MATH SOLVE

2 months ago

Q:
# Find the real-valued general solution to the system x' = (-1 -2 4 3)x

Accepted Solution

A:

Answer:The real-valued general solution to the system is:[tex]x=\left[\begin{array}{ccc}x_1\\x_2\\x_3\\x_4\end{array}\right]=\left[\begin{array}{ccc}b_1e^{-t}\\b_2e^{-2t}\\b_3e^{4t}\\b_4e^{3t}\end{array}\right][/tex]Step-by-step explanation:We are given a system as:[tex]x'=(-1\ \ -2\ \ 4\ \ 3)x[/tex]i.e.we have:[tex]\left[\begin{array}{ccc}x_1'\\x_2'\\x_3'\\x_4'\end{array}\right]=[-1\ \ -2\ \ 4\ \ 3]\left[\begin{array}{ccc}x_1\\x_2\\x_3\\x_4\end{array}\right][/tex]i.e.[tex]\left[\begin{array}{ccc}x_1'\\x_2'\\x_3'\\x_4'\end{array}\right]=\left[\begin{array}{ccc}-x_1\\-2x_2\\4x_3\\3x_4\end{array}\right][/tex]Let the variables x_i's be differentiated with respect to t.i.e. we have:[tex]x_1'=-x_1\\\\i.e.\\\\\dfrac{x_1'}{x_1}=-1[/tex]on integrating both side of the equation we have:[tex]\log x_1=-t+c_1\\\\i.e.\\\\x_1=e^{-t+c_1}\\\\i.e.\\\\x_1=e^{-t}e^{c_1}\\\\i.e.\\\\x_1=b_1e^{-t}[/tex]Similarly,[tex]x_2'=-2x_2\\\\i.e.\\\\\dfrac{x_2'}{x_2}=-2\\\\\\\text{on\ integrating\ both\ side}\\\\\log x_2=-2t+c_2\\\\i.e.\\\\x_2=b_2e^{-2t}[/tex]Similarly we get:[tex]x_3=b_3e^{4t}[/tex]and[tex]x_4=b_4e^{3t}[/tex]