Question

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

Answer 1

Answer:

The real-valued general solution to the system is:

$x=\left[\begin{array}{ccc}x_1\\x_2\\x_3\\x_4\end{array}\right]=\left[\begin{array}{ccc}b_1e^{-t}\\b_2e^{-2t}\\b_3e^{4t}\\b_4e^{3t}\end{array}\right]$

Step-by-step explanation:

We are given a system as:

$x'=(-1\ \ -2\ \ 4\ \ 3)x$

i.e.

we have:

$\left[\begin{array}{ccc}x_1'\\x_2'\\x_3'\\x_4'\end{array}\right]=[-1\ \ -2\ \ 4\ \ 3]\left[\begin{array}{ccc}x_1\\x_2\\x_3\\x_4\end{array}\right]$

i.e.

$\left[\begin{array}{ccc}x_1'\\x_2'\\x_3'\\x_4'\end{array}\right]=\left[\begin{array}{ccc}-x_1\\-2x_2\\4x_3\\3x_4\end{array}\right]$

Let the variables x_i's be differentiated with respect to t.

i.e. we have:

$x_1'=-x_1\\\\i.e.\\\\\dfrac{x_1'}{x_1}=-1$

on integrating both side of the equation we have:

$\log x_1=-t+c_1\\\\i.e.\\\\x_1=e^{-t+c_1}\\\\i.e.\\\\x_1=e^{-t}e^{c_1}\\\\i.e.\\\\x_1=b_1e^{-t}$

Similarly,

$x_2'=-2x_2\\\\i.e.\\\\\dfrac{x_2'}{x_2}=-2\\\\\\\text{on\ integrating\ both\ side}\\\\\log x_2=-2t+c_2\\\\i.e.\\\\x_2=b_2e^{-2t}$

Similarly we get:

$x_3=b_3e^{4t}$

and

$x_4=b_4e^{3t}$