Example 1: Find the general solution of xdx + sin²2 (ydx – xdy) = 0. Solution: dx cos 2v) dv dx 1 dv + 1 cos 2v dv = So - 1 - 5v +- sin 2v = ln c 2 1 1 In x Inz - Ine -im2) - 0 1 SE) + isin 2 () : (Y) 2 \x. Note: y = VX X 4x In- 4 2y – 2y + x sin– = 0 C X

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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I don't know how it became 1/4 instead of 1/2. Help... Thank you.

Example 1: Find the general solution of xdx + sin2 2 (ydx – xdy) = 0.
Solution:
dx
(1
cos 2v) dv
「告如。
dx
1
dv +
2
So
cos 2v dv =
1
-zv +-sin 2v = ln c
4
1
+
4
1
In x
2
In x – In c
sin 2
= 0
Note: y = vx
-
2y
4x In
C
2y + x sin
- -
- = 0
X
Transcribed Image Text:Example 1: Find the general solution of xdx + sin2 2 (ydx – xdy) = 0. Solution: dx (1 cos 2v) dv 「告如。 dx 1 dv + 2 So cos 2v dv = 1 -zv +-sin 2v = ln c 4 1 + 4 1 In x 2 In x – In c sin 2 = 0 Note: y = vx - 2y 4x In C 2y + x sin - - - = 0 X
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