AMS 361: Applied Calculus IV
(DE & BVP)
Outline for Lecture 4
4.
Separable equations and applications
The following equation is called Separable equation
One can write it as
So, this equation can be integrated on both sides. Although we probably most of the time to get the y=y(x) form, we can get the solution in implicit form.
is the simple (but could be implicit) form of the IVP
e.g. 1. Solve
The solution is y(x)=7*e^(-3x^2)
What if y(0) = -4? We can get the solution of y(x)=-4*e^(-3x^2).
e.g. 2. Solve
The solution is (y^3-5y)=4x-x^2+C
It’s not possible, not practical, and not necessary to find the explicit form.
If we implement IC, y(1) = 3, then we can get C=9.
Thus, the solution becomes (y^3-5y)=4x-x^2+9.
DEF:
General solution: Solution with an arbitrary constant
Particular solution: Solution with without the arbitrary constant
Singular solution: Solution that can’t be obtained by setting the arbitrary constant. This is possible for nonlinear DE.
e.g.
We can get the following as
Now, if at t=0, T(0)=50, A=375, we can get
Now, we can play a look games with this equation:
For example, at t=75, we now T=125 è k=0.0035, we can find the time needed to get to T=150.
The answer is t=105.
What’s time needed to reach 375
e.g.3. Revisit of Torricelli’s
Law of Draining Sink: The draining rate of volume of water in a draining
sink is proportional to the square root of height of water in the sink:
where V is the volume, y is the height, while k is a positive constant.
In fact, we can “derive” this equation from lower-level
“principles”. First, assuming the hole area “a” and the
water “drop” leaking out of the hole at a speed of 
Which is obtained by
remembering y = ½ g t^2 and v = g t.
Thus the volume rate of water leak is
Second, assuming the container is NOT a cylinder and the cross-section area A is a function of height y, i.e., A(y),
We an write the volume as
Therefore,
Finally, we then have
For the special case of a bow initially filled with liquid, we can compute
Thus, we can formulate one eq
Solving this equation by separation of variables, we get
Remarks:
(1)
When t=0, we have y=R;
(2)
To empty the bow, we reach y=0, so the time is 
Obviously, (2a) the bigger the R, the longer the time to empty; (2b) the
bigger the “k”, the shorter the time to empty.
One numerical example:
We now use the separable equations technique to solve this equation... (Example 8 on page 39)
According to figure 1.4.8 (on page 40), a hemisphere bowl of top radius (a=4) initially is full of water, compute the time needed to empty the bow. The draining hole’s diameter is 1 inch.
A(y) = Pi r^2 = Pi * [4^2 – (4-y)^2] = Pi*[8y-y^2]
The gravity constant g = 32 ft/s^2. Now, let’s solve the equation,
Pi*[8y-y^2] dy/dt = -Pi*(1/24)^2 sqrt(2*32*y)
[16/3] y^(3/2) – [2/5] y^(5/2) = -(1/72) t +C
Initial condition: y(t=0) = 4 ft, then we can determine C=448/115. Thus, when we set y(t) = 0, we can find the time needed to empty the bowl is t = 2150 seconds.
There are a lot of concepts including
First order, nonlinear, separable,
homogenous, Bernoulli, etc…