Page 56 - Schaum's Outline of Differential Equations

CHAP. 5] EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS 39

Integrating both sides of this last equation, we find

which is the solution in implicit form.

5.20. Solve

Rewriting this equation in differential form, we have

which is not exact. Furthermore, no integrating factor is immediately apparent. We can, however, rearrange this
equation as

2
2
The group in parentheses is of the form ay dx + bx dy, where a = 3 and b = —l, which has an integrating factor x y .
2
2
Since the expression in parentheses is already multiplied by x , we try an integrating factor of the form I(x, y) = y" .
2
Multiplying (_/) by y" , we have
which can be simplified (see Table 5-1) to

Integrating both sides of (2), we obtain

as the solution in implicit form.
5.21. Convert y' = 2xy - x into an exact differential equation.
Rewriting this equation in differential form, we have

Here M(x, y) = -2xy + x and N(x, y) = l. Since

and

are not equal, (_/) is not exact. But

is a function of x alone. Using Eq. (5.8), we have I(x,y) = xdx = e * as an integrating factor. Multiplying (_/)e
2
by e~* , we obtain

which is exact.

5.22. Convert y dx + xy dy = 0 into an exact differential equation.
Here M(x, y) = y 2 and N(x, y) = xy. Since

and