Final answer:
To solve the initial value problem y"-8y'+16y=0, y(0)=4, y'(0)=82/5, we use the method of solving a homogeneous second order linear differential equation. By substituting the given initial values, we find the values of a, b, c, and d.
Step-by-step explanation:
To solve the initial value problem y"-8y'+16y=0, y(0)=4, y'(0)=82/5, we can use the method of solving a homogeneous second order linear differential equation. This type of equation has the general solution y = aebt + cedt. By substituting the given initial values, we can find the values of a, b, c, and d.
Given y(0) = 4, we have 4 = aeb*0 + ced*0 = a + c. (1)
Given y'(0) = 82/5, we have (82/5) = abeb*0 + cded*0 = ab + cd. (2)
To solve the system of equations (1) and (2), we can substitute a = 4 - c from equation (1) into equation (2) to get (82/5) = (4 - c)b + cd. Rearranging this equation gives us:
0 = cb - cd - (82/5 - 4)b
0 = cb - cd - (82b - 20b)/5
0 = cb - cd - (62b)/5
Adding cb to both sides:
cb = cd - (62b)/5
Dividing every term by b:
c = d - 62/5
Now, we can substitute the value of c in terms of d into equation (1) to solve for a:
4 = a + (d - 62/5)
4 = a + d - 62/5
20/5 + 62/5 = a + d
82/5 = a + d
Therefore, the solution to the initial value problem is y = aebt + cedt, where a = 82/5 - d, b = 0, c = 4 - (82/5 - d), and d can take any real value.