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Unformatted text preview: EXAMPLE Exact Equations and Integrating Factors 99 Thus, if (MM )1v is to equal (/VLNLC, it is necessary that dtt My — N x —— = ——’——— . 27
dx N M ( >
If (M y — Nx) /N is a function of x only, then there is an integrating factor [1, that also
depends only on x; further, tux) can be found by solving Eq. (27), which is both linear and separable. 7
A similar procedure can be used to determine a condition under which Eq. (23) has an integrating factor depending only on y; see Problem 23. Find an integrating factor for the equation
oxy + yo + (x2 + my 2 0 (19) and then solve the equation.
In Example 3 we showed that this equation is not exact. Let us determine whether it has an integrating factor that depends on x only. On computing the quantity (My — Nx)/N, we ﬁnd
that Mytxy) —Nx(x,y) _ 3X+2y— (2x+y) _}
N(x,y) — xzl—xy —x' Thus there is an integrating factor it that is a function of x only, and it satisﬁes the differential
equation (28) du _ M a; __ 3; . (29)
Hence u(x) = x. (30) Multiplying Eq. (19) by this integrating factor, we obtain
(3x2y + xyz) + (x3 + xzwy’ = 0 (31)
The latter equation is exact, and it is easy to show that its solutions are given implicitly by
x3y + %Jc2y2 = c. ' (32) Solutions may also be readily found in explicit form since Eq. (32) is quadratic in y.
You may also verify that a second integrating factor of Eq. (19) is 1
x7 : _———_a
M y) xy(2x+y) and that the same solution is obtained, though with much greater difficulty, if this integrating
factor is used (see Problem 32). WW PROBLEMS Determine whether each of the equations in Problems 1 through 12 is exact. If it is exact, ﬁnd
the solution. 1. (2x+3) + 2I—2)y” :0
3. (3x2—2xy+2)dx+(6y2—x2+3)dy20
4. (2x322 + 2y) + (2x23) + 2x)y’ : O 2. (2x + 4y) + (2x + 2y)y’ = O dy ax + by 6 dy ax — by D‘EZ—bx‘tcy 'E‘_bx—cy
7. (ex siny — 2ysinx) dx + (e’C cosy +2cosx) dy = O 8. (e‘siny+3y)dx — (3x — e‘siny)dy =0 9. 026’” cos 2x — 23“ sin 2x + 2x) dx + (xe"«v cos 2x — 3) dy = 0
10. (y/x+6x)dx+(lnx—2)dy=0, x>0
11. (xlny +xy)dx + (ylnx +xy) dy = O; x dx I y dy In each of Problems 13 and 14 so
approximately Where the solution is valid. 13. (2x — y) dx + (2y — x) dy : 0, y(1) = 3
14. (9x2+y—1)dx—~(4y—x)dy=O, y(1)=0 x>0, y>O 12 lve the given initial value problem and determine at least In each of Problems
then solve it using that value of b. 15. (ch2 + bxzy) dx + (x + y)x2 dy = O 16. 07er + x) dx + bxezxy dy = 0
17. Assume that Eq. (6) meets the requirements of Theorem 2.6.1 in a rectangle R and is therefore exact. Show that a possible function iMx, y) is y
N (x, t) dt, YO my) = / Mme) ds +
X0 Where (x0,y0) is a point in R.
18. Show that any separable equation Mtx)+Nty>y’=0 is also exact. s in Problems 19 through 22 are not exact but become exact when Show that the equation
solve the equations. multiplied by the given integrating factor. Then 19 x2y3 + xtl + y2)y’ i 0, MXJ) = 1/1013
. 2 _x 20' — 26—1 sinx) dx + (COSy + e 008x) dy = O, pt(x,y) : ye)“
.V y 21. ydx+(2x—yey)dy=O, u(x,y) =y 22. (x+2)siny dx+xcosy dyzO, it(x,y)=xe"
23. Show that if (Nx — My)/M : Q, Where Q is a function of y only, then the differential equation M+Ny":0 has an integrating factor of the form My) =6Xp/dey 24. Show that if (Nx — My)/ (xM — yN) = R, Where R depends on the quantity xy only, then the differential equation
M + Ny’ = O .
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This note was uploaded on 02/11/2012 for the course MTH 2201 taught by Professor Kigaradze during the Spring '08 term at FIT.
 Spring '08
 Kigaradze
 Equations

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