define matrix exponential

Consider a square matrix A of size $\mathrm{n}\times \mathrm{n}$. 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 $e^{At}$. 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,

$X\text{'}=\left(\begin{array}{cc}2& 8\\ 0& 4\end{array}\right)X+\left(\begin{array}{c}2\\ 16t\end{array}\right)$

where,

$A=\left(\begin{array}{cc}2& 8\\ 0& 4\end{array}\right)$

Now we find the characteristic equation of the coefficient matrix,

$$\begin{gathered} det(A-\lambda l)=0\to \left|\begin{array}{cc}2-\lambda & 8\\ 0& 4-\lambda \end{array}\right| \\ \left(2-\lambda \right)\left(4-\lambda \right)=0 \end{gathered}$$

So we have eigenvalues of ${\lambda }_1=2,{\lambda }_2=4$.

Find the eigenvector and solution vector of the given linear system

For${\lambda }_1=2$

$\begin{array}{rcl}(A-2\overline{)I}0)& =& \left(\begin{array}{ccc}2-2& 8& 0\\ 0& 4-2& 0\end{array}\right)\\ & =& \left(\begin{array}{ccc}0& 8& 0\\ 0& 2& 0\end{array}\right)\end{array}$

Choosing $k_2=0$we have $k_1=1$as arbitrary value.

This gives an eigenvector and a corresponding solution vector respectively.

$K_1=\left(\begin{array}{c}1\\ 0\end{array}\right),x_1=\left(\begin{array}{c}1\\ 0\end{array}\right)e^{2t}$

Find the complementary function of the given linear system

For ${\lambda }_2=4$,

role="math" localid="1668058263117" $\begin{array}{rcl}(A-2\overline{)I}0)& =& \left(\begin{array}{ccc}2-4& 8& 0\\ 0& 4-4& 0\end{array}\right)\\ & =& \left(\begin{array}{ccc}-2& 8& 0\\ 0& 2& 0\end{array}\right)\\ -2k_1+8k_2& =& 0\\ k_1& =& 4k_2\end{array}$

Choosing $k_2=1$we have$k_1=4$.

This gives an eigenvector $K_2=\left(\begin{array}{c}4\\ 1\end{array}\right),X_2=\left(\begin{array}{c}4\\ 1\end{array}\right)e^{4t}$.

Hence the complementary function is$x=c_1\left(\begin{array}{c}1\\ 0\end{array}\right)e^{2t}+c_2\left(\begin{array}{c}4\\ 1\end{array}\right)e^{4t}$

Find the inverse matrix of ϕt

Using the eigenvectors construct the matrix of ${\varphi }_t$.

$\mathrm{\varphi }\left(\mathrm{t}\right)=\left(\begin{array}{cc}{e}^{2t}& 4e^{4t}\\ 0& e^{4t}\end{array}\right)$

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

$\begin{array}{rcl}\left|\mathrm{\varphi }\left(\mathrm{t}\right)\right|& =& \left|\begin{array}{cc}{e}^{2t}& 4e^{4t}\\ 0& e^{4t}\end{array}\right|\\ & =& e^{2t}{e}^{4t}\\ & =& e^{6t}\ne 0\\ & & \\ & & \end{array}$

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

So now,

role="math" localid="1668059161055" ${\mathrm{\varphi }}^{-1}\left(\mathrm{t}\right)=\frac{1}{e^{6t}}\left(\begin{array}{cc}{e}^{4t}& -4e^{4t}\\ 0& e^{2t}\end{array}\right)=\left(\begin{array}{cc}{e}^{-2t}& -4e^{-2t}\\ 0& e^{-4t}\end{array}\right)$

Hence the inverse matrix of $\varphi \left(t\right)$is:

${\mathrm{\varphi }}^{-1}\left(\mathrm{t}\right)=\left(\begin{array}{cc}{e}^{-2t}& -4e^{-2t}\\ 0& e^{-4t}\end{array}\right)$

Find the particular solution of the given linear system by integration

Obtaining the solution,

$$\begin{gathered} X_p=\varphi \left(t\right){\int }_{}^{}{\varphi }^{-1}\left(t\right)F\left(t\right)dt \\ \left(\begin{array}{cc}{e}^{2t}& 4e^{4t}\\ 0& e^{4t}\end{array}\right){\int }_{}^{}\left(\begin{array}{cc}{e}^{-2t}& -4e^{-2t}\\ 0& e^{-4t}\end{array}\right)\left(\begin{array}{c}2\\ 16t\end{array}\right)dt \\ \left(\begin{array}{cc}{e}^{2t}& 4e^{4t}\\ 0& e^{4t}\end{array}\right){\int }_{}^{}\left(\begin{array}{c}\begin{array}{cc}2e^{-2t}& -64t{e}^{-2t}\end{array}\\ 16t{e}^{-4t}\end{array}\right)dt \end{gathered}$$

Integrate,

$\begin{array}{ll}64{\int }_{}^{}t{e}^{-2t}dt& \\ u=t& v=e^{-2t}\\ du=1& dv=-\frac{1}{2}{e}^{-2t}\end{array}$

where,

$\begin{array}{rcl}64{\int }_{}^{}t{e}^{-2t}dt& =& 64\left[uv-{\int }_{}^{}vdu\right]\\ & =& 64\left[-\frac{1}{2}t{e}^{-2t}+\frac{1}{2}{\int }_{}^{}{e}^{-2t}dt\right]\\ & =& 64\left[-\frac{1}{2}t{e}^{-2t}-\frac{1}{4}{e}^{-2t}\right]\\ & =& -32t{e}^{-2t}-16e^{-2t}\end{array}$

Integrate,

$\begin{array}{ll}16{\int }_{}^{}t{e}^{-4t}dt& \\ u=t& v=e^{-4t}\\ du=1& dv=-\frac{1}{4}{e}^{-4t}\end{array}$

where,

role="math" localid="1668060006249" $\begin{array}{rcl}16{\int }_{}^{}t{e}^{-4t}dt& =& 16\left[uv-{\int }_{}^{}vdu\right]\\ & =& 16\left[-\frac{1}{4}t{e}^{-4t}+\frac{1}{4}{\int }_{}^{}{e}^{-4t}dt\right]\\ & =& 16\left[-\frac{1}{4}t{e}^{-4t}-\frac{1}{16}{e}^{-4t}\right]\\ & =& -4t{e}^{-4t}-e^{-4t}\end{array}$

Hence the value after integration is $-4{\mathrm{te}}^{-4t}-e^{-4t}$.

Find the general solution of the given linear system by integration

To get the general solution, solve,

$\begin{array}{rcl}{X}_p& =& \left(\begin{array}{cc}{e}^{2t}& 4e^{4t}\\ 0& e^{4t}\end{array}\right){\int }_{}^{}\left(\begin{array}{c}\begin{array}{cc}2e^{-2t}& -64t{e}^{-2t}\end{array}\\ 16t{e}^{-4t}\end{array}\right)dt\\ & =& \left(\begin{array}{cc}{e}^{2t}& 4e^{4t}\\ 0& e^{4t}\end{array}\right)\left(\begin{array}{c}\begin{array}{cc}-e^{-2t}& +32t{e}^{-2t}\end{array}+16t{e}^{-2t}\\ -4t{e}^{-4t}-e^{-4t}\end{array}\right)\\ & =& \left(\begin{array}{c}-1+32t+16-16t-4\\ -4t-1\end{array}\right)\\ & =& \left(\begin{array}{c}-11+16t\\ -1-4t\end{array}\right)\\ & & \end{array}$

Use the formula,

$$\begin{gathered} X=X_c+X_p \\ X=c_1\left(\begin{array}{c}1\\ 0\end{array}\right)e^{2t}+c_2\left(\begin{array}{c}4\\ 1\end{array}\right)e^{4t}+\left(\begin{array}{c}-11+16t\\ -1-4t\end{array}\right) \end{gathered}$$

Therefore the general solution of the system is $X=c_1\left(\begin{array}{c}1\\ 0\end{array}\right)e^{2t}+c_2\left(\begin{array}{c}4\\ 1\end{array}\right)e^{4t}+\left(\begin{array}{c}-11+16t\\ -1-4t\end{array}\right)$.