Calculus-Sec-8-4-Solution

8.4   Differential Equations                         by SGLee - HSKim, JYLee , JHLee

1. Find the solutions of the given initial-value problem

The given equation is separable, and therefore, it can be rewritten as

Using the initial condition, we become to know that , and therefore,

2. Find the orthogonal trajectories of the family of the following curves: .

Since , the orthogonal trajectories must have their derivative perpendicular to

the given family of curves, namely

(1)

Note that the given curves never touch x-axis, which means that ,

and so the above differential equation (1) makes sense. Now noting that  is of separable type, we get

.

3. Solve the differential equation .

Separation of variables gives  By the integration, we have  .

So, by simplification, we obtain

4. Solve the differential equation  that satisfies the initial condition .

We first note that the given one is a separable differential equation.

Therefore,  after splitting it into variables and using the integration in each variable, we  have

Since , we have . Summing up, we obtain

5. A tank contains 1000L of brine with 15kg of dissolved salt.

Brine that contains 0.07kg of salt per liter of water enters the tank at a rate of

and another brine that contains 0.02kg of salt per liter of water enters the tank at a rate of .

The solution is kept thoroughly mixed and drains from the tank at a rate of .

How much salt is in the tank after  tends to ?

Let  be the amount of salt (in kilograms) after  minutes.

The rate at which enters the tank is

The tank always contains  of liquid and so the rate at which leaves the tank

is

Thus, the differential equation is    <-- 0.1-y/40=(4-y)/40 으로 고쳐야 함

Solving this separable differential equation, we obtain

Since , we have . As  tends to  converges to .

6. Let  and  be the number of preys and predators at time , respectively.

We suppose that these two species satisfy the following system of differential equations:

.

(a) Find all equilibrium solutions and the expression for .

(b)Solve  with given initial conditions  and  by the method of separable variables.

(a) The equilibrium are steady state solutions, and so they must satisfy

Using  the chain rule, we obtain

(1)

(b) Note that the initial condition  when  and by solving the  equation (1),

we get

It  remains to specify the constant . Indeed, due to initial condition,

we can  see

Hence, the solution satisfies

7. The half-life of cesium-137 is 30 years. Suppose that we have a 100mg sample.

After how long will only 1mg remain?

Let  be the mass of cesium-137 (in milligrams) that remains after  years.

Then  and , so we have .

In order to  determine the value of , we use the fact that .

Thus we get

We want to find the value of  such that .

Solving the following equation for  and taking the natural  logarithm of both sides, we have

8. Let  be a positive constant. We consider the following differential equation

.                                       (1)

For what value of  does the solution of (1) exist for all time (global existence or blow up in a finite time)?

Since the equation is separable, we get .

We consider first the case  . Keeping in mind that , we obtain

In this  case, a solution exists globally for all time. Next we consider the case

Since  and , we obtain , which again exists globally.

Finally, in case that , as in the previous case of , we have

.

The  above solution is well-defined as long as  but its limit, goes  to  as  approaches .

Therefore, the solution blows up in a finite time  for the case .

9. Solve the differential equation , with the initial condition  and

determine the interval in which the solution exists.

(Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent)

(You may do it with Sage at http://math1.skku.ac.kr/. Open resources in http://math1.skku.ac.kr/pub/)

Since   and , we obtain the following integral curve.

Let us enlarge it.

The integral curve has a vertical tangent around . Therefore, the desired interval is .

Now, we can consider another solution.

Since  and , we obtain .

To find the interval of definition, look for points where the integral curve has a vertical tangent.

The integral curve has a vertical tangent around . Therefore,  the desired interval is .

10. Solve the differential equation , with the initial condition .

Hence plot the solution curve. Also plot the solution curves for a range of initial conditions.

Since   and , we obtain the following integral curve.

The required solution is given by .

We plot the solutions curves for a range of initial values.