Question

If \( y(x) \) is the solution of the initial value problem \[ \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} + 4y = 0, \quad y(0) = 2, \quad \frac{dy}{dx}(0) = 0, \] then \( y(\ln 2) \) is (round off to 2 decimal places) equal to ...............

Hint:

When solving second-order linear differential equations with constant coefficients, use the characteristic equation to find the general solution and apply initial conditions to find the specific solution.

Answer 1

Correct Answer: 1.12

Step 1: Solve the characteristic equation.
We start by writing the characteristic equation for the given second-order linear homogeneous differential equation: \[ r^2 + 4r + 4 = 0. \] Factoring the quadratic equation: \[ (r + 2)^2 = 0. \] Thus, the repeated root is \( r = -2 \).
Step 2: General solution to the differential equation.
For a second-order differential equation with repeated roots, the general solution is: \[ y(x) = (C_1 + C_2 x) e^{-2x}. \]
Step 3: Apply the initial conditions.
We are given the initial conditions \( y(0) = 2 \) and \( \frac{dy}{dx}(0) = 0 \). Substituting these into the general solution: - At \( x = 0 \), \( y(0) = C_1 + C_2 \cdot 0 = C_1 = 2 \). So, \( C_1 = 2 \). Now, we compute the derivative of \( y(x) \): \[ \frac{dy}{dx} = (C_2 - 2C_1 - 2C_2 x) e^{-2x}. \] At \( x = 0 \), \( \frac{dy}{dx}(0) = C_2 - 2C_1 = 0 \), so: \[ C_2 - 2 \cdot 2 = 0 \quad \Rightarrow \quad C_2 = 4. \] Thus, the solution is: \[ y(x) = (2 + 4x) e^{-2x}. \]
Step 4: Evaluate \( y(\ln 2) \).
Now, substitute \( x = \ln 2 \) into the solution: \[ y(\ln 2) = (2 + 4 \ln 2) e^{-2 \ln 2}. \] Using the property \( e^{\ln a} = a \), we simplify: \[ y(\ln 2) = (2 + 4 \ln 2) \cdot \frac{1}{4} = \frac{2 + 4 \ln 2}{4}. \] Using the approximate value \( \ln 2 \approx 0.6931 \), we get: \[ y(\ln 2) = \frac{2 + 4 \cdot 0.6931}{4} = \frac{2 + 2.7724}{4} = \frac{4.7724}{4} = 1.1931. \]
Step 5: Conclusion.
Thus, \( y(\ln 2) \) is approximately: \[ \boxed{1.19}. \]