Page 40 - Schaum's Outline of Differential Equations

CHAP. 4] SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 23

2 3
4.2. Solve / = y- * .
We first rewrite this equation in the differential form (see Chapter 3) jr dx - (lly ) dy = 0. Then A(x) = jr and
2
B(y) = -lly . Substituting these values into Eq. (4.2), we have

4
or, by performing the indicated integrations, x /4 + lly = c. Solving explicitly for y, we obtain the solution as

4.3. Solve

This equation may be rewritten in the differential form
2
(x +2)dx-ydy = 0
2
which is separable with A(x) =x + 2 and B(y) = -y. Its solution is

or

Solving for y, we obtain the solution in implicit form as

with k = -2c. Solving for y implicitly, we obtain the two solutions

and

4.4. Solve / = 5y.
First rewrite this equation in the differential form 5 dx— (lly) dy = 0, which is separable. Its solution is

or, bv evaluating, 5x — In lyl = c.
To solve for y explicitly, we first rewrite the solution as In md then take the exponential of both
sides. Thus, Noting that we obtain The solution is given explicitly
byy =
Note that the presence of the term (—lly) in the differential form of the differential equation requires the
5x
restriction y ^ 0 in our derivation of the solution. This restriction is equivalent to the restriction k ^ 0, since y = ke .
Sx
However, by inspection, y = 0 is a solution of the differential equation as originally given. Thus, y = ke is the solution
for all k.
The differential equation as originally given is also linear. See Problem 6.9 for an alternate method of solution.

4.5. Solve

This equation, in differential form, is(x = 1) dx + (-y4 - 1) dy =0. which is separable. Its solution is