Solve the initial value problem y′′+4y′+40y=0y(0)= 3y′(0)=5.

Step-by-step explanation:

Given the initial value problem y′′+4y′+40y=0 where y(0)= 3 and y′(0)=5.

Find the solution in the attached file.

Since the solution is a complex number, the solution to the second order differential equation is given as shown;

y(x) =

from the complex solution D = -2+6i

y(x) =

Substituting the initial values

when y(0) = 3, the solution becomes;

From y(x) =

at y'(0) = 5, the solution becomes;

Since c₁ = 3;

5 = 6c₂ - 2(3)

5 = 6c₂ - 6

6c₂ = 11

c₂ = 11/6

Substituting the constant into the solution to the differential equation

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)