Solve: (D + 2) ^ 2 * y = e ^ (2x) - e ^ (-2x)

The given equation is:

\[
(D + 2)^2 \cdot y = e^{2x} - e^{-2x}
\]

### Step 1: Expand the left-hand side
We need to first expand \((D + 2)^2\), where \(D = \frac{d}{dx}\) is a differential operator:

\[
(D + 2)^2 = (D + 2)(D + 2) = D^2 + 4D + 4
\]

Now the equation becomes:

\[
(D^2 + 4D + 4) \cdot y = e^{2x} - e^{-2x}
\]

### Step 2: Solve for \(y\)
Now we solve the differential equation:

\[
(D^2 + 4D + 4)y = e^{2x} - e^{-2x}
\]

We will first find the particular solution \(y_p\) for the non-homogeneous equation, then find the general solution.

### Step 3: Find the particular solution
The right-hand side of the equation is \(e^{2x} - e^{-2x}\), so we try a particular solution of the form:

\[
y_p = Ae^{2x} + Be^{-2x}
\]

Now, compute the derivatives:

\[
y_p' = 2Ae^{2x} - 2Be^{-2x}
\]
\[
y_p'' = 4Ae^{2x} + 4Be^{-2x}
\]

Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the left-hand side of the equation:

\[
(D^2 + 4D + 4) y_p = (4Ae^{2x} + 4Be^{-2x}) + 4(2Ae^{2x} - 2Be^{-2x}) + 4(Ae^{2x} + Be^{-2x})
\]
\[
= (4A + 8A + 4A)e^{2x} + (4B - 8B + 4B)e^{-2x}
\]
\[
= 16Ae^{2x} + 0 \cdot e^{-2x}
\]

For this to match the right-hand side \(e^{2x} - e^{-2x}\), we must have:

\[
16A = 1 \quad \Rightarrow \quad A = \frac{1}{16}
\]
\[
0 = -1 \quad \Rightarrow \quad B = -\frac{1}{16}
\]

Thus, the particular solution is:

\[
y_p = \frac{1}{16}e^{2x} - \frac{1}{16}e^{-2x}
\]

### Step 4: Find the homogeneous solution
The corresponding homogeneous equation is:

\[
(D^2 + 4D + 4)y_h = 0
\]

The characteristic equation is:

\[
r^2 + 4r + 4 = 0
\]
\[
(r + 2)^2 = 0
\]

So, the solution to the homogeneous equation is:

\[
y_h = (C_1 + C_2x)e^{-2x}
\]

### Step 5: General solution
The general solution is the sum of the particular and homogeneous solutions:

\[
y = y_h + y_p = (C_1 + C_2x)e^{-2x} + \frac{1}{16}(e^{2x} - e^{-2x})
\]

This is the solution to the differential equation.

---

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

1. What are the characteristic roots of second-order linear differential equations?
2. How does one approach non-homogeneous differential equations with exponential terms?
3. Can you solve a differential equation with trigonometric functions on the right-hand side?
4. How do we apply the method of undetermined coefficients?
5. How does the solution change if the differential operator has complex roots?

**Tip:** The method of undetermined coefficients works well when the non-homogeneous term involves exponentials, polynomials, or trigonometric functions.

This page provides a step-by-step solution to the differential equation (D + 2)^2 * y = e^(2x) - e^(-2x), covering concepts such as differential operators, homogeneous and particular solutions, and the method of undetermined coefficients. Ideal for college-level students learning about advanced differential equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation