The auxiliary equation of y”” – 4y” – 4y’ + 16y = 0 is r³ – 4r² – 4r + 16 = 0

The auxiliary equation of y”” – 4y” – 4y’ + 16y = 0 is r³ – 4r² – 4r + 16 = 0. The auxiliary equation has a root in the interval [-0.75, 3.75]. Let f(r) = r³ – 4r² – 4r + 16 so that the roots ” of the auxiliary quation can be determined by solving f(r) = 0. Then, applying four iterations of the Bisection Method to f(r), using the 55

The Correct Answer and Explanation is :

To solve ( f(r) = r^3 – 4r^2 – 4r + 16 = 0 ) using the Bisection Method in the interval ([-0.75, 3.75]), we iterate as follows:

Steps:

1. Initial interval: ([a, b] = [-0.75, 3.75]).

• Compute ( f(a) = f(-0.75) = (-0.75)^3 – 4(-0.75)^2 – 4(-0.75) + 16 = 15.078125 > 0 ).
• Compute ( f(b) = f(3.75) = (3.75)^3 – 4(3.75)^2 – 4(3.75) + 16 = -6.828125 < 0 ).
• Since ( f(a) \cdot f(b) < 0 ), the root lies in this interval.

1. First iteration:

• Midpoint: ( c_1 = \frac{a + b}{2} = \frac{-0.75 + 3.75}{2} = 1.5 ).
• Compute ( f(c_1) = f(1.5) = (1.5)^3 – 4(1.5)^2 – 4(1.5) + 16 = 1.375 > 0 ).
• Update: Since ( f(c_1) > 0 ), set ( a = c_1 = 1.5 ), ( b = 3.75 ).

1. Second iteration:

• Midpoint: ( c_2 = \frac{a + b}{2} = \frac{1.5 + 3.75}{2} = 2.625 ).
• Compute ( f(c_2) = f(2.625) = (2.625)^3 – 4(2.625)^2 – 4(2.625) + 16 = -3.78515625 < 0 ).
• Update: Since ( f(c_2) < 0 ), set ( a = 1.5 ), ( b = c_2 = 2.625 ).

1. Third iteration:

• Midpoint: ( c_3 = \frac{a + b}{2} = \frac{1.5 + 2.625}{2} = 2.0625 ).
• Compute ( f(c_3) = f(2.0625) = (2.0625)^3 – 4(2.0625)^2 – 4(2.0625) + 16 = -0.31494140625 < 0 ).
• Update: Since ( f(c_3) < 0 ), set ( a = 1.5 ), ( b = c_3 = 2.0625 ).

1. Fourth iteration:

• Midpoint: ( c_4 = \frac{a + b}{2} = \frac{1.5 + 2.0625}{2} = 1.78125 ).
• Compute ( f(c_4) = f(1.78125) = (1.78125)^3 – 4(1.78125)^2 – 4(1.78125) + 16 = 0.41693115234375 > 0 ).
• Update: Since ( f(c_4) > 0 ), set ( a = c_4 = 1.78125 ), ( b = 2.0625 ).

Result:

After 4 iterations, the root lies in ([1.78125, 2.0625]). The approximate root is the midpoint ( c_4 = 1.78125 ).

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Explanation:

The Bisection Method is a numerical approach to find a root of a function by iteratively halving the interval where the root lies. At each step:

1. Compute the midpoint of the interval.
2. Evaluate the function at the midpoint.
3. Update the interval based on the sign of the function at the midpoint.

Since the function ( f(r) ) is continuous, and ( f(a) \cdot f(b) < 0 ), the Intermediate Value Theorem guarantees a root exists in the interval. Repeated halving reduces the interval’s size, narrowing closer to the root.