The Initial Value Problem y' = f(x, y) ; y(c) = d 61
2 y 2 yx -K K x-2 -K - 1 (^0) K (^2) -2 -1 0 2
-1 -1
-2 - 2
(a) (b)
Figure 2.10 Graph of the Explicit Solutions y 1 (x) and y 2 (x) of the DE (12)
In the previous example, it was fairly easy to det ermine an explicit solution
from the implicit solution and to determine the interval on which the solution
exists. However , this will not be the case in general. Normally, we will not be
a ble to solve the given relation in x and y explicitly for either x or y. Therefore,
usually we will obtain a relation in x and y by some means, verify that this
relation formally satisfies the differential equation under consideration, and
say that the relation is an impli cit solution. For example, we will say that the
relation
(16) y^3 + 2 .xy - x^2 = C ,
where C is a constant, is an implicit solution of the differential equation
(17)
I 2X - 2y
y =.
3y^2 + 2x
In order to verify formally that (16) is a solution of (17), we implicitly differ-
entiate (16) with resp ect to x and obtain
3y^2 y' + 2(xy' + y) - 2x = 0.
Solving this equation for y' yields the differential equation (17).
We will soon discover that, in theory, we will be able to explicitly solve
linear differential equa tions and determine the interval on which the solution
exists directly from the differential equation itself. However, the b est we will
b e able to accomplish usually for nonlinear differentia l equations is to obtain
an implicit solution, a series solution, or a numerical solution. This is one
of the primary differences between the kinds of res ults that we can exp ec t to
obtain for linear differential equations versus nonlinear differential equations.