# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

**Download**(direct link)

**:**

**179**> 180 181 182 183 184 185 .. 609 >> Next

The graph of y = ô(³) is shown in Figure 6.4.3. Observe that it has the qualitative form that we indicated earlier. To find the amplitude of the eventual steady oscillation it is sufficient to locate one of the maximum or minimum points for t > 10. Setting the derivative of the solution (23) equal to zero, we find that the first maximum is located approximately at (10.642, 0.2979), so the amplitude ofthe oscillation is approximately 0.0479.

FIGURE 6.4.3 Solution of the initial value problem (16), (17), (18).

Note that in this example the forcing function g is continuous but g is discontinuous at t = 5 and t = 10. It follows that the solution ô and its first two derivatives are continuous everywhere, but ô"' has discontinuities at t = 5 and at t = 10 that match the discontinuities in g at those points.

PROBLEMS In each of Problems 1 through 13 find the solution of the given initial value problem. Draw the graphs of the solution and of the forcing function; explain how they are related.

> 1 y' + y = f (0; y<0) = y<0) = l; f(l) = j ^ nj° IJ <lj2

> 2. / + 2y + 2y = h(,y. y(0) = 0. Ó(0) = 1; HO =1 J I/^fand t , 2n

Chapter 6. The Laplace Transform

> 3. y" + 4y = sin t - u2n(t) sin(t - 2n); y(0) = 0, Ó(0) = 0

> 4. Ó + 4y = sin t + un(t) sin(t - n); y(0) = 0, Ó(0) = 0

> 5. / + 3Ó + 2y = f (t); y(0) = 0, /(0) = 0; f (t) ={ 0, 0 J f0< 10

> 6. y" + 3y + 2y = u2(t); y(0) = 0, /(0) = 1

> 7. y' + y = u3n(t); y(0) = 1, y(0) = 0

> 8. y'+ y + 5y = t - un/2(t)(t - n/2); y(0) = 0, /(0) = 0

> 9. Ó + y = g(t); y(0) = 0, y(0) = 1; g(t) = j 32, 0° J ^ < 6

> 10. Ó + y+ 4y = g(t); y(0) = 0, y(0) = 0; g(t) ^ 0“ ^ 0 J Ï < ^

> 11. Ó + 4y = u„(t) - U3n(t); y(0) = 0, j/(0) = 0

> 12. yv - y = U1(t) - u2(t); y(0) = 0, Ó (0) = 0, /(0) = 0, /'(0) = 0

> 13. Ó + 5Ó + 4y = 1 - un(t); y(0) = 0, /(0) = 0, /(0) = 0, /"(0) = 0

14. Find an expression involving uc(t) for a function f that ramps up from zero at t = t0 to

the value h at t = t0 + k.

15. Find an expression involving uc(t) for a function gthat ramps up from zero at t = t0 to the value h at t = t0 + k and then ramps back down to zero at t = t0 + 2k.

> 16. A certain spring-mass system satisfies the initial value problem

u" + 1 u + u = kg(t), u(0) = 0, u'(0) = 0,

where g(t) = u3/2(t) - u5/2(t) and k > 0 is a parameter.

(a) Sketch the graph of g(t). Observe that it is a pulse of unit magnitude extending over one time unit.

(b) Solve the initial value problem.

(c) Plot the solution for k = 1/2, k = 1, and k = 2. Describe the principal features of the solution and how they depend on k.

(d) Find, to two decimal places, the smallest value of k for which the solution u(t) reaches the value 2.

(e) Suppose k = 2. Find the time ò after which | u(t) | < 0.1 for all t > ò.

> 17. Modify the problem in Example 2 in the text by replacing the given forcing function g(t)

by

f (t) = lus(t)(t - 5) - u5+k(t)(t - 5 - k)]/k-

(a) Sketch the graph of f (t) and describe how it depends on k. For what value of k is f (t) identical to g(t) in the example?

(b) Solve the initial value problem

Ó + 4y = f (t), y(0) = 0, y(0) = 0.

(c) The solution in part (b) depends on k, but for sufficiently large t the solution is always

a simple harmonic oscillation about y = 1/4. Try to decide how the amplitude of this

eventual oscillation depends on k. Then confirm your conclusion by plotting the solution for a few different values of k.

> 18. Consider the initial value problem

fk (t), y(0) = 0, y (0) = 0,

4 - k < t < 4 + k 0 < t < 4 - k and t > 4 + k

Ó + 3 y + 4y =

where

k )=| 0/2k

and 0 < k < 4.

6.4 Differential Equations with Discontinuous Forcing Functions

323

(a) Sketch the graph of fk (t). Observe that the area under the graph is independent of k. If fk(t) represents a force, this means that the product of the magnitude of the force and the time interval during which it acts does not depend on k.

(b) Write fk(t) in terms of the unit step function and then solve the given initial value problem.

**179**> 180 181 182 183 184 185 .. 609 >> Next