Transcribed Image Text: 25. Riccati Equations. The equation
dy
dt
is known as a Riccati25 equation. Suppose that some particular solution
y of this equation is known. A more general solution containing one
arbitrary constant can be obtained through the substitution
Ba
1
v(t)
Show that v(t) satisfies the first-order linear equation
dv
dt
Note that v(1) will contain a single arbitrary constant.
26. Verify that the given function is a particular solution of the given
Riccati equation. Then use the method of Problem 25 to solve the
following Riccati equations:
a. y' = 1+1²-2ty + y²; y(t) = t
1
1
b. y' = -√2
= 9₁(1) +92(1)y+q3(t) y²
C.
dy
dt
y = y₁(t) +
= -(92 +2931) V - 93.
2
t
2 cos² t - sin²t + y²
2 cost
+y²; y(t) =
t
; yı(t) = sint
27. The propagation of a single action in a large population (for
example, drivers turning on headlights at sunset) often depends
partly on external circumstances (gathering darkness) and partly on
a tendency to imitate others who have already performed the action
in question. In this case the proportion y(t) of people who have
performed the action can be described 26 by the equation
dy/dt = (1-y)(x(t) +by),
29.
30. y" +t(y)² = 0
31. 2t2y" + (y')³ = 2ty', t > 0
Equations with the Independent Variable Missing. Consider
second-order differential equations of the form y" = f(y, y'), in
which the independent variable t does not appear explicitly. If
we let v = y', then we obtain dv/dt = f(y, v). Since the right-
hand side of this equation depends on y and v, rather than on
t and v, this equation contains too many variables. However, if
we think of y as the independent variable, then by the chain
rule, dv/dt = (dv/dy) (dy/dt) = v(dv/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
mation and that of y. A relation between y and t results from solving dy/dt = v(y),
which is a separable equation. Again, there are two arbitrary constants
in the final result. In each of Problems 32 through 35, use this method
to solve the given differential equation.
32. yy" + (y)² = 0
ve the form 33.
33.
34.
(28)
where x(t) measures the external stimulus and b is the imitation
coefficient.
a. Observe that equation (28) is a Riccati equation and that
y₁ (t) = 1 is one solution. Use the transformation suggested in
Problem 25, and find the linear equation satisfied by v(t).
b. Find v(t) in the case that x(t) = at, where a is a constant.
Leave your answer in the form of an integral.
***************
25 Riccati equations are named for Jacopo Francesco Riccati (1676-1754), a
Venetian nobleman, who declined university appointments in Italy, Austria,
and Russia to pursue his mathematical studies privately at home. Riccati
studied these equations extensively; however, it was Euler (in 1760) who
discovered the result stated in this problem.
Some Special Second-Order Differential Equations. Second-order
differential equations involve the second derivative of the unknown
function and have the general form y" = f(t, y, y'). Usually,
such equations cannot be solved by methods designed for first-order
equations. However, there are two types of second-order equations
that can be transformed into first-order equations by a suitable change
of variable. The resulting equation can sometimes be solved by the
methods presented in this chapter. Problems 28 through 37 deal with
these types of equations.
26 See Anatol Rapoport, "Contribution to the Mathematical Theory of Mass
Behavior: I. The Propagation of Single Acts," Bulletin of Mathematical
Biophysics 14 (1952), pp. 159-169, and John Z. Hearon, "Note on the Theory
of Mass Behavior," Bulletin of Mathematical Biophysics 17 (1955), pp. 7-13.
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 differential 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
31, use this substitution to solve the given equation.
28. t2y" +2ty' - 1 = 0, t> 0
ty"+y' = 1, t> 0
y" + y = 0
yy"-(y')³ = 0
35. y" + (y)² = 2e-y
Hint: In Problem 35 the transformed equation is a Bernoulli equation.
See Problem 23 in Section 2.4.
In each of Problems 36 through 37, solve the given initial value
problem using the methods of Problems 28 through 35.
36. y'y" = 2, y(0) = 1, y'(0) = 2
37. (1+²) y" + 2ty' + 3t2 = 0, y(1) = 2, y'(1) = -1