Page 57 - Schaum's Outline of Differential Equations

40 EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAR 5

are not equal, (1) is not exact. But

(ll
d
is a function of y alone. Using Eq. (5.9), we have as an integrating factor I(x, y) = e y~> y = e ln)I = 1/y. Multiplying
'
the given differential equation by I(x, y) = lly, we obtain the exact equation y dx + x dy = 0.
5.23. Convert y — into an exact differential equation.
Rewriting this equation in differential form, we have

Here M(x, y)=y(i— xy) and N(x, y) = x. Since

are not equal, (1) is not exact. Equation (5.10), however, is applicable and provides the integrating factor

Multiplying (1) by I(x, y), we obtain

which is exact.

Supplementary Problems

In Problems 5.24 through 5.40, test whether the differential equations are exact and solve those that are.

3
2 2
5.24. (y + 2xy ) dx+(l+ 3x y + x)dy = 0 5.25. (jcy + 1) att+(jty-l) rfy = 0
3
2
2 2
3
x
2
5.26. e** (3x y -x )dx + e ' dy = 0 5.27. 3x y dc + (2X ); + 4y ) dy = 0
5.28. ydx + xdy = 0 5.29. (jc-y)dc+(jc + y)rfy = 0
5.30. (y sin x + xy cos x) dx + (x sin x + 1) dy = 0
2
2
5.33. y dt + t dy = 0
4 2
3 3
2
5.34. (4t y -2ty)dt+(3t y -t )dy = 0
2
2
2
5.36. (t -x) dt-tdx = 0 5.37. (t + x ) dt + (2tx-x)dx = 0
2
2
5.38. 2xe ' dt+(l + e ') dx = 0 5.39. sin t cos x dt— sin x cos t dx = 0
5.40. (cos x + x cos t) dt + (sin t - t sin x) dx = 0