Answer:
y(t) = 2.5 e⁶ᵗ + 2.5 e⁻⁶ᵗ
Or
y(t) = 5 e⁻⁶ᵗ
Step-by-step explanation:
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ
Let us find our value for y(t) that satisfies the conditions
1) y" - 36y = 0
y" = (d²y/dt²)
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ
y' = (dy/dt) = 6c₁ e⁶ᵗ - 6c₂ e⁻⁶ᵗ
y" = (d/dt)(dy/dt) = 36c₁ e⁶ᵗ + 36c₂ e⁻⁶ᵗ
y" - 36y = 36c₁ e⁶ᵗ + 36c₂ e⁻⁶ᵗ - 36(c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ) = 36c₁ e⁶ᵗ + 36c₂ e⁻⁶ᵗ - 36c₁ e⁶ᵗ - 36c₂ e⁻⁶ᵗ = 0.
The function satisfies this condition.
2) y(0) = 5
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ
At t = 0
y(0) = c₁ e⁰ + c₂ e⁰ = 5
c₁ + c₂ = 5 (e⁰ = 1)
3) lim t→+[infinity] y(t)=0
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ = 0 as t→+[infinity]
c₁ e⁶ᵗ = - c₂ e⁻⁶ᵗ as t→+[infinity]
c₁ = - c₂ e⁻¹²ᵗ as t→+[infinity]
e⁻¹²ᵗ = 0 as t→+[infinity]
c₁ = c₂ or c₁ = 0
Recall c₁ + c₂ = 5
If c₁ = 0, c₂ = 5
If c₁ = c₂, c₁ = c₂ = 2.5
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ = 2.5 e⁶ᵗ + 2.5 e⁻⁶ᵗ
Or
y(t) = 5 e⁻⁶ᵗ
Answer:
1)(0.75,55)
2)(1.5,28)
3)(2,0)
(2.25,-17)
(3,-80)
(4,-192)
Step-by-step explanation:
1) substitute "t" values in table into the equation:
64-16(0.75)^2=
55
64-16(1.5)^2=
28
64-16(2)^2=
0
64-16(2.25)^2=
-17(ball already reached ground and went underground)
64-16(3)^2=
-80
64-16(4)^2=
-192