Solve differential equation (2y-e^x)dx+xdy=0, x>0

melodykap 2020-11-10 Answered
Solve differential equation (2yex)dx+xdy=0, x>0
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Expert Answer

Nathanael Webber
Answered 2020-11-11 Author has 117 answers

(2yex)dx+xdy=0
2yex+xdy/dx=0
(2y)/xex/x+dy/dx=0
dy/dx+(2/x)y=ex/x
Above differential equation is form of dy/dx+Py=Q
Integrating factor will be
I.F.=ePdx
=e2/xdx
=e2lnx =elnx2
=x2
Solution will be given as
I.Fy=I.FQdx
x2y=x2e2/xdx
x2y=xexdx
x2y=[xexdx[d/dx(x)exdx]]dx
x2y=[xexex]+c
y=1/x2(xexex+c)

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