# Thread: First order DE

1. ## First order DE

Let

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

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

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

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

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

Help!

2. Originally Posted by Cairo
Let

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

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

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

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

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

Help!

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

and with your IC

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

solving for z gives

$
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

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

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

$
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

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

and with your IC

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

solving for z gives

$
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

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

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

$
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

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

from

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

4. Originally Posted by Cairo
Thanks for this.

How did you get

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

from

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

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

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

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

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.

Thanks for your help.