Question 899908
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Hi
ye^x  Use Chain Rule for this term
2yy' -(1y'e^x + ye^x) = 0
 2yy' - y'e^x - ye^x = 0
 2yy' - y'e^x = ye^x 
 y'(2y-e^x) = ye^x
  y' = ye^x/(2y-e^x)