The solution $y(x)$ of the differential equation $\frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+4y=0$ satisfying the conditions $y(0)=4, \frac{dy}{dx}(0)=8$ is :

Question

Accepted Answer

To solve the given differential equation and verify that Option 2 is the correct solution, we start by analyzing the differential equation:

$$
\frac{d^{2}y}{dx^{2}} + 4\frac{dy}{dx} + 4y = 0
$$

This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation for this differential equation is obtained by assuming a solution of the form $ y = e^{rx} $, leading to:

$$
r^2 + 4r + 4 = 0
$$

Solving this quadratic equation using the quadratic formula $ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = 4 $, and $ c = 4 $, we find:

$$
r = \frac{-4 \pm \sqrt{16 - 16}}{2} = \frac{-4 \pm 0}{2} = -2
$$

Since we have a repeated root $ r = -2 $, the general solution of the differential equation is:

$$
y(x) = (C_1 + C_2 x)e^{-2x}
$$

We now apply the initial conditions to determine the constants $ C_1 $ and $ C_2 $.

1. **Initial Condition 1:** $ y(0) = 4 $

$$
   y(0) = (C_1 + C_2 \cdot 0)e^{0} = C_1 = 4
   $$

2. **Initial Condition 2:** $ \frac{dy}{dx}(0) = 8 $

First, find the derivative of $ y(x) $:

$$
   \frac{dy}{dx} = \frac{d}{dx}[(C_1 + C_2 x)e^{-2x}] = (C_2 - 2(C_1 + C_2 x))e^{-2x}
   $$

Evaluate at $ x = 0 $:

$$
   \frac{dy}{dx}(0) = (C_2 - 2C_1)e^{0} = C_2 - 2C_1 = 8
   $$

Substitute $ C_1 = 4 $:

$$
   C_2 - 2 \times 4 = 8 \implies C_2 - 8 = 8 \implies C_2 = 16
   $$

Thus, the specific solution satisfying the initial conditions is:

$$
y(x) = (4 + 16x)e^{-2x}
$$

This matches Option 2: $(16x + 4)e^{-2x}$.

**Explanation for Incorrect Options:**

1. **Option 1: $4e^{-2x} + 16e^{2x}$**

This solution does not satisfy the initial conditions. Specifically, the term $16e^{2x}$ does not fit the form of solutions for repeated roots, and it does not satisfy the initial condition for the derivative.

2. **Option 3: $4e^{2x}$**

This solution does not satisfy the differential equation, as it does not account for the repeated root structure of the characteristic equation. It also fails to satisfy the initial conditions.

3. **Option 4: $4e^{-2x} + 16xe^{2x}$**

This solution incorrectly includes a term $16xe^{2x}$, which is not consistent with the solution form for repeated roots. It does not satisfy the initial conditions or the differential equation.

Therefore, Option 2 is the correct solution that satisfies both the differential equation and the given initial conditions.