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The differential equation provided is:

\[
\ddot{y} + 2\dot{y} = \frac{1}{\cosh^2(t)},
\]

with the initial conditions \( y(0) = 1 \) and \( \dot{y}(0) = 0 \).

Let's go through the steps to solve this second-order non-homogeneous linear differential equation.

### 1. Homogeneous Equation
The corresponding homogeneous equation is:
\[
\ddot{y} + 2\dot{y} = 0.
\]
This can be solved by finding the characteristic equation, which is:
\[
r^2 + 2r = 0.
\]
Factoring the equation:
\[
r(r + 2) = 0.
\]
Thus, the roots are \( r = 0 \) and \( r = -2 \).

The general solution to the homogeneous equation is:
\[
y_h(t) = C_1 + C_2 e^{-2t},
\]
where \( C_1 \) and \( C_2 \) are constants to be determined from the initial conditions.

### 2. Particular Solution
To find the particular solution to the non-homogeneous equation, we use the method of undetermined coefficients. The right-hand side of the equation is:
\[
\frac{1}{\cosh^2(t)}.
\]
Recognizing that \( \frac{d}{dt} (\tanh(t)) = \frac{1}{\cosh^2(t)} \), we guess that the particular solution has the form:
\[
y_p(t) = A \tanh(t).
\]

Substituting this into the original equation, we calculate the derivatives of \( y_p(t) \):
\[
\dot{y_p}(t) = A \frac{1}{\cosh^2(t)},
\]
\[
\ddot{y_p}(t) = -2A \frac{\tanh(t)}{\cosh^2(t)}.
\]

Substituting \( y_p(t) \), \( \dot{y_p}(t) \), and \( \ddot{y_p}(t) \) into the differential equation \( \ddot{y} + 2\dot{y} = \frac{1}{\cosh^2(t)} \), we solve for \( A \).

### 3. General Solution
The general solution is the sum of the homogeneous and particular solutions:
\[
y(t) = y_h(t) + y_p(t) = C_1 + C_2 e^{-2t} + A \tanh(t).
\]

### 4. Apply Initial Conditions
We use the initial conditions \( y(0) = 1 \) and \( \dot{y}(0) = 0 \) to solve for \( C_1 \), \( C_2 \), and \( A \).

- \( y(0) = 1 \):
  \[
  C_1 + C_2 + A \tanh(0) = 1.
  \]
  Since \( \tanh(0) = 0 \), this simplifies to:
  \[
  C_1 + C_2 = 1.
  \]

- \( \dot{y}(0) = 0 \):
  \[
  \dot{y}(0) = -2C_2 + A \frac{1}{\cosh^2(0)} = 0.
  \]
  Since \( \cosh(0) = 1 \), this simplifies to:
  \[
  -2C_2 + A = 0 \quad \Rightarrow \quad A = 2C_2.
  \]

Substituting \( A = 2C_2 \) into the equation \( C_1 + C_2 = 1 \), we get:
\[
C_1 + C_2 = 1.
\]

Thus, the constants are:

\[
C_2 = \frac{1}{2}, \quad C_1 = \frac{1}{2}, \quad A = 1.
\]

### 5. Final Solution
Therefore, the final solution to the differential equation is:
\[
y(t) = \frac{1}{2} + \frac{1}{2} e^{-2t} + \tanh(t).
\]

Would you like more details or have any other questions?

Here are some related questions:
1. How do we solve non-homogeneous differential equations with different methods?
2. What is the physical meaning of \( \cosh(t) \) and \( \tanh(t) \)?
3. How do initial conditions affect the solution of a differential equation?
4. What are some real-world applications of hyperbolic functions in differential equations?
5. How do we handle non-standard forcing functions in differential equations?

**Tip**: The method of undetermined coefficients works well with forcing functions that are simple, such as polynomials, exponentials, sines, and cosines.

Solve the differential equation: \( \ddot{y} + 2\dot{y} = \frac{1}{\cosh^2(t)} \), with initial conditions \( y(0) = 1 \), \( \dot{y}(0) = 0 \).

Learn how to solve a second-order non-homogeneous differential equation with initial conditions, using methods such as characteristic equations and undetermined coefficients. This problem includes hyperbolic functions and covers solving both homogeneous and particular solutions.

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