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Question : 80 of 160
Marks:
+1,
-0
Solution:
Given, differential equation
‌2‌dx+dy=(6xy+4x−3y)‌dx‌⇒‌‌dy=(6xy+4x−3y−2)‌dx‌⇒‌‌‌=6xy+4x−3y−2‌⇒‌‌‌=6xy+4x−2−3y‌⇒‌‌2x(3y+2)−1(3y+2)‌⇒‌‌(3y+2)(2x−1)‌⇒‌‌‌=dx(2x−1)Integrating to both sides w.r.t
x, we get
‌∫‌=∫(2x−1)‌dx‌⇒‌‌log|3y+2|=‌−x+C1‌⇒‌‌log|3y+2|=x2−x+C1‌⇒log|3y+2|=3x2−3x+3C1‌=3x2−3x+C(where
C=3C1 )
So,
log|3y+2|=3x2−3x+C
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