Page 42 - Schaum's Outline of Differential Equations

CHAP. 4] SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS 25

4.8. Solve e f dx - y dy = 0; y(0) = 1.
The solution to the differential equation is given by Eq. (4.2) as

x
or, by evaluating, as y = 2e +k, k = —2c. Applying the initial condition, we obtain (1) = 2e + k, 1 = 2 + k, or
k = —l. Thus, the solution to the initial-value problem is

[Note that we cannot choose the negative square root, since then y(0)=-l, which violates the initial
condition.]
To ensure that y remains real, we must restrict x so that 2e* — 1 > 0. To guarantee that / exists [note that
x
x
x
y'(x) = dyldx = e /y], we must restrict x so that 2e - 1^0. Together these conditions imply that 2e - 1 > 0, or
x>lni.
4.9. Use Eq. (4.4) to solve Problem 4.8.
For this problem, x 0= 0, y 0= 1, A(x) = e", and B(y) = —y. Substituting these values into Eq. (4.4), we obtain

Evaluating these integrals, we have

x
2
Thus, y = 2e — 1, and, as in Problem 4.8,
4.10. Solve x cos xdx+(i- 6/) dy = 0; y(ri) = 0.
5
Here, x 0 = n,y 0= 0, A(x) = x cos x, and B(y) = 1 - 6y . Substituting these values into Eq. (4.4), we obtain

Evaluating these integrals (the first one by integration by parts), we find

or
Since we cannot solve this last equation for y explicitly, we must be content with the solution in its present
mplicit form.

4.11. Solve

This differential equation is not separable, but it is homogeneous as shown in Problem 3.9(a). Substituting
Eqs. (4.6) and (4.7) into the equation, we obtain

which can be algebraically simplified to