Question

I AM REALLY STRUGGLING ON THIS PROBLEM PLEASE HELP ME CORRECT AND NEAT WORK IS MUCH APPRECIATED THANKS

(1 point) Consider the linear system 3'=[} }); a. Find the eigenvalues and eigenvectors for the coefficient matrix. EL and 12 = b. Find the real-valued solution to the initial value problem y! 3yı + 2y2, -5yı - 3y2, yı(O) = 5, y2(0) = -5. Use t as the independent variable in your answers. yi(t) = y2(t) =

Answer 1

a.

3 - 2 :-(3-)(3+) + 10 = 22 +1 -5 -3 - So, eigen values are given by roots of r2 +1 = 0 i.e. i, -i 3+ i 2 [1 }(3-1) now, for 11 -i: So, take eigen vector v1 = -5 -3+ 0 0 (-3+i) 1-6 :-) t [3 - then, for x2 =i : -5 2 -3-1 -.]-[8 468,7"] So, take the eigen vector v2 (-+(3+i)) t

b.$the \ general\ solution\ is,\\ y(t)=c_1v_1e^{x_1t}+c_2v_2e^{x_2t}=c_1\begin{pmatrix} \frac{1}{5}(-3+i)\\ 1\\ \end{pmatrix}e^{-it}+c_2\begin{pmatrix} -\frac{1}{5}(3+i)\\ 1\\ \end{pmatrix}e^{it}\\ Now,y(t)=\begin{pmatrix} y_1(t)\\ y_2(t) \end{pmatrix}\\ given, that \ y(0)=\begin{pmatrix} y_1(0)\\ y_2(0) \end{pmatrix}=\begin{pmatrix} 5\\ -5 \end{pmatrix}\\ \therefore ,\begin{bmatrix} \frac{1}{5}(-3+i)&-\frac{1}{5}(3+i) \\ 1&1 \end{bmatrix}\begin{pmatrix} c_1\\ c_2 \end{pmatrix}=\begin{pmatrix} 5\\ -5 \end{pmatrix}\\\Rightarrow \begin{pmatrix} c_1\\ c_2 \end{pmatrix}=\begin{pmatrix} -\frac{5i}{2} &\frac{1}{2}-\frac{3i}{2} \\ \frac{5i}{2}& \frac{1}{2}(1+3i) \end{pmatrix}\begin{pmatrix} 5\\ -5 \end{pmatrix}=\begin{pmatrix} -\frac{5}{2}-5i\\ -\frac{5}{2}+5i\\ \end{pmatrix}\\ \therefore y_1(t)=(-\frac{5}{2}-5i)(\frac{1}{5}(-3+i))e^{-it}+(-\frac{5}{2}+5i)(-\frac{1}{5}(3+i))e^{it} \\=\frac{5}{2}(1+i)e^{-it}+\frac{5}{2}(1-i)e^{it}=5cos(t)+5sin(t) \\$$y_2(t)=(-\frac{5}{2}-5i)e^{-it}+(-\frac{5}{2}+5i)e^{it}=-5cos(t)-10sin(t)$