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Question : 80 of 160
Marks:
+1,
-0
Solution:
We have,
x‌cos(‌)(ydx+xdy) =y‌sin‌(xdy−ydx) ⇒‌‌xy‌sin(‌)‌d‌y−y2‌sin(‌)‌d‌x =xy‌cos(‌)‌d‌x+x2‌cos(‌)‌d‌y ⇒‌‌[xy‌sin(‌)−x2‌cos(‌)]dy =[xy‌cos(‌)+y2‌sin(‌)]dx ⇒‌‌‌=‌| xy‌cos()+y2‌sin() |
| xy‌sin(‌)−x2‌cos(‌) |
⇒‌‌‌=‌| ‌‌cos()+()2‌sin() |
| ‌sin(‌)−cos(‌) |
Put,
y=vx⇒‌=v+x‌ ∴‌‌v+x‌=‌| v‌cos‌v+v2‌sin‌v |
| v‌sin‌v−cos‌v |
⇒‌‌‌‌x‌=‌| 2v‌cos‌v |
| v‌sin‌v−cos‌v |
⇒‌‌‌‌‌| v‌sin‌v−cos‌v |
| v‌cos‌v |
dv=2‌ ⇒‌‌‌‌∫(tan‌v−‌)dv=2‌∫‌ ⇒‌‌‌‌log|sec‌v|−log|v|=2‌log|x|+log|C1| ⇒‌‌‌‌log|‌|−log|x|2=log|C1| ⇒‌‌‌‌log|‌|=log|C1| ⇒‌‌‌‌‌=C1⇒‌=C1 ⇒‌‌‌=C1⇒sec(‌)=C1xy ⇒‌‌‌=‌‌‌(∵ where
‌=C)
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