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.