Question

Verify that the vector Xa is a particular solution of the given nonhomogeneous linear system. ,_21 et+ 1 tef, one has 1 te is a particular solution of the given system Since the above expressions--Select Need Help?Read It Talk to a Tutor

Answer 1

Given system of diff eq is

$x'=x+2y\\ y'=4x+3y\\ \sim \\ \begin{bmatrix} x'\\ y'\end{bmatrix}=\begin{bmatrix} 1 & 2\\ 4 & 3 \end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}$

To solve above diff eq , first will find eigen values of coef matrix and corresponding eigen vector .

$det\begin{bmatrix} 1 -c& 2\\ 4 & 3-c \end{bmatrix}=0\\ (1-c)(3-c)-8=0\\ c^2-4c-5=0\\ (c+1)(c-5)=0\\ c=-1,5$

Let u be an eigen vector corr to -1,

$\begin{bmatrix} 1& 2\\ 4 & 3 \end{bmatrix}u=-u$

$\begin{bmatrix} 1+1& 2\\ 4 & 3+1 \end{bmatrix}u=0$

$\begin{bmatrix} 2& 2\\ 4 & 4 \end{bmatrix}\begin{bmatrix} u_1\\ u_2\end{bmatrix}=0\\ u_1=-u_2\\ u=\begin{bmatrix} -1\\ 1\end{bmatrix}$

Let v be an eigen vector corr to eigen value 5,

$\begin{bmatrix} 1& 2\\ 4 & 3 \end{bmatrix}\begin{bmatrix} v_1\\ v_2\end{bmatrix}=5v$

$\begin{bmatrix} 1-5& 2\\ 4 & 3-5 \end{bmatrix}\begin{bmatrix} v_1\\ v_2\end{bmatrix}=0$

$\begin{bmatrix} -4& 2\\ 4 & -2 \end{bmatrix}\begin{bmatrix} v_1\\ v_2\end{bmatrix}=0$

$-2v_1=v_2\\ v=\begin{bmatrix} 1\\ -2\end{bmatrix}$

Hence general solution is

$\begin{bmatrix} x(t)\\ y(t)\end{bmatrix}=Ae^{-t}\begin{bmatrix} -1\\ 1\end{bmatrix}+Be^{5t}\begin{bmatrix} 1\\ -2\end{bmatrix}$

                 $=\begin{bmatrix} -Ae^{-t}+Be^{5t}\\ Ae^{-t}-2Be^{5t}\end{bmatrix}$

$x(t)= -Ae^{-t}+Be^{5t},y(t)=Ae^{-t}-2Be^{5t}$