Question

Find the General solution of the given system dx/dt=x+2y and dy/dt=4x+3

Answer 1

We have the following system: $\begin{cases} \frac{dx}{dt}=x+2y \\ \frac{dy}{dt}=4x+3y \end{cases}$ , or in the matrix form: $X^{\prime}=AX,$ where $X=\begin{bmatrix} x \\ y \end{bmatrix}, \ A=\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$ .

The characteristic equation: $\det (A-\lambda I)=\begin{vmatrix} 1-\lambda & 2 \\ 4 & 3-\lambda \end{vmatrix}=(1-\lambda )(3-\lambda )-4\times 2=\lambda ^2-4\lambda -5=0.$

Eigenvalues: $\lambda _1=-1,\ \lambda _2=5.$

Let $v_1$ be eigen vector, associated with the eigenvalue $\lambda_1$ :

$\begin{bmatrix} 1 -\lambda_1& 2 \\ 4 & 3-\lambda_1 \end{bmatrix} v_1=\begin{bmatrix} 2 & 2 \\ 4 & 4 \end{bmatrix}v_1= \begin{bmatrix} 2v_{11} +2 v_{12}\\ 4 v_{11}+4v_{12} \end{bmatrix}=0,$ then $v_1= \begin{bmatrix} 1 \\ -1 \end{bmatrix}$ .

Let $v_2$ be eigen vector, associated with the eigenvalue $\lambda_2$ :

$\begin{bmatrix} 1 -\lambda_2& 2 \\ 4 & 3-\lambda_2 \end{bmatrix} v_2=\begin{bmatrix} -4 & 2 \\ 4 & -2 \end{bmatrix}v_2= \begin{bmatrix} -4v_{21} +2 v_{22}\\ 4 v_{21}-2v_{22} \end{bmatrix}=0,$ then $v_2= \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ .

Then solution has form $X=c_1e^{\lambda_1 t} v_1+ c_2e^{\lambda_2 t} v_2 = \begin{bmatrix} c_1 e^{-t} +c_2 e^{5t}\\ -c_1e^{-t}+2c_2e^{5t} \end{bmatrix}$ .

Answer: $\begin{cases} x=c_1 e^{-t} +c_2 e^{5t}\\ y=-c_1e^{-t}+2c_2e^{5t} \end{cases}$ .