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One can estimate the coordinates of the maximum point from the graph, but to find them more precisely we seek the point where the solution has a horizontal tangent line. By differentiating the solution (30), y = 9e—2t — 7e—3t, with respect to t we obtain
y' = -18e~2t + 21e—3t. (34)
Setting y' equal to zero and multiplying by e3t, we find that the critical value tc satisfies et = 7/6; hence
tc = ln(7/6) = 0.15415. (35)
The corresponding maximum value yM is given by
2t 3t 108
yM = 9e c — 7e c = _ = 2.20408. (36)
In this example the initial slope is 3, but the solution of the given differential equation behaves in a similar way for any other positive initial slope. In Problem 26 you are asked to determine how the coordinates of the maximum point depend on the initial slope.
Returning to the equation ay" + by' + cy = 0 with arbitrary coefficients, recall that when r1 = r2, its general solution (17) is the sum of two exponential functions. Therefore the solution has a relatively simple geometrical behavior: as t increases, the magnitude of the solution either tends to zero (when both exponents are negative) or else grows rapidly (when at least one exponent is positive). These two cases are illustrated by the solutions of Examples 2 and 3, which are shown in Figures 3.1.1 and 3.1.2, respectively. There is also a third case that occurs less often; the solution approaches a constant when one exponent is zero and the other is negative.
Chapter 3. Second Order Linear Equations
PROBLEMS In each of Problems 1 through 8 find the general solution of the given differential equation.
7. y" — 9y' + 9y = 0
1. y" + 2y; — 3y = 0
3. 6y" — y ' — y = 0
5. y" + 5y' = 0
H O,/ I o,. ___ n
2. y" + 3y; + 2y = 0
4. 2y" — 3y' + y = 0
6. 4y" — 9y = 0
8. y" — 2y' — 2y = 0
In each of Problems 9 through 16 find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior as t increases.
9. y"+ y' — 2y = 0, y(0) = 1, y'(0) = 1
10. y" + 4y' + 3y = 0, y(0) = 2, y'(0) = -1
11. 6y" — 5y'+ y = 0, y(0) = 4, y'(0) = 0
12. y" + 3y'= 0, y(0) = -2, y'(0) = 3
13. y" + 5y' + 3y = 0, y(0) = 1, y'(0) = 0
14. 2y" + y' — 4y = 0, y(0) = 0, y'(0) = 1
15. y" + 8y' — 9y = 0, y(1) = 1, y'(1) = 0
16. 4y" — y = 0, y(—2) = 1, y'(—2) = -1
17. Find a differential equation whose general solution is y = c1e2t + c2e—3t.
18. Find a differential equation whose general solution is y = c1e—t/2 + c2e—2t.
? 19. Find the solution of the initial value problem
Then determine the maximum value of the solution and also find the point where the solution is zero.
21. Solve the initial value problem y" — y' — 2y = 0, y (0) = a, y'(0) = 2. Thenfind a so
that the solution approaches zero as t ^ ?.
22. Solve the initial value problem 4y" — y = 0, y (0) = 2, y'(0) = f. Then find fi so that the solution approaches zero as t ^?.
In each of Problems 23 and 24 determine the values of a, if any, for which all solutions tend to zero as t ^ ?; also determine the values of a, if any, for which all (nonzero) solutions become unbounded as t ^?.
23. y" — (2a — 1)y' + a(a — 1)y = 0 24. y" + (3 — a)y' — 2(a — 1)y = 0
? 25. Consider the initial value problem
where f > 0.
(a) Solve the initial value problem.
(b) Plot the solution when f = 1. Find the coordinates (t0, y0) of the minimum point of the solution in this case.
(c) Find the smallest value of f for which the solution has no minimum point.
? 26. Consider the initial value problem (see Example 4)
y"— y = 0, y(0) = 5, y'(0) = -|.
Plot the solution for 0 < t < 2 and determine its minimum value.
20. Find the solution of the initial value problem
2y" — 3y'+ y = 0, y(0) = 2, y'(0) = 2.
2y" + 3 y' — 2y = 0, y(0) = 1, y'(0) = -?,
y" + 5y' + ^y = 0, y(0) = 2, y' (0) = ?,
where ? > 0.
(a) Solve the initial value problem.
3.2 Fundamental Solutions of Linear Homogeneous Equations
(b) Determine the coordinates t and y of the maximum point of the solution as functions
v j m s m *
(c) Determine the smallest value of P for which ym > 4.
(d) Determine the behavior of t and y as P -> oo.
27. Find an equation of the form ay" + by' + cy = 0 for which all solutions approach a multiple of e— as t ^x.
Equations with the Dependent Variable Missing. For a second order differential equation of the form y" = f (t, y'), the substitution v = y', v' = y" leads to a first order equation of the form v' = f (t, v). If this equation can be solved for v, then y can be obtained by integrating dy/dt = v. Note that one arbitrary constant is obtained in solving the first order equation for v, and a second is introduced in the integration for y. In each of Problems 28 through 33 use this substitution to solve the given equation.
28. 12y" + 2ty' — 1 = 0, t > 0 29. ty" + y' = 1, t > 0
30. y" + t(y'f = 0 31. 2t2y" + (y'f = 2ty', t > 0
32. y" + y' = e—t 33. 12y" = (y')2, t > 0
Equations with the Independent Variable Missing. If a second order differential equation has the form y" = f (y, y'), then the independent variable t does not appear explicitly, but only through the dependent variable y. If we let v = y', then we obtain d v/dt = f (y,v). Since the right side of this equation depends on y and v, rather than on t and v, this equation is not of the form of the first order equations discussed in Chapter 2. However, if we think of y as the independent variable, then by the chain rule d v/dt = (d v/dy)(dy/dt) = v(d v/dy). Hence the original differential equation can be written as v(dv/dy) = f (y,v). Provided that this first order equation can be solved, we obtain v as a function of y. A relation between y and t results from solving dy/dt = v(y). Again, there are two arbitrary constants in the final result. In each of Problems 34 through 39 use this method to solve the given differential equation.