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Question Number 55939 by MJS last updated on 06/Mar/19
y′y=1−cotxy=?
Answered by kaivan.ahmadi last updated on 06/Mar/19
1ydydx=1−cotx⇒1ydy=(1−cotx)dx⇒∫1ydy=∫(1−cotx)dx⇒lny=x−ln∣sinx∣⇒y=ex−ln∣sinx∣⇒y=ex∣sinx∣+C
Commented by MJS last updated on 06/Mar/19
thankyou
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