Is it possible to find the equilibrium of these equations

Y= X
X = e^{-bY/X}

b is a constant... setting both equations to 0, we have

0 = X
0 = e^{-bY/X}

Question

Is it possible to find the equilibrium of these equations

Y= X
X = e^{-bY/X}

b is a constant... setting both equations to 0, we have

0 = X
0 = e^{-bY/X}

usukidoll · Answer

now can we really find an equilibrium if we set these equations to 0 or do we have to substitute y = x into the second equation and then take the natural log.

usukidoll · Answer

@rational

usukidoll · Answer

the thing is .... if -bY/X is a regular fraction.. it's undefined unless y = 0

but since there is an exponential if y =0 and x = 0 we will only have the first equation satisfied, but the second equation will be 0=1 which is not true. 
Maybe taking the natural log of the second equation will work but ln 0 is also undefined so we can't do that. The same result happens when you take a natural log of a negative number...undefined.

usukidoll · Answer

these two equations are actually difference equations

$$Y_{t+1}=X_t, x_{t+1}=e^{\frac{-bY_t}{x_t}}$$

usukidoll · Answer

subsitution values were given from this http://i.stack.imgur.com/MaKfP.jpg

usukidoll · Answer

I am on 2d. I know what 2c is already... I just need to find the equilibrium of those equations.

usukidoll · Answer

question should be does the equilibrium really exist with this? but there has to be one... otherwise what's the point in asking 2e if there's no equilibrium which is what x and y values make the equation 0 besides (0,0) which isn't working here

usukidoll · Answer

@ganeshie8