Given that y(t)=c1e^(6t)+c2e^(−6t) is a solution to the differential equation y′′−36y=0, where c1 and c2 are arbitrary constants, find a function y(t) that satisfies the conditions.

y′′−36y=0,

y(0)=5,

lim t→+[infinity] y(t)=0.

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: howdy

step-by-step explanation: