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

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)