Ordinary Differential Equations

Transcription

1 Ordinary Differential Equations Swaroop Nandan Bora Department of Mathematics Indian Institute of Technology Guwahati Guwahati

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3 A first-order differential equation is an equation in which dy dx = f(x,y) f(x,y) is a function of two variables defined on a region in the xy-plane. A solution of the given equation is a differentiable function defined on an interval I of x-values (perhaps infinite) such that on that interval. d y(x) = f(x,y(x)) dx The general solution to a first-order differential equation is a solution that contains all possible solutions.

4 The general solution always contains an arbitrary constant but having this property doesnt mean a solution is the general solution. Establishing that a solution is the general solution may require deeper results from the theory of differential equations.

5 We often need a particular rather than the general solution to a first-order differential equation dy dx = f(x,y). The particular solution satisfying the initial condition y(x 0 ) = y 0 is the solution whose value is y 0 when x = x 0. Thus the graph of the particular solution passes through the point (x 0,y 0 ) in the xy-plane. A first-order initial-value problem is a differential equation dy dx = f(x,y) whose solution must satisfy an initial condition y(x 0) = y 0.

6 Every differential equation need not have a solution. Even if solution exists, that may not be unique. Picard s theorem: Suppose f(x,y) and f are continuous on the interior of a rectangle R, and that y (x 0,y 0 ) is an interior point of R, then the IVP dy dx = f(x,y), y(x 0) = y 0 has unique solution y = y(x) for x in some open interval containing x 0.

7 Applications Orthogonal trajectories Let F(x,y,c) = 0 be a given one-parameter family of curves in the xy-plane. A curve that intersects the curves of the above family at right angles is called an orthogonal trajectory of the given family. Main point: Suppose that the differential equation of above family can be obtained as dy dx = f(x,y). Then the orthogonal trajectory must have a slope 1/(f(x,y)).

8 Examples Example 1 Given a family of circles: x 2 +y 2 = a 2. Its differential equation is obtained as dy dx = x y Differential equation of the orthogonal trajectory Giving the orthogonal trajectory as dy dx = y x y = kx, all are straight lines passing through the origin.

9 Example 2 Given a family of parabolas: y = cx 2. The differential equation is dy dx = 2y x. The differential equation of the orthogonal trajectory which gives us dy dx = x 2y. x 2 +2y 2 = k 2.

10 Problems in Mechanics Newton s 2nd law of motion m dv dt = F. Example A body of weight 8 pounds falls from rest toward the earth from a great height. As it falls, air resistance acts upon it which is numerically equal to 2v, where v is the velocity in feet per second. To find the velocity and distance covered in time t. Differential equation is g = 32,w = 8,m = 8/32 = 1/4 m dv dt = F 1 +F 2. 1dv 4 dt = 8 2v. Since the body falls from rest, initial condition is v(0) = 0.

11 Example 2 Solving it and using the initial condition v = 4(1 e 8t ) For finding the distance, the differential equation is Solving and using the initial condition dx dt = 4(1 e 8t ) x = 4(t+ 1 8 e 8t 1 8 )

12 Interpretation of results The solution for v tells us that as t, then v reaches the limiting velocity 4 (ft/sec). Again the solution for x tells us that as t, then x also. Does this imply that the body will pierce through the earth and continue forever?? Of course not. Because when the body reaches the earth s surface, the above differential equations no longer apply.

13 Rate of growth and decay Example The rate at which a radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. Half of the original number of radioactive nuclei have undergone disintegration in a period of 1500 years. What percentage of the original radioactive nuclei will remain after 4500 years? Formulation: Let x be the amount of radioactive nuclei present after t years. The differential equation is dx dt = kx. The initial condition is, given x 0 is the amount present initially, x(0) = x 0. But according to the given condition x(1500) = 1 2 x 0.

14 Rate of growth and decay Solution: The solution is: x = ce kt. By using the condition of half-life e k = ( 1 2 )1/1500. Therefore the solution is x = x 0 [( 1 2 )1/1500 ] t. For (i) x = x 0 ( 1 2 )3 = 1 8 x 0. that is, 12.5 percent of original number remain after 4500 years.

15 Population growth Growth of population, such as human, animal species, bacteria colony. Assumptions: The population increase is approximately continuous. In fact this increase is also a differentiable function of time. Given a population, we let x be the number of individuals at time t. Assuming that the rate of change of population is proportional to the number of individuals, we get the differential equation dx dt = kx. If the number of individuals was x 0 at time t 0, we have the initial condition: x(t 0 ) = x 0.

16 Population growth The solution is given by x(t) = x 0 e k (t t 0 ). The population governed by this model (increasing exponentially with time) is known as Malthusian law. But more realistically in many cases the number of individuals x at time t is described by a differential equation dx dt = kx λx2. The additional term λx 2 is the result of some cause that tends to limit the ultimate growth of the population.

17 Mixture problem A substance S is allowed to flow into a certain mixture in a container at a certain rate. The mixture is kept uniform by stirring. Further, in one such situation, this uniform mixture simultaneously flows out of the container at another (generally different rate. Letting x denote the amount of S present at time t, the derivative dx/dt denotes the rate of change of x with respect to t. If IN denotes the rate at which S enters the mixture and OUT denotes the rate at which it leaves, Then we have the differential equation dx dt = IN OUT.

18 Example A tank initially contains 50 gallons of pure water. Starting at time t = 0, a brine containing 2 lb of dissolved salt per gallon flows into the tank at the rate of 3 gal/min. The mixture is kept uniform by stirring and the well-mixed mixture simultaneously flows out of the tank at the same rate. To find the amount of salt at time t. Formulation: dx = IN OUT. dt IN= (2 lb/gal)(3 gal/min)= 6 lb/min The initial value problem is ( x ) OUT = 50 lb/gal (3gal/min) = 3x 50 lb/min. dx dt 3x = 6, x(0) = 0. 50

19 Mixture problem Solution x = 100+ce 3t/50. Using the initial condition x = 100(1 e 3t/50 ). How much salt is present at the end of 25 minutes? x(25) = 100(1 e 1.5 ) 78lb. What happens when t?