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Question Number 126229 by MathSh last updated on 18/Dec/20
(1+y)x′=x
Answered by Dwaipayan Shikari last updated on 18/Dec/20
(1+y)dxdy=x⇒∫dy1+y=∫dxx⇒1+y=Cx
Answered by ebi last updated on 18/Dec/20
(y+1)x′=x(1x)x′=1y+1∫(1x)dx=∫1y+1dyln∣x∣=ln∣y+1∣+cln∣x∣=ln∣y+1∣+ln∣C∣ln∣x∣=ln∣C(y+1)∣x=C(y+1)
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