Page 64 - Advanced engineering mathematics

44 CHAPTER 2 Linear Second-Order Equations

and then once again to get
3
y(x) = 2x + cx + k
with c and k as arbitrary constants. It seems natural that the solution of a second-order differ-
ential equation, which involves two integrations, should contain two arbitrary constants. For any
choices of c and k, we can graph the corresponding solution, obtaining integral curves. Figure 2.1
shows integral curves for several choices of c and k.
Unlike the first-order case, there may be many integral curves through a given point in the
plane. In this example, if we specify that y(0) = 3, then we must choose k = 3, leaving c still
arbitrary. These solutions through (0,3) are
3
y(x) = 2x + cx + 3.
Some of these curves are shown in Figure 2.2.

10

5

x
–1.5 –1 –0.5 0 0.5 1 1.5

–5

–10

–15
3
FIGURE 2.1 Graphs of some functions y = 2x +
cx + k.

15

10

5

x
–1.5 –1 –0.5 0 0.5 1 1.5
–5

–10

3
FIGURE 2.2 Graphs of some functions y = 2x +
cx + 3.

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