If xy + 6ey = 6e, find the value of y'' at the point where x = 0.
Asked by maham237 @ in Mathematics viewed by 417 People
Answered by maham237 @
Xy + 6e^y = 6e
taking first derivative :
xy' + y + 6y(e^y)(y') = 0
xy" + y' + y' + 6y'(e^y)(y') + 6yy(e^y)y' + 6y(e^y)y" = 0
Put x = 0,
0 + 6e^y = 6e,
y = 1
0 + 1 + 6(1)(y')e = 0,
Put y' = 0
0 + 0 + 0 + 0 + 0 +6 ey" = 0,
y" = 0
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