# Find the general solution $\left[\begin{smallmatrix}x \\ y \end{smallmatrix}\right]$

Let $$\begin{bmatrix} \dot x \\ \dot y \end{bmatrix}=\begin{bmatrix}-5 & -3 \\ 3 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$$.

a) Find the general solution $$\begin{bmatrix}x \\ y \end{bmatrix}$$.

b)Find $$λ$$ and a basis for which the system reduces to ̇$$\dot u=λu+v$$ and $$̇\dot v=λv$$.

The eigenvalue $$λ=-2$$ gives us the eigenvector $$\begin{bmatrix}-1 \\ 1 \end{bmatrix}$$ and generalized eigenvector $$\begin{bmatrix}1/3 \\ 0 \end{bmatrix}$$.

For part a), the general solution $$\begin{bmatrix}x \\ y \end{bmatrix}=c_1 e^{-2t}\begin{bmatrix}-1 \\ 1 \end{bmatrix}+c_2e^{-2t}\Bigg(t\begin{bmatrix}-1 \\ 1 \end{bmatrix}+\begin{bmatrix}1/3 \\ 0 \end{bmatrix}\Bigg)$$

I'm not sure about part b) can anyone explain it?

• its jordan form I guess write the system with matrices with u and v Commented Nov 7, 2018 at 23:59

$$\begin{bmatrix} \dot x \\ \dot y \end{bmatrix}=\begin{bmatrix}-5 & -3 \\ 3 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} \implies X'=AX$$ You have that $$\begin{cases} u'=\lambda u + v \\ v'=\lambda v \end{cases}$$ $$\implies \pmatrix { u \\ v}' =\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}\pmatrix {u \\ v } \implies U'=BU$$
B is just the Jordan matrix. Try to evaluate it now $$J=H^{-1}AH$$ Where H is the matrix with eigenvectors $$H=\pmatrix {-1 & 1/3 \\1 &0}$$
• If you evaluate $H^{-1}AH$ you will have a diagonilzied matrix...Like B ...Just do the calculation it takes 5 minues. Commented Nov 8, 2018 at 2:09
• H is not diagonal thats J or B that is diagonal as it's the matrix of the system with u and v variables... I got for J$$J=\pmatrix {-2 &1 \\0 &-2}$$ H is the matrix that changes the basis that takes from A to matrix B @MANONMARS45 more information here en.wikipedia.org/wiki/Matrix_similarity Commented Nov 8, 2018 at 2:20