Question

2.2 Find the solution of the following couple of differential equations: x^ prime =4x-y and y^ prime =-4x+4y When t = 0, x(0) = 1 and y(0) = 1 .

Answer 1

$\begin{cases} \frac{dx}{dt}=4x-y\\ \frac{dy}{dt}=-4x+4y \end{cases}$

Characteristic equation:

$\begin{vmatrix} 4-\lambda & -1\\ -4 & 4-\lambda \end{vmatrix}=(4-\lambda)^2-4=$

$=(4-\lambda-2)(4-\lambda+2)=(2-\lambda)(6-\lambda)$

Eigen values:

$\lambda_1=2$

$\lambda_2=6$

Then we find eigen vectors:

for $\lambda_1$ :

$\begin{pmatrix} 2 & -1\\ -4 & 2 \end{pmatrix} \begin{pmatrix} x_{01}\\ y_{01} \end{pmatrix}=0$

$x_{01}=1$

$y_{01}=2$

for $\lambda_2$ :

$\begin{pmatrix} -2 & -1\\ -4 & -2 \end{pmatrix} \begin{pmatrix} x_{02}\\ y_{02} \end{pmatrix}=0$

$x_{02}=1$

$y_{02}=-2$

Common solution:

$\begin{pmatrix} x\\ y \end{pmatrix}=C_1 \begin{pmatrix} 1\\ 2 \end{pmatrix}*e^{2t}+C_2 \begin{pmatrix} 1\\ -2 \end{pmatrix}*e^{6t}$

When t=0, x(0)=1 and y(0)=1, so

$\begin{pmatrix} 1\\ 1 \end{pmatrix}=C_1 \begin{pmatrix} 1\\ 2 \end{pmatrix}+C_2 \begin{pmatrix} 1\\ -2 \end{pmatrix}$

Solution: $C_1=\frac{3}{4}$ $C_2=\frac{1}{4}$

Solution of the system:

$\begin{pmatrix} x\\ y \end{pmatrix}=\frac{3}{4} \begin{pmatrix} 1\\ 2 \end{pmatrix}*e^{2t}+\frac{1}{4} \begin{pmatrix} 1\\ -2 \end{pmatrix}*e^{6t}$