Intertwined differential equation Given $$\frac{dx(t)}{dt}=x(t)+y(t),\quad \frac{dy(t)}{dt}=x(t)-y(t),\quad \frac{dx(t=0)}{dt}=0$$. What is  $ x(\infty)/y(\infty)?$
I differentated first equation, do proper substitutions and obtained solutions as $$x(t)=A\cosh (t\sqrt{2})\quad y(t)=A(\sqrt{2}\sinh (t\sqrt{2})-\cosh (t\sqrt{2})$$. My evaluation of   $ x(\infty)/y(\infty)=1+\sqrt{2}$. My question is (1). Is the answer correct?
(2).Is there a simpler method of evaluating $ x(\infty)/y(\infty)$?
In the same spirit as @Aryadeva, you also have
$$x'+y'=2(x+y) \implies x+y=Ce^{2t}$$ that is to say
$$x=\frac{1}{2} \left(C e^{2t}+K\right)\qquad \text{and} \qquad y=\frac{1}{2} \left(C e^{2t}-K\right)$$ The condition for $x$ gives $C=0$; so $x=K$ and $y=-K$.
Your answer is totally correct, but I'm surprised nobody has mentioned this method yet and I think it is interesting.
One can represent this system using a matrix. Let $\underline{r}(t)=[x(t),y(t)]^{\mathrm{T}}$. Then,
$$\dot{\underline{r}}=\begin{bmatrix}
1 &amp; 1\\
1 &amp; -1
\end{bmatrix}\underline{r}$$
First let's define the matrix exponential: For a square matrix $\mathbf{X}$, define
$$\exp(\mathbf{X})=e^\mathbf{X}=\sum_{k=0}^\infty\frac{1}{k!}\mathbf{X}^k$$
It indeed follows the usual notion of the derivative, that is for a scalar $t$,
$$\frac{\mathrm{d}}{\mathrm{d}t}\exp(\mathbf{X}t)=\mathbf{X}\exp(\mathbf{X}t)$$
So, let
$$\mathbf{M}=\begin{bmatrix}
1 &amp; 1\\
1 &amp; -1
\end{bmatrix}$$
Then we can restate our system as
$$\dot{\underline{r}}=\mathbf{M}\underline{r}$$
We make the guess solution $\underline{r}(t)=\underline{c}\exp(\mathbf{M}t)$, with $\underline{c}$ being some constant vector. Then, we can see that
$$\dot{\underline{r}}=\mathbf{M}\underline{c}\exp(\mathbf{M}t)=\mathbf{M}\underline{r}$$
As required. Furthermore, we may notice that $\underline{r}(0)=\underline{c}\exp(0\mathbf{M})=\underline{c}$. Defining $\underline{r}_0=\underline{r}(0)$, we have
$$\underline{r}(t)=\underline{r}_0\exp(\mathbf{M}t).$$
It can be shown that this can equivalently be written as
$$\underline{r}(t)=c_1 e^{\lambda_1 t}\underline{u}_1+c_2 e^{\lambda_2 t}\underline{u}_2$$
With $\lambda_1,\lambda_2$ being the eigenvalues of the matrix $\mathbf{M}$ and $\underline{u}_1,\underline{u}_2$ being the respective eigenvectors. And of course $c_1,c_2$ are some constants to be determined by the initial conditions. One can compute the eigenvalues/eigenvectors of $\mathbf{M}$ to be:
$$\lambda_1=-\sqrt{2} ~;~ \lambda_2=\sqrt{2}$$
$$\underline{u}_1=\begin{bmatrix}
1-\sqrt{2}\\
1
\end{bmatrix} ~;~ \underline{u}_2=\begin{bmatrix}
1+\sqrt{2}\\
1
\end{bmatrix}$$
Which leads to
$$x(t)=d_1 e^{-\sqrt{2}t}+d_2 e^{\sqrt{2}t}$$
$$y(t)=c_1 e^{-\sqrt{2}t}+c_2 e^{\sqrt{2}t}$$
With of course $d_1=(1-\sqrt{2})c_1$ and $d_2=(1+\sqrt{2})c_2$. What happens when we impose the condition $\dot{x}(0)=0?$ First we compute:
$$\dot{x}(t)=-\sqrt{2}~d_1 e^{-\sqrt{2}t}+\sqrt{2}~d_2 e^{\sqrt{2}t}$$
So,
$$\dot{x}(0)=\sqrt{2}(d_2-d_1)$$
Thus we require $d_2=d_1$ or $(1-\sqrt{2})c_1=(1+\sqrt{2})c_2$. Let $d_1=d_2=C/2$ thus
$$x(t)=C\cosh(\sqrt{2}t)$$
$$y(t)=C\left(\sqrt{2}\sinh(\sqrt{2}t)-\cosh(\sqrt{2}t)\right)$$
So
$$\lim_{t\to\infty}\frac{y(t)}{x(t)}=\lim_{t\to\infty}\frac{\sqrt{2}\sinh(\sqrt{2}t)}{\cosh(\sqrt{2}t)}-1=\sqrt{2}-1$$
$$\left(\lim_{t\to\infty}\frac{y(t)}{x(t)}\right)^{-1}=\lim_{t\to\infty}\frac{x(t)}{y(t)}=\frac{1}{\sqrt{2}-1}=1+\sqrt{2}.$$