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

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet