Question

Let \( \varphi : \mathbb{R} \to \mathbb{R} \) be the solution of the differential equation \[ 4 \frac{d^2 y}{dx^2} + 16 \frac{dy}{dx} + 25y = 0 \] satisfying \( \varphi(0) = 1 \) and \( \varphi'(0) = -\frac{1}{2} \). Then, the value of \( \lim_{x \to \infty} e^{2x} \varphi(x) \) is equal to ............ (rounded off to two decimal places).

Hint:

For second-order linear differential equations with constant coefficients, use the characteristic equation to find the general solution. For complex roots, the solution involves exponential decay and oscillations.

Answer 1

Step 1: Solve the differential equation.
The given second-order linear differential equation is: \[ 4 \frac{d^2 y}{dx^2} + 16 \frac{dy}{dx} + 25y = 0. \] Dividing the equation by 4 to simplify: \[ \frac{d^2 y}{dx^2} + 4 \frac{dy}{dx} + \frac{25}{4}y = 0. \] This is a homogeneous linear differential equation with constant coefficients. The characteristic equation is: \[ r^2 + 4r + \frac{25}{4} = 0. \] The discriminant of this quadratic is: \[ \Delta = 4^2 - 4 \cdot 1 \cdot \frac{25}{4} = 16 - 25 = -9. \] Thus, the roots are complex: \[ r = \frac{-4 \pm \sqrt{-9}}{2} = -2 \pm \frac{3i}{2}. \] Therefore, the general solution is: \[ \varphi(x) = e^{-2x} \left( C_1 \cos\left( \frac{3x}{2} \right) + C_2 \sin\left( \frac{3x}{2} \right) \right). \] Step 2: Apply initial conditions.
Using \( \varphi(0) = 1 \), we get: \[ 1 = C_1 e^0 \cos(0) + C_2 e^0 \sin(0) \implies C_1 = 1. \] Next, using \( \varphi'(0) = -\frac{1}{2} \), we differentiate \( \varphi(x) \) to get: \[ \varphi'(x) = e^{-2x} \left( -2 \cos\left( \frac{3x}{2} \right) + C_2 \cdot \frac{3}{2} \cos\left( \frac{3x}{2} \right) - C_2 \cdot \frac{3}{2} \sin\left( \frac{3x}{2} \right) \right). \] At \( x = 0 \), this becomes: \[ -\frac{1}{2} = -2C_1 \implies C_2 = -\frac{1}{2}. \] Thus, the solution is: \[ \varphi(x) = e^{-2x} \left( \cos\left( \frac{3x}{2} \right) - \frac{1}{2} \sin\left( \frac{3x}{2} \right) \right). \] Step 3: Evaluate the limit.
Now, we evaluate the limit \( \lim_{x \to \infty} e^{2x} \varphi(x) \): \[ \lim_{x \to \infty} e^{2x} e^{-2x} \left( \cos\left( \frac{3x}{2} \right) - \frac{1}{2} \sin\left( \frac{3x}{2} \right) \right) = \lim_{x \to \infty} \left( \cos\left( \frac{3x}{2} \right) - \frac{1}{2} \sin\left( \frac{3x}{2} \right) \right) = 0. \] Final Answer: \[ \boxed{0}. \]