Question

Find a fundamental set of solutions for the system: x' = (2 0 3 4)x

Answer 1

Answer:

The fundamental set of solutions are:

$x_1=Ae^{2t}$

$x_2=B$

$x_3=Ce^{3t}$

$x_4=De^{4t}$

Hence, the fundamental set of solutions are:

$\{e^{2t},B,e^{3t},e^{4t}\}$

Step-by-step explanation:

We are given a set as:

$x'=(2\ \ 0\ \ 3\ \ 4)x$

which in matrix form could be given by:

$\left[\begin{array}{ccc}x_1'\\x_2'\\x_3'\\x_4'\end{array}\right]=[2\ \ 0\ \ 3\ \ 4]\left[\begin{array}{ccc}x_1\\x_2\\x_3\\x_4\end{array}\right]$

Now, this could also be written in the form:

$\left[\begin{array}{ccc}x_1'\\x_2'\\x_3'\\x_4'\end{array}\right]=\left[\begin{array}{ccc}2\cdot x_1\\0\cdot x_2\\3\cdot x_3\\4\cdot x_4\end{array}\right]\\\\i.e.\\\\\left[\begin{array}{ccc}x_1'\\x_2'\\x_3'\\x_4'\end{array}\right]=\left[\begin{array}{ccc}2x_1\\0\\3x_3\\4x_4\end{array}\right]$

Now, in order to find each of the x_i's we have:

$x_1'=2x_1\\\\i.e.\\\\\dfrac{x_1'}{x_1}=2\\\\\text{on\ integrating\ both\ side\ we\ have}\\\\\log x_1=2t+c\\\\\text{where\ c\ is\ a\ constant}\\\\x_1=e^{2t+c}\\\\i.e.\\\\x_1=e^{2t}e^c\\\\i.e.\\\\x_1=Ae^{2t}$

where A is a constant.

Similarly we have:

$x_2'=0\\\\i.e.\\\\\text{on\ integrating\ both\ side\ we\ have}\\\\x_2=B$

where B is a constant.

$x_3'=3x_3$

As done for x_1 we have:

$x_3=Ce^{3t}$

and

$x_4'=4x_4$

Hence,

$x_4=De^{4t}$