Section 1.3 Integration

Transcription

1 Section 1.3 Integration Key terms: Integral Constant of integration Fundamental theorem of calculus First order DE One parameter family of solutions General solution Initial value problem Particular solution Motion of a ball

2 Integrations play an important role in determining a closed form (a formula) for the solution of first order DEs. What is the integral? There are several ways to answer. 1. Area under the graph of a function. The definite integral b f(x)dx is interpreted a as the area under the graph of the function f between x = a and x = b. It represents the area of the shaded region in 2. The antiderivative. This answer emphasizes the indefinite integral. If the function g is continuous, then The C refers to the arbitrary constant of integration. Thus the process of indefinite integration involves finding antiderivatives. Given a function g, we want to find a function f such that f ꞌ = g.

3 The connection between area and antiderivative is the Fundamental Theorem of Calculus. This says that if f ꞌ = g, then 3. Tables of integration formulas. Using such tables is handy, but it does not lead to any deep understanding of the integral. In our course it is finding antiderivatives and using tables (when needed) that is most important.

4 Solution of DEs by Integration. The solution of an important class of differential equations amounts to finding antiderivatives. A first-order differential equation can be written as y' = f(t,y) where the right-hand side is a function of the independent variable t and the unknown function y. Suppose that the right-hand side is a function only of t and does not depend on y. Then equation the previous equation becomes y' = f(t) In this case we see immediately that the solution is y(t) = f(t)dt

5 Example: Solve the DE y ꞌ = 2t. The solution is y(t) = t 2 + C. Graph several solution for different values of C MATLAB code: C =[-2:1:2]; t=[-2:0.01:2]; y=t.^2;figure, for jj=1:length(c), yy=y+c(jj);plot(t,yy),hold on, end,hold off, addaxes t 2 + C Not distributed with MATLAB C = C = 1 We say y(t) = t 2 + C is a one parameter family of solutions. They are defined on (-, ). WHY? C = 0 C = -1 This an example of a general solution. These curves are vertical translates of one another. -1 C = This is always the case for solution curves of a DE of the form y ꞌ = f (t).

6 The constant of integration allows us to put an extra condition on a solution. This is illustrated in the next example. Example: Find the solution of yꞌ(t) = tsin(t) that satisfies y(π/2) = 3. This is an example of an initial value problem (IVP). It requires a particular solution that satisfies the initial condition y(π/2) = 3. The general solution is given by y(t) = t sin(t) dt = sin(t) - t cos(t) + C So we have a one parameter family of solutions. These are defined for all t in (-, ). The initial condition y(π/2) = 3 is used to find a particular value of C. Use integration by parts or a table of integrals. y(π/2) = sin(π/2) - π/2cos(π/2) + C= 1 + C = 3, so C = 2 So the particular solution of the IVP is y(t) =sin(t) + tcos(t) + 2.

7 Solution of the IVP y(t) = sin(t) - t cos(t) + C Point (pi/2, 3) MATLAB code: C =[-1:1:3];t=[-pi:0.01:pi]; y=sin(t) - t.*cos(t); figure, for jj=1:length(c), yy=y+c(jj);plot(t,yy), hold on,end,hold off C = -1 The particular solution curve passes through the point (π/2, 3) Warning: the domain of the solution will not always be (-, ). Example: Solve IVP y ꞌ = 1/x, y(1) = 4. the general solution is y(x) = ln( x ) + C. To satisfy the initial condition the point (1, 4) must lie on the solution curve. Since y(x) is not defined at x = 0 either x < 0 or x > 0. Because of the initial condition x > 0. A bit of work shows C = 4 so the solution of the IVP is y(x) = ln(x) + 4, with x > 0.

8 Example: Motion of a ball. Previously we showed the DE is where x(t) is the height of the ball above the surface of the earth and g is the acceleration due to gravity. If we measure x in feet and time in seconds, g = 32 ft/s 2. Suppose a ball is thrown into the air with initial velocity v 0 = 20 ft/s. Assuming the ball is thrown from a height of x 0 = 6 feet, how long does it take for the ball to hit the ground? First we introduce the velocity to reduce the second-order equation to a system of two first-order equations: Solving the second equation by integration, we get Evaluating this at t = 0, we see that the constant of integration is C 1 = v(0) = v 0 = 20, the initial velocity. Hence, the velocity is v(t) = gt + v 0 = 32t + 20, and the first DE becomes Solving by integration, we get Once more we evaluate this at t = 0 using that C 2 = x(0) = x 0 = 6, the initial elevation of the ball. Hence, our final solution is Set x(t) = 0 and solve for t to find that t = 1.5 sec.