y = -2 + Ce^{x}
Substituting in the initial value, The solution isy = -2 + 5e^{x-2}
x_{h} = e^{4t}
is a homogeneous solution.
Solving gives A = 1 and B = -1/2. Therefore
The solutions will be
for two different constants C_{1} and C_{2}.
Their difference is
x_{2}(t)-x_{1}(t) = (C_{2}-C_{1})x_{h}(t) = (C_{2}-C_{1})e^{4t}
Since the exponential gets large as t gets large, the solutions get further apart.
Define:
But initially, V = 400 and the rate of leakage is 4, so k = 0.01.
Amount added is .
Amount lost is .
Change over is
The differential equation is Including the initial value, the model is
The free response solves
V' + 0.01V = 0; V(0) = 400
It is the amount of water there would be if you started with 400 liters, added none, and allowed the water to leak out as before.The forced response solves
V' + 0.01V = 3; V(0) = 400
It is the amount of water there would be if you started with an empty tank, added water as in the original problem, and allowed water to leak out as in the original problem.
If u>2, then -2-u+u^{2}>0.
If -1<u<2, then -2-u+u^{2}<0.
If u<-1, then -2-u+u^{2}>0.
The solution is increasing if u>2 or u<-1. It is decreasing if -1<u<2.
The equilibrium at -1 is stable. The equilibrium at 2 is unstable.