Solution Of First Order Equations Worksheets With Answers - S. Ghorai Page 3
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5S. Ghorai
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Example 4. Solve (2x + sin x tan y)dx
cos x sec
y dy = 0
2
Solution: Here M = 2x + sin x tan y and N =
cos x sec
y. Hence, M = N . Hence,
2
the solution is u = C, where u = x
cos x tan y
4
Reduction to exact equation: integrating factor
An integrating factor µ(x, y) is a function such that
M (x, y) dx + N (x, y) dy = 0
(2)
becomes exact on multiplying it by µ. Thus,
µM dx + µN dy = 0
is exact. Hence
∂(µM )
∂(µN )
=
.
∂y
∂x
Comment: If an equation has an integrating factor, then it has infinitely many inte-
grating factors.
Proof: Let µ be an integrating factor. Then
µM dx + µN dy = du
Let g(u) be any continuous function of u. Now multiplying by µg(u), we find
µg(u)M dx + µg(u)N dy = g(u)du
µg(u)M dx + µg(u)N dy = d
g(u) du
Thus,
µg(u)M dx + µg(u)N dy = dv,
whare v =
g(u) du
Hence, µg(u) is an integrating factor. Since, g is arbitrary, there exists an infinite
number of integrating factors.
Example 5. xdy
ydx = 0.
2
Solution: Clearly 1/x
is an integrating factor since
xdy
ydx
= 0
d(y/x) = 0
2
x
Also, 1/xy is an integrating factor since
xdy
ydx
= 0
d ln(y/x) = 0
xy
2
2
2
Similarly it can be shown that 1/y
, 1/(x
+ y
) etc. are integrating factors.
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