$$ \left\{ \begin{array} { l } { y ^ { \prime \prime } - 2 y ^ { \prime } + 5 y = 29 e ^ { - 4 x } } \ { y ( 0 ) = 2 , y ^ { \prime } ( 0 ) = 2 } \end{array} ight. $$

Question

$$ \left\{ \begin{array} { l } { y ^ { \prime \prime } - 2 y ^ { \prime } + 5 y = 29 e ^ { - 4 x } } \ { y ( 0 ) = 2 , y ^ { \prime } ( 0 ) = 2 } \end{array} ight. $$

UpStudy AI · Accepted Answer

To solve the differential equation

$$
y'' - 2y' + 5y = 29e^{-4x}
$$

with the initial conditions $$ y(0) = 2 $$ and $$ y'(0) = 2 $$, we will follow these steps:

1. **Find the complementary solution** $$ y_c $$.
2. **Find a particular solution** $$ y_p $$.
3. **Combine the solutions** and apply the initial conditions.

### Step 1: Find the Complementary Solution

The complementary solution is found by solving the homogeneous equation:

$$
y'' - 2y' + 5y = 0
$$

The characteristic equation is:

$$
r^2 - 2r + 5 = 0
$$

Using the quadratic formula:

$$
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = 1 \pm 2i
$$

Thus, the roots are $$ r = 1 + 2i $$ and $$ r = 1 - 2i $$. The complementary solution is:

$$
y_c = e^{x} (C_1 \cos(2x) + C_2 \sin(2x))
$$

### Step 2: Find a Particular Solution

To find a particular solution $$ y_p $$, we can use the method of undetermined coefficients. Since the right-hand side is $$ 29e^{-4x} $$, we will try a particular solution of the form:

$$
y_p = Ae^{-4x}
$$

Calculating the derivatives:

$$
y_p' = -4Ae^{-4x}
$$
$$
y_p'' = 16Ae^{-4x}
$$

Substituting $$ y_p $$, $$ y_p' $$, and $$ y_p'' $$ into the original equation:

$$
16Ae^{-4x} - 2(-4Ae^{-4x}) + 5(Ae^{-4x}) = 29e^{-4x}
$$

This simplifies to:

$$
16A + 8A + 5A = 29
$$
$$
29A = 29 \implies A = 1
$$

Thus, the particular solution is:

$$
y_p = e^{-4x}
$$

### Step 3: Combine the Solutions

The general solution is:

$$
y = y_c + y_p = e^{x} (C_1 \cos(2x) + C_2 \sin(2x)) + e^{-4x}
$$

### Step 4: Apply Initial Conditions

Now we apply the initial conditions $$ y(0) = 2 $$ and $$ y'(0) = 2 $$.

1. **Using $$ y(0) = 2 $$**:

$$
y(0) = e^{0}(C_1 \cos(0) + C_2 \sin(0)) + e^{0} = C_1 + 1 = 2
$$
$$
C_1 + 1 = 2 \implies C_1 = 1
$$

2. **Using $$ y'(0) = 2 $$**:

First, we need to find $$ y' $$:

$$
y' = \frac{d}{dx} \left( e^{x} (C_1 \cos(2x) + C_2 \sin(2x)) + e^{-4x} \right)
$$

Using the product rule and chain rule:

$$
y' = e^{x} (C_1 \cos(2x) + C_2 \sin(2x)) + e^{x} (-2C_1 \sin(2x) + 2C_2 \cos(2x)) - 4e^{-4x}
$$

At $$ x = 0 $$:

$$
y'(0) = e^{0}(C_1 \cos(0) + C_2 \sin(0)) + e^{0}(-2C_1 \sin(0) + 2C_2 \cos(0)) - 4e^{0}
$$
$$
= C_1 + 2C_2 - 4
$$

Substituting $$ C_1 = 1 $$:

$$
1 + 2C_2 - 4 = 2 \implies 2C_2 - 3 = 2 \implies 2C_2 = 5 \implies C_2 = \frac{5}{2}
$$

### Final Solution

Thus, the solution to the differential equation is:

$$
y = e^{x} \left( \cos(2x) + \frac{5}{2} \sin(2x) \right) + e^{-4x}
$$
The solution to the differential equation is $$ y = e^{x} \left( \cos(2x) + \frac{5}{2} \sin(2x) \right) + e^{-4x} $$.

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