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Solve the initial value problem

y"-8y'+16y=0, y(0)=4, y'(0)=82/5
y = aeᵇᵗ+ceᵈᵗ
a =
b =
c =
d =
(enter integers or fractions)

1 Answer

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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.

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