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).

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

Correct Answer: 5.7

The given equation is \( y'' + 0.8y' + 0.16y = 0 \). This is a second-order linear homogeneous differential equation. First, find the characteristic equation: \[ m^2 + 0.8m + 0.16 = 0. \] Solving for \( m \), we get: \[ m = -0.4, -0.4 \quad {(repeated root)}. \] Thus, the solution is of the form: \[ y = (c_1 + c_2x)e^{-0.4x}. \] Given \( y(0) = 3 \) and \( y'(0) = 4.5 \), we can solve for \( c_1 \) and \( c_2 \): \[ y(0) = c_1 = 3, \] and \[ y'(0) = c_1(-0.4) + c_2 = 4.5 \quad \Rightarrow \quad -1.2 + c_2 = 4.5 \quad \Rightarrow \quad c_2 = 5.7. \] Thus, \[ y = (3 + 5.7x)e^{-0.4x}. \] Now, to find \( y(1) \), substitute \( x = 1 \): \[ y(1) = (3 + 5.7 \times 1)e^{-0.4} \approx 5.83. \] Thus, the correct answer is \( y(1) \approx 5.83 \).