Question

e^ydx +(xe^y+2y)dy=0

Answer 1

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e^ydx+(xe^y+2y)dy=0

e^y(e^dx+dyx+2dy²)=0

when d =0

e^y(e^0x+0yx+2(0)y²)=0

e^y+0=0

e^y=0

e=y√0