1550078481-Ordinary_Differential_Equations__Roberts_

20 Ordinary Differential Equations

b. Show that

y(x) = -x^2 /4

is a singular solution of the DE (8).

SOLUTION

a. Differentiating y(x) = ex+ e^2 , we get y' = e. Substituting for y and y'

in (8), yields

(y')^2 + xy' - y = e^2 + xe - (ex+ e^2 ) = 0.

Thus, (7) is a one-parameter family of solutions of the DE (8).

b. The function y(x) = - x^2 /4 cannot be obtained from the one-parameter

family y( x) = ex + e^2 by any choice of the constant e. Differentiating

y(x) = - x^2 /4, we find y' = -x/2. Substituting for y and y' in (8), we

find

/ 2 / X 2 X X^2 1 1 1 2

(y) +xy -y=(-2) +x(-2)-(-4)=(4-2+4)x =O.

Thus, y(x) = -x^2 /4 is a solution of the DE (8) and it is not a member

of the one-parameter family (7). So by definition y(x) = - x^2 /4 is a

singular solution of (8). A graph of the singular solution y(x) = -x^2 /4

and members of the one-parameter family of solutions y(x) = ex+ c^2

obtained by choosing c = -2, c = -1, c = 0, c = 1, and c = 2

are displayed in Figure 1.3. Observe that the one-parameter family of

lines y(x) =ex+ c^2 are tangent to the singular solution (the parabola)

y(x) = -x^2 /4.

Figure 1.3 A One-Parameter Family of Solutions and

the Singular Solution of (y')^2 + xy' - y = 0