y´´+6y´+13y=0 ; y1= e−3xcos2x, y2= e−3xsen2x; y(0)=2,y(0)=0

We are given the second-order linear homogeneous differential equation:

\[
y'' + 6y' + 13y = 0
\]

We also know the general solution to the corresponding homogeneous equation, where \( y_1 = e^{-3x} \cos 2x \) and \( y_2 = e^{-3x} \sin 2x \) are two linearly independent solutions. The general solution to the differential equation can be written as:

\[
y(x) = c_1 y_1 + c_2 y_2 = c_1 e^{-3x} \cos 2x + c_2 e^{-3x} \sin 2x
\]

where \( c_1 \) and \( c_2 \) are constants that we need to determine using the initial conditions. The initial conditions provided are:

1. \( y(0) = 2 \)
2. \( y'(0) = 0 \)

### Step 1: Apply the initial condition \( y(0) = 2 \)

Substitute \( x = 0 \) into the general solution:

\[
y(0) = c_1 e^0 \cos 0 + c_2 e^0 \sin 0 = c_1 (1) (1) + c_2 (1) (0) = c_1
\]

So, \( c_1 = 2 \).

### Step 2: Differentiate the general solution to find \( y'(x) \)

Now, differentiate \( y(x) \) with respect to \( x \):

\[
y'(x) = \frac{d}{dx} \left( c_1 e^{-3x} \cos 2x + c_2 e^{-3x} \sin 2x \right)
\]

Using the product rule and chain rule:

\[
y'(x) = c_1 \left( -3e^{-3x} \cos 2x - 2e^{-3x} \sin 2x \right) + c_2 \left( -3e^{-3x} \sin 2x + 2e^{-3x} \cos 2x \right)
\]

Simplifying:

\[
y'(x) = e^{-3x} \left[ c_1 (-3 \cos 2x - 2 \sin 2x) + c_2 (-3 \sin 2x + 2 \cos 2x) \right]
\]

### Step 3: Apply the initial condition \( y'(0) = 0 \)

Substitute \( x = 0 \) into \( y'(x) \):

\[
y'(0) = e^0 \left[ c_1 (-3 \cos 0 - 2 \sin 0) + c_2 (-3 \sin 0 + 2 \cos 0) \right]
\]

\[
y'(0) = \left[ c_1 (-3 \cdot 1 - 2 \cdot 0) + c_2 (-3 \cdot 0 + 2 \cdot 1) \right]
\]

\[
0 = -3c_1 + 2c_2
\]

Substitute \( c_1 = 2 \) from the earlier step:

\[
0 = -3(2) + 2c_2
\]

\[
0 = -6 + 2c_2
\]

\[
2c_2 = 6 \quad \Rightarrow \quad c_2 = 3
\]

### Final Solution

Thus, the particular solution to the differential equation is:

\[
y(x) = 2e^{-3x} \cos 2x + 3e^{-3x} \sin 2x
\]

This is the solution that satisfies the given initial conditions.

---

Would you like further details or have any questions? Here are five related problems for you to explore:

1. Solve \( y'' + 4y' + 5y = 0 \) with initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \).
2. Find the general solution to \( y'' - 2y' + 5y = 0 \).
3. Use the method of undetermined coefficients to solve \( y'' + 6y' + 13y = 10e^{-3x} \).
4. Solve \( y'' + 9y = 0 \) and find the solution if \( y(0) = 0 \) and \( y'(0) = 1 \).
5. Show that \( y_1 = e^{-3x} \cos 2x \) and \( y_2 = e^{-3x} \sin 2x \) are linearly independent.

**Tip:** To determine whether two solutions are linearly independent, you can check their Wronskian determinant.

Learn how to solve the second-order linear homogeneous differential equation y'' + 6y' + 13y = 0 using initial conditions y(0) = 2 and y'(0) = 0. This solution explains key mathematical concepts, such as product and chain rules, and applies the existence and uniqueness theorem.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation