Question

Let y be the solution of the initial value problem \( y'' + 0.8y' + 0.16y = 0 \) where \( y(0) = 3 \) and \( y'(0) = 4.5 \). Then, \( y(1) \) is equal to ......... (rounded off to 1 decimal place).

• A. \( 5.83 \)
• B. \( 5.7 \)
• C. \( 5.6 \)
• D. \( 5.5 \)

Hint:

For repeated roots in a differential equation, the solution will include both a constant term and a term involving \( x \), both multiplied by an exponential function.

Answer 1

The Correct Option is A

Given Differential Equation:

\[ y'' + 0.8y' + 0.16y = 0 \]

Step 1: Solve the Characteristic Equation

The characteristic equation is:

\[ m^2 + 0.8m + 0.16 = 0 \]

Solving for \( m \) using the quadratic formula:

\[ m = \frac{-0.8 \pm \sqrt{(0.8)^2 - 4(0.16)}}{2} \]

\[ m = \frac{-0.8 \pm \sqrt{0.64 - 0.64}}{2} = \frac{-0.8 \pm 0}{2} = -0.4, -0.4 \]

Since we have a repeated root \( m = -0.4 \), the general solution is:

\[ y(x) = (c_1 + c_2x)e^{-0.4x} \]

Step 2: Apply Initial Conditions

Given \( y(0) = 3 \) and \( y'(0) = 4.5 \), we substitute \( x = 0 \):

\[ y(0) = (c_1 + c_2(0))e^{0} = c_1 = 3 \]

To find \( c_2 \), first compute \( y'(x) \):

\[ y'(x) = \left( c_1 + c_2x \right)(-0.4e^{-0.4x}) + c_2e^{-0.4x} \]

Substituting \( x = 0 \):

\[ y'(0) = (-0.4c_1 + c_2)e^{0} = -0.4(3) + c_2 = 4.5 \]

\[ -1.2 + c_2 = 4.5 \quad \Rightarrow \quad c_2 = 5.7 \]

Step 3: Compute \( y(1) \)

Substituting \( x = 1 \) into the solution:

\[ y(1) = (3 + 5.7(1))e^{-0.4} \]

\[ y(1) = (3 + 5.7)e^{-0.4} = 8.7 \times 0.6703 \approx 5.83 \]

Final Answer:

\( y(1) \approx 5.83 \).