# Which of these equations have an equilibrium solution. What are they? How are their DE's different?

dy/dx=6y-y^2

dy/dx 15-3y

## To determine which of these equations have equilibrium solutions, we need to set the derivatives equal to zero and solve for y.

1. dy/dx = 6y - y^2

Setting dy/dx equal to zero:

0 = 6y - y^2

Rearranging the equation:

y^2 - 6y = 0

Factoring out y:

y(y - 6) = 0

We can see that there are two possible equilibrium solutions: y = 0 and y = 6.

2. dy/dx = 15 - 3y

Setting dy/dx equal to zero:

0 = 15 - 3y

Rearranging the equation:

3y = 15

Dividing by 3:

y = 5

We find that the only equilibrium solution for this equation is y = 5.

The differential equations (DE) for the two equations are different. The first equation, dy/dx = 6y - y^2, is a non-linear differential equation because of the presence of the y^2 term. The second equation, dy/dx = 15 - 3y, is a linear differential equation because the y term is not squared or multiplied by any other function of y.