8.4 Differential equations

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. Solve the differential equation .

 Separation of variables gives By the integration, we have So, by simplification, we obtain


3. 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

 .


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 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 .

 The equilibrium are steady state solutions, and so they must satisfy  Using  the chain rule, we obtain

             (1)

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

 Note that the initial condition when and by solving the  equation (13), we get It  remains to specif   y 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)? Verify your answer.

 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)

http://matrix.skku.ac.kr/cal-lab/cal-8-4-9.html 

 Since   and , we obtain the following integral curve.


 

f = lambda x,y: y^2-2*y-2*x^3-4*x^2-2*x-3

contour_plot(f, (-5, 5), (-5, 5), fill=0, contours=20,plot_points=200)

 

 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 .