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

Answer:The fundamental set of solutions are:[tex]x_1=Ae^{2t}[/tex][tex]x_2=B[/tex][tex]x_3=Ce^{3t}[/tex][tex]x_4=De^{4t}[/tex]Hence, the fundamental set of solutions are:[tex]\{e^{2t},B,e^{3t},e^{4t}\}[/tex]Step-by-step explanation:We are given a set as:[tex]x'=(2\ \ 0\ \ 3\ \ 4)x[/tex]which in matrix form could be given by:[tex]\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][/tex]Now, this could also be written in the form:[tex]\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][/tex]Now, in order to find each of the x_i's we have:[tex]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}[/tex]where A is a constant.Similarly we have:[tex]x_2'=0\\\\i.e.\\\\\text{on\ integrating\ both\ side\ we\ have}\\\\x_2=B[/tex]where B is a constant.[tex]x_3'=3x_3[/tex]As done for x_1 we have:[tex]x_3=Ce^{3t}[/tex]and [tex]x_4'=4x_4[/tex]Hence,[tex]x_4=De^{4t}[/tex]