1. ## First order DE

Let

$\displaystyle \frac{dx}{dt}=-x-x^{2}+y^{2} and \frac{dy}{dt}=-y-2xy.$

I have shown that the two equations are the real and imaginary parts of a single differential equation for the complex variable
$\displaystyle z(t), \frac{dz}{dt}=-z-z^{2}.$

I have evaluated the differential equation for z(t) to get

$\displaystyle t+C=\ln\left(\frac{z+1}{z}\right).$

However, I need to determine the solution to the DE with initial condition $\displaystyle z=i at t=0.$ I then need to use this to show that the solution of the original equations for x(t) and y(t) with initial conditions
$\displaystyle x(0)=0$ and $\displaystyle y(0)=1$ has both x(t) and y(t) approaching $\displaystyle \exp\left(\frac{-t}{2}\right)/2$ as $\displaystyle t\rightarrow\infty.$

Help!

2. Originally Posted by Cairo
Let

$\displaystyle \frac{dx}{dt}=-x-x^{2}+y^{2} and \frac{dy}{dt}=-y-2xy.$

I have shown that the two equations are the real and imaginary parts of a single differential equation for the complex variable
$\displaystyle z(t), \frac{dz}{dt}=-z-z^{2}.$

I have evaluated the differential equation for z(t) to get

$\displaystyle t+C=\ln\left(\frac{z+1}{z}\right).$

However, I need to determine the solution to the DE with initial condition $\displaystyle z=i at t=0.$ I then need to use this to show that the solution of the original equations for x(t) and y(t) with initial conditions
$\displaystyle x(0)=0$ and $\displaystyle y(0)=1$ has both x(t) and y(t) approaching $\displaystyle \exp\left(\frac{-t}{2}\right)/2$ as $\displaystyle t\rightarrow\infty.$

Help!

$\displaystyle \frac{z+1}{z} = k e^t$

$\displaystyle \frac{z+1}{z} = \frac{i+1}{i} e^t = (1 - i) e^t$

solving for z gives

$\displaystyle z = \frac{1}{\left( e^t - 1 \right) - i e^t} = \frac{\left( e^t - 1 \right) + i e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}$

giving

$\displaystyle x = \frac{\left( e^t - 1 \right) }{\left( e^t - 1 \right) ^2 + e^{2t}}$ , $\displaystyle y = \frac{ e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}$

the solutions of your DE noting that $\displaystyle x(0) = 0, y(0) = 1$. Now when $\displaystyle t \to \infty$ then x and y approach

$\displaystyle x,y \to \frac{e^t}{2 e^{2t}} = \frac{e^{-t}}{2}$ (not quite what you got but close.)

3. Originally Posted by Danny

$\displaystyle \frac{z+1}{z} = k e^t$

$\displaystyle \frac{z+1}{z} = \frac{i+1}{i} e^t = (1 - i) e^t$

solving for z gives

$\displaystyle z = \frac{1}{\left( e^t - 1 \right) - i e^t} = \frac{\left( e^t - 1 \right) + i e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}$

giving

$\displaystyle x = \frac{\left( e^t - 1 \right) }{\left( e^t - 1 \right) ^2 + e^{2t}}$ , $\displaystyle y = \frac{ e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}$

the solutions of your DE noting that $\displaystyle x(0) = 0, y(0) = 1$. Now when $\displaystyle t \to \infty$ then x and y approach

$\displaystyle x,y \to \frac{e^t}{2 e^{2t}} = \frac{e^{-t}}{2}$ (not quite what you got but close.)
Thanks for this.

How did you get

$\displaystyle x,y \to \frac{e^t}{2 e^{2t}} = \frac{e^{-t}}{2}$

from

$\displaystyle x = \frac{\left( e^t - 1 \right) }{\left( e^t - 1 \right) ^2 + e^{2t}}$ , $\displaystyle y = \frac{ e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}$

4. Originally Posted by Cairo
Thanks for this.

How did you get

$\displaystyle x,y \to \frac{e^t}{2 e^{2t}} = \frac{e^{-t}}{2}$

from

$\displaystyle x = \frac{\left( e^t - 1 \right) }{\left( e^t - 1 \right) ^2 + e^{2t}}$ , $\displaystyle y = \frac{ e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}$
I'll do $\displaystyle x, y$ follows similarly. Expanding $\displaystyle x$ gives

$\displaystyle x = \frac{e^t - 1}{ 2 e^{2t} - 2 e^t + 1}$

Hand waving. In the numerator, as $\displaystyle t \to \infty$, the $\displaystyle e^t$ is more dominate than the $\displaystyle -1$. In the denominator, the $\displaystyle 2e^{2t}$ is more dominate than $\displaystyle 2e^t$ and $\displaystyle 1$. Hence

$\displaystyle x \to \frac{e^t}{2e^{2t}}$ as $\displaystyle t \to \infty$.

You could be a little more formal using limits but this gets the idea across.

5. Thanks. It suddenly dawned on me to use L'Hopital's rule and the result follows.