60 Ordinary Differential Equationsconstants can be combined by subtraction into the single constant, C = c2 -c1.
In addition, since a multiple of a constant, combinations of constants, and
functions of constants are constant, when solving differential equations we
will often replace such expressions by a new constant.)Multiplying the equation (13) by 2 and then adding x^2 to the resulting
equation, we obtain the following relation for the solution to the DE (12)(14) y^2 + x^2 = 2C.
For C < 0, equation (14) yields no real solution to the DE (12). For C = 0,
equation (14) produces the point (0, 0) which is not a solution to the DE (12).Assuming C > 0 and replacing 2C in equation (14) by the new constant K^2 ,
where we may also assume K > 0, equation (14) becomes
(15) y2 +x2 = K2.From equation (15) we see that geometrically the general solution the DE (12)is a family of concentric circles with centers at the origin and radii K > 0.
To verify that (15) is actu ally an implicit solution, we must show that (15)
defines at least one function y 1 (x) which is an explicit solution of (12). Solvingequation (15) for y , we get y(x) = ±.JK^2 - x^2. The functions
Y1(x) = JK^2 - x^2 and Y2(x) = -JK^2 - x^2
are both defined and real for x E [-K , K]. Differentiating y 1 ( x) and y 2 ( x), we
obtain
11 (x)- - x
Y1 - -y-;:::f{===.2;;:=_=x::;;:2 and y~(x) = .JK2 x - x2Both y~(x) and y~(x) are defined and real for x E (-K,K). Substituting y 1
and y~ into the DE (12) y' = -x/y, we obtain the identity-x = -x I v~K- 2 ___ x_2
.JK2 - x2for x E (-K,K). Thus, y 1 (x ) is an expli cit solution of (12) on the interval
(-K, K). And consequently, by definition, equation (15) is an implicit solution
of (12) on (-K,K). Likewise, y 2 (x) can be shown to b e an explicit solution
of (12) on the interval (-K, K). Hence, the implicit solution (15) defines at
least two expli cit solutions of (12) on the interval (-K, K). The graph of the
solution Y1 (x) is shown in Figure 2.10 (a) and the graph of the solution y 2 (x)
is shown in Figure 2.10 (b).