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

something that characterises a right triangle is the fact that we may use pythagoras' theorem to find any one of its sides given that we have the length of the other two sides. here, we have the length of the hypotenuse, √(87) yd, and one other side, √(23) yd, thus we can easily find the length of the remaining side using pythagoras' theorem as defined below:

c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

thus, given that c = √(87) and a = √(23), we may substitute these into the formula we defined above to get:

c^2 = a^2 + b^2

(√87)^2 = (√23)^2 + b^2 (substitute c = √87 and a = √23 into the equation)

87 = 23 + b^2 (evaluate (√87)^2 = 87 and (√23)^2 = 23 - remember that the square root of a number, squared will equal that very same number)

64 = b^2 (subtract 23 from both sides)

√64 = b (take the square root of both sides)

8 = b (evaluate √64 = 8)

therefor, the length of the unknown leg in the right triangle is 8 yd (first answer).

the answer is simply just 6n add the two together as they have an n