If xy + 6ey = 6e,
find the value of y”
at the point where
x = 0.
Update:
If
xy + 6e^y = 6e, find the value of y”
at the point where x = 0.
If
xy + 6e^y = 6e, find the value of y”
at the point where x = 0.
-
xy + 6e^y = 6e passes through (0,1)
1st DI: x y’ + y + y’ 6e^y = 0 → y’ = -y/(x + 6e^y)
2nd DI: x y” + y’ + y’ + y” 6e^y + (y’)² 6e^y = 0
→ y” = -(2y’ + (y’)² 6e^y)/(x + 6e^y)
→ y” = -(2(-y/(x + 6e^y)) + (-y/(x + 6e^y))² 6e^y)/(x + 6e^y)
→ y”(0) = -(2(-1/(0 + 6e^1)) + (-1/(0 + 6e^1))² 6e^1)/(0 + 6e^1) → y”(0) = -(-2/(6e) + (1/(6e))/(6e) → y”(0) = -((-2+1)/(6e))/(6e) → y”(0) = 1/(36e²)
-
Assuming you meant xy + 6e^y = 6e.
Use implicit differentiation (each side). What does that give?
Answer that and we will proceed.
Do you have questions that need to be resolved? Ask to hear answers from experts here!
Leave a Reply