Linear DE systems with constant coefficients.

Question

System given: $\begin{cases} x'=2x-y \\ y'=x+2y \end{cases} $

My solution: $$x''=2x'-y'=2x'-x-2y=2x'-x-2(2x-x')=6x'-5x \Rightarrow 6\lambda -5=0 \Rightarrow x(t)=C_1e^{\frac{5t}{6}}$$ I know already that this is not the right soultion, but I cannot understand why?

Answer 1

You can simplify a little this system: $$\begin{cases} x'=2x-y \\ y'=x+2y \end{cases}\implies \begin{cases} x'-2x=-y \\ y'-2y=x \end{cases}$$ $$\begin{cases} (xe^{-2t})'=-ye^{-2t} \\ (ye^{-2t})'=x e^{-2t}\end{cases}\implies \begin{cases} u'=-v \\ v'=u\end{cases}$$ $$\begin{cases} u''+u=0 \\ v=-u'\end{cases} \text { ,where } \begin{cases} u=xe^{-2t} \\ v=ye^{-2t}\end{cases}$$ $$u''+u=0 \implies r^2+1=0 \implies r=\pm i $$ $$ \begin{cases} u(t)=c_1\cos t + c_2 \sin t \\ v(t)=c_1\sin t - c_2 \cos t \end{cases} $$ $$ \implies \begin{cases} x(t)=e^{2t}(c_1\cos t + c_2 \sin t ) \\ y(t)=e^{2t}(c_1\sin t - c_2 \cos t ) \end{cases} $$

Answer 2

Given system of differential equation $$\begin{cases} x'=2x-y \\ y'=x+2y \end{cases}$$ can be written as $$\dfrac{dX}{dt}=AX\qquad\text{where}\quad A=\begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix}\quad\text{and}\quad X=\begin{pmatrix} x \\ y \end{pmatrix}~.$$ Consider the solution of the differential equation is of the form $~x=\bar \alpha ~e^{\lambda~t}~~$where $~\bar \alpha~$ is the eigen-vector corresponding to the eigen-value $~\lambda~$.

For non trivial solution $$\begin{vmatrix} 2-\lambda & -1 \\ 1 & 2-\lambda \end{vmatrix}=0$$ $$\implies (2-\lambda)^2+1=0$$ $$\implies \lambda=2\pm~ i$$ Now we have to find the eigen-vector corresponding to $~\lambda=2\pm ~i~$.

For $~\lambda=2-i~$, $$AX=(2-i)X\implies \left(\begin{matrix} 2 & -1 \\ 1 & 2 \end{matrix}\right)\left(\begin{matrix} x \\ y \end{matrix}\right)=(2-i)~\left(\begin{matrix} x \\ y \end{matrix}\right)\implies i~x-y=0$$ So $~\bar\alpha^{(1)}=\left(\begin{matrix} i \\ -1 \end{matrix}\right)~$.

For $~\lambda=2+i~$, $$AX=(2+i)X\implies \left(\begin{matrix} 2 & -1 \\ 1 & 2 \end{matrix}\right)\left(\begin{matrix} x \\ y \end{matrix}\right)=(2+i)~\left(\begin{matrix} x \\ y \end{matrix}\right)\implies x-iy=0$$ So $~\bar\alpha^{(2)}=\left(\begin{matrix} i \\ 1 \end{matrix}\right)~$.

So the general solution is $$X=A~\bar\alpha^{(1)}~e^{(2-i)~t}+B~\bar\alpha^{(2)}~e^{(2+i)~t}$$where $~A,~B~$are constants.

So $$ X=A~\left(\begin{matrix} i \\ -1 \end{matrix}\right)~e^{(2-i)~t}+B~\left(\begin{matrix} i \\ 1 \end{matrix}\right)~e^{(2+i)~t}$$ $$\implies \begin{pmatrix} x \\ y \end{pmatrix}=A~\left(\begin{matrix} i \\ -1 \end{matrix}\right)~e^{(2-i)~t}+B~\left(\begin{matrix} i \\ 1 \end{matrix}\right)~e^{(2+i)~t}$$ $$\implies\begin{cases} x=iAe^{(2-i)t}~+~iBe^{(2+i)t} \\ y=-Ae^{(2-i)t}~+~Be^{(2+i)t} \end{cases}$$where $~A,~B~$are constants.