Integral Curve (generic name for a member of the collection of known as the general solution)

Transcription

1 Section 1.2 Solutions of Some Differential Equations Key Terms/Ideas: Phase Line (This topic is not in this section of the book.) Classification of Equilibrium Solutions: Source, Sink, Node SPECIAL CASE: First order autonomous DEs of the form Initial condition: additional information y(t 0 ) = y 0, a number Independent variable t does not appear explicitly. Initial Value Problem: (abbreviated IVP) dy DE : f(t,y), dt with initial condition y(t ) y (some numerical value) 0 0 General Solution (formula representing all solutions of a DE) Integral Curve (generic name for a member of the collection of known as the general solution)

2 First order autonomous DEs and the phase line: Recall that an autonomous ODE has the form so the expression for the derivative does not explicitly involve the independent variable t. Equilibrium solutions are constant functions, y = constant, that make the derivative equal to zero. To find the equilibrium solutions we set f(y) = 0 and solve for y. Examples: y' = y(1 y) Equilibrium solutions y = 0, y = 1 y' = y 2 4y + 3 Equilibrium solutions y = 1, y = 3 y' = 4y y 3 Equilibrium solutions y = 0 y = 2, y = -2 The equilibrium solutions divide the direction field into horizontal strips. The graphs of the equilibrium solution will be horizontal lines which form the boundary of the strips.

3 y' = y 2 4y + 3 Equilibrium solutions y = 1, y = 3 Observe that in each strip the slope lines are rising or falling. (At the equilibrium solutions the slope lines are horizontal.)

4 When we can t easily draw the direction field we can analyze the behaviors in the strips using what is known as a phase line. A phase line is a vertical bar divided into strips using the equilibrium solutions. For y' = y 2 4y + 3 we have Equilibrium solutions y = 1, y = 3 To illustrate the same behavior as the slope lines in the direction field we use arrows (up or down) in the strips. y' = y 2 4y + 3 = (y 1)(y 3) Just choose a value in each strip, substitute into the expression for the derivative and determine the SIGN. If positive put in the strip and if negative put in the strip. Use y = 4, then y' > 0 so we use. Use y = 2, then y' < 0 so we use. Use y = 0, then y' > 0 so we use. Warning: The phase line is not discussed in this section.

5 Phase line. Direction field.

6 Example: Draw the phase line for ODE y' = y(y 2 4.) Find the equilibrium solutions: y(y 2 4 )= y(y 2)(y + 2) equilibrium solutions y = 0, y = 2 and y = -2. Use y = 3, the y' > 0 so we use. Use y = 1, the y' < 0 so we use. Use y = -1, the y' > 0 so we use. Use y = -4, the y' < 0 so we use. Compare the phase line to the direction field.

7 Example: Draw the phase line for ODE y' = y(y 2 4.) Find the equilibrium solutions: y(y 2 4 )= y(y 2)(y + 2) equilibrium solutions y = 0, y = 2 and y = -2. Use y = 3, the y' > 0 so we use. Use y = 1, the y' < 0 so we use. Use y = -1, the y' > 0 so we use. Use y = -4, the y' < 0 so we use. Compare the phase line to the direction field.

8 Classification of Equilibrium Solutions The equilibrium solutions divide the direction field into horizontal strips. The graphs of the equilibrium solution will be horizontal lines which form the boundary of the strips. The behavior of the slope lines on opposite sides of the horizontal line that represents the equilibrium solution is described as follows: Sink: An equilibrium solution for which all nearby solutions tend toward this solution. Source: An equilibrium solution for which all nearby solutions tend away from this solution. Node: An equilibrium solution that has the property that it is neither a sink nor a source.

9 The behaviors around an equilibrium solution are illustrated in the following direction fields.

10 Solving some autonomous ODEs: the object is to obtain a formula for the general solution. Here we will develop an expression for the solution of an autonomous DE of the form a 0 The solution technique is to rearrange terms so that we can perform an integration. Basically we separate the variables as shown next. Next integrate both sides: Solve for y = y(t): at ay - b = ±Ce (Rename ±C to be C and solve for y.)

11 This is called the general solution of the DE and it represents the set of all possible solutions to the DE. The geometrical representation of the general solution is an infinite family of curves called integral curves. Frequently, we want to focus our attention on a single member of the infinite family of solutions by specifying the value of the arbitrary constant. Most often, we do this indirectly by specifying instead a point that must lie on the graph of the solution. This is called specifying an initial condition. For example we require the solution of the DE so that y(0) = y 0 The DE with the initial condition is called an initial value problem (IVP): In general, we can specify the initial condition at any value of t in the domain of y(t) for an initial condition.

12 Radioactive Waste Radioactive (or nuclear) waste is a byproduct from nuclear reactors, fuel processing plants, hospitals and research facilities. Radioactive waste is also generated while decommissioning and dismantling nuclear reactors and other nuclear facilities. There are two broad classifications: high-level or low-level waste. High-level waste is primarily spent fuel removed from reactors after producing electricity. Low-level waste comes from reactor operations and from medical, academic, industrial, and other commercial uses of radioactive materials. cesium-137, is a radioactive isotope of caesium which is formed as one of the more common fission products by the nuclear fission of uranium-235 and other fissionable isotopes in nuclear reactors and nuclear weapons. Cesium-137 has a number of practical uses. In small amounts, it is used to calibrate radiation-detection equipment; in medicine, it is used in radiation therapy. In industry, it is used in flow meters, thickness gauges, moisture-density gauges, etc. Radioactive material decays at a rate proportional to the amount present. The DE for k > 0 is dq(t) dt kq

13 If 120 mg of cesium-137 decays to 107 mg in 5 years determine the rate constant k and an expression for Q(t), the amount of cesium present at time t. Separating the variables we have dq(t) dq(t) dt kq Integrating both sides we have ln Q(t) k t C Solving for Q we get So we have Q(t) Ce kt Q k t C k t C k t Q(t) e e e Ce kdt Rename e C to be C, an arbitrary constant. At the beginning (t = 0), Q(0) = 120 so here C = 120; Q(t) 120e kt To find rate constant k we set t = 5 and Q(5) = 107. Then we have Q( 5) e k Using algebra we get ln 120 k t Thus Q(t) 120e The half-life of a radio active material is the time required for an amount of the material to decay to one-half the original amount. Show that the halflife of cesium-137 is about 30.2 years.

14 Mice and Owls Model Our Mice and Owls model has an initial value problem (IVP). Initial condition Rewrite the equation a bit and then separate the variables: p(0) = 600 Separate to get Now integrate both sides: Remember to put + C Solving for p we get Apply the initial condition: Rename the arbitrary constant. The general solution. Set t = 0 and p = 600 and solve for C. 600 = C C = -300 Solution of the IVP is p = e 0.5t How do we find the time (in months) that the population of mice becomes extinct?

15 Falling Body Model Our falling body model has an initial value problem (IVP) Rewrite the DE as dv 1-1 = 49 - v = v - 49 dt 5 5 dv v = dt 5 v(0) = 0 Separating the variables we have Integrating we get Solving for v we get We omitted some algebra steps to get this expression. Applying the initial condition we have So the solution of the IVP is This gives the velocity of the falling object at any positive time. What is the limiting (terminal) velocity (the fastest the body could travel)? How do you determine the time at which the body is 80% of its limiting velocity?

16 To find the velocity of the object when it hits the ground we need to know the time of impact. Let x(t) be the distance that the object has fallen at time t. Suppose the object is dropped from a height of 300 meters then we have another DE, with initial condition x(0) = 0 m. Separating the variables gives us and integrating we get Applying the initial condition gives So Next set x = 300 and solve for t, the time it hits the ground.

17 To solve for t we need a numerical method. Note that t appears in two places. It is not possible in this case to algebraically solve for t. It can be shown that in this case the time the object hits the ground is approximately sec. The velocity of impact can be computed from Convert the meters per second to miles per hour. 1 meter per second = miles per hour So the impact velocity is about mi/hr.

18 Newton s Law of Cooling Suppose that a building loses heat in accordance with Newton's law of cooling and that the rate constant k has the value per hour. Assume that the interior temperature is 68 F at the time the heating system fails. If the external temperature is 15 F, how long will it take for the interior temperature to fall to 40 F? Newton s Law of Cooling says that the temperature of an object changes at a rate proportional to the difference between the temperature of the object itself and the temperature of its surroundings (the ambient temperature). Let Q(t) be the temperature of the object at time t. The ambient temperature is 15 F and the rate constant is per hour (since we are cooling). We get differential equation dq = 0.19(Q - 15), Q(0) = 68 dt The solution of this IVP is Q(t) = e -0.19t. (Perform separation of variables) How do you determine the time (in hours) it takes for the interior temperature to reach 40 F?

19 dq Details on solving = 0.19(Q - 15), Q(0) = 68 dt dq Separate the variables: = -0.19dt Q - 15 Integrate both sides: Solve for Q: dq = -0.19dt Q - 15 ln Q - 15 = -0.19t +C -0.19t+C -0.19t C -0.19t Q = 15 + e = 15 + e e = 15 +Ce Next apply initial condition Q(0) = 68 to find C. We renamed e C to be C. 68 = 15 +C C = 53 So we have Q(t) = e -0.19t