Question

Given a function \( y(x) \) satisfying the differential equation \[ y'' - 0.25y = 0, \] with initial conditions \( y(0) = 1 \); \( y'(0) = 1 \), what is the value of \( y(\log_e 100) \)? 
here, y' and y'' are the first and second derivatives of y, respectively.
 

• A. 14.05
• B. 14.25
• C. 14.95
• D. 14.65

Hint:

When solving second-order linear differential equations, use the characteristic equation to find the general solution and apply initial conditions to solve for constants.

Answer 1

The Correct Option is C

We are given the second-order linear homogeneous differential equation: \[ y'' - 0.25y = 0 \] Step 1: Solving the differential equation. The characteristic equation for the differential equation is: \[ r^2 - 0.25 = 0 \] Solving for \( r \): \[ r = \pm \frac{1}{2} \] Thus, the general solution to the differential equation is: \[ y(x) = A e^{x/2} + B e^{-x/2} \] Step 2: Using the initial conditions. From the initial condition \( y(0) = 1 \): \[ A e^{0} + B e^{0} = 1 \quad \Rightarrow \quad A + B = 1 \quad {(1)} \] From the initial condition \( y'(0) = 1 \), we first compute the derivative: \[ y'(x) = \frac{A}{2} e^{x/2} - \frac{B}{2} e^{-x/2} \] At \( x = 0 \): \[ \frac{A}{2} - \frac{B}{2} = 1 \quad \Rightarrow \quad A - B = 2 \quad {(2)} \] Now, solving the system of equations (1) and (2): \[ A + B = 1 \] \[ A - B = 2 \] Adding these two equations: \[ 2A = 3 \quad \Rightarrow \quad A = \frac{3}{2} \] Substitute \( A = \frac{3}{2} \) into equation (1): \[ \frac{3}{2} + B = 1 \quad \Rightarrow \quad B = -\frac{1}{2} \] Step 3: Finding \( y(\log_e 100) \). Now that we have the values of \( A \) and \( B \), the solution to the differential equation is: \[ y(x) = \frac{3}{2} e^{x/2} - \frac{1}{2} e^{-x/2} \] Substitute \( x = \log_e 100 \): \[ y(\log_e 100) = \frac{3}{2} e^{\log_e 100 / 2} - \frac{1}{2} e^{-\log_e 100 / 2} \] Since \( \log_e 100 = 2 \log_e 10 \), we have: \[ y(\log_e 100) = \frac{3}{2} \times 10 - \frac{1}{2} \times \frac{1}{10} \] Simplifying: \[ y(\log_e 100) = 15 - 0.05 = 14.95 \] Step 4: Conclusion. The value of \( y(\log_e 100) \) is 14.95, so the correct answer is (C).