Page 55 - Schaum's Outline of Differential Equations

38 EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAR 5

5.17. Solve Problem 5.15 using the integrating factor given in Problem 5.16.
Using the results of Problem 5.16, we can rewrite the given differential equation as

which is exact. Equation (1) can be solved using the steps described in Eqs. (5.4) through (5.6).
Alternatively, we note from Table 5-1 that (1) can be rewritten as d[ln (y/x)] = 0. Then, by direct integration,
1
In (y/x) = Cj. Taking the exponential of both sides, we find ylx = e , or finally,

5.18. Solve (y2-y) dx + x dy = 0
This differential equation is not exact, and no integrating factor is immediately apparent. Note, however, that
if terms are strategically regrouped, the differential equation can be rewritten as

The group of terms in parentheses has many integrating factors (see Table 5-1). Trying each integrating factor
2
separately, we find that the only one that makes the entire equation exact is I(x, y) = lAy . Using this integrating factor,
we can rewrite (_/) as

Since (2) is exact, it can be solved using the steps described in Eqs. (5.4) through (5.6).
Alternatively, we note from Table 5-1 that (2) can be rewritten as -d(xly) + ldx = 0, or as d(xly)= Idx
Integrating, we obtain the solution

5.19 Solve (y2-xy) dx +( x+x2y2) dy = 0.
This differential equation is not exact, and no integrating factor is immediately apparent. Note, however, thai
the differential equation can be rewritten as

2
The first group of terms has many integrating factors (see Table 5-1). One of these factors, namely I(x, y) = l/(xy) ,
2
is an integrating factor for the entire equation. Multiplying (1) by l/(xy) , we find

or equivalently,

Since (2) is exact, it can be solved using the steps described in Eqs. (5.4) through (5.6).
Alternatively, we note from Table 5-1

so that (2) can be rewritten as