Page 53 - Schaum's Outline of Differential Equations

36 EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAR 5

5.10. Solve the differential equation given in Problem 5.9.
This equation was shown to be exact, so the solution procedure given by Eqs. (5.4) through (5.6), with t and x
replacing x and y, respectively, is applicable. We seek a function g(t, x) having the property that dg is the right side
of the given differential equation. Here

Integrating both sides with respect to t, we have

or
Differentiating (1) with respect to x, we obtain

Hence,

where the right side of this last equation is the coefficient of dx in the original differential equation. It follows that

4
Now h(x) = x + GI, and (1) becomes

The solution to the differential equation is given implicitly by (5.6) as

or, by taking the square roots of both sides of this last equation, as

We can solve for x explicitly with the quadratic formula, whence

5.11. Solve

The differential equation has the differential form given in Problem 5.1. Its solution is given in (2) of Problem 5.2
2
2
as x y + y = c 2. Using the initial condition, y = —5 when x = 2, we obtain (2) (-5) + (-5) = c 2, or c 2 = -25. The solution
2
2
to the initial-value problem is therefore x y + y = -25 or y = -25/(x + 1).
5.12. Solve

This differential equation in standard form has the differential form of Problem 5.7. Its solution is given in (2)
2
2
of Problem 5.8 as y t + y = c 2. Using the initial condition y = -2 when t = 1, we obtain (-2) (1) + (-2) = c 2, or c 2 = 2.