James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 9.3 Separable Equations 601
(b) If we put x − 0 in the general solution in part (a), we get ys0d − s 3 K . To satisfy
the initial condition ys0d − 2, we must have s 3 K − 2 and so K − 8. Thus the solution
of the initial-value problem is
y − s 3 x 3 1 8
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Some computer software can plot
curves defined by implicit equations.
Figure 2 shows the graphs of several
members of the family of solutions of
the differential equation in Example 2.
As we look at the curves from left
to right, the values of C are 3, 2, 1, 0,
21, 22, and 23.
Example 2 Solve the differential equation dy
dx − 6x 2
2y 1 cos y .
SOLUtion Writing the equation in differential form and integrating both sides, we have
s2y 1 cos yddy − 6x 2 dx
y s2y 1 cos yddy − y 6x 2 dx
4
3
y 2 1 sin y − 2x 3 1 C
_2 2
where C is a constant. Equation 3 gives the general solution implicitly. In this case it’s
impossible to solve the equation to express y explicitly as a function of x.
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FIGURE 2
_4
Example 3 Solve the equation y9 − x 2 y.
SOLUtion First we rewrite the equation using Leibniz notation:
dy
dx − x 2 y
If a solution y is a function that satisfies
ysxd ± 0 for some x, it follows from
a uniqueness theorem for solutions of
differential equations that ysxd ± 0 for
all x.
If y ± 0, we can rewrite it in differential notation and integrate:
dy
y − x 2 dx y ± 0
y dy
y − y x 2 dx
ln | y | − x 3
3 1 C
This equation defines y implicitly as a function of x. But in this case we can solve
explicitly for y as follows:
| y | − e ln | y | − e
sx 3 y3d1C − e C e x 3 y3
so
y − 6e C e x 3 y3
We can easily verify that the function y − 0 is also a solution of the given differential
equation. So we can write the general solution in the form
y − Ae x 3 y3
where A is an arbitrary constant (A − e C , or A − 2e C , or A − 0).
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