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Question Number 145300 by imjagoll last updated on 04/Jul/21

       yy′ = x e^(x/y)

yy=xexy

Answered by puissant last updated on 04/Jul/21

y′=(x/y)e^(x/y)   put   t=(x/y)  ⇒(dy/dt) = te^t ⇒ dy = te^t dt  ⇒∫ dy = ∫ te^t dt  ∫te^t dt = te^t −e^t +k = e^t (t−1)+k  ⇒ y= e^t (t−1)+γ....

y=xyexyputt=xydydt=tetdy=tetdtdy=tetdttetdt=tetet+k=et(t1)+ky=et(t1)+γ....

Commented by imjagoll last updated on 04/Jul/21

if t=(x/y) ⇒y=(x/t)   ⇒(dy/dx) = (1/t)−(x/t^2 ) (dt/dx)

ift=xyy=xtdydx=1txt2dtdx

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