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The given differential equation is:

\[
\frac{dy}{dx} + y \tanh(x) = \text{sech}^3(x)
\]

We are also provided with the initial condition \( y(0) = 0 \) and are asked to find \( y(\ln(2)) \).

### Step 1: Solve the Differential Equation
This is a first-order linear ordinary differential equation (ODE) of the form:

\[
\frac{dy}{dx} + P(x) y = Q(x)
\]

where \( P(x) = \tanh(x) \) and \( Q(x) = \text{sech}^3(x) \).

The solution to a first-order linear ODE can be found using an integrating factor \( \mu(x) \), which is given by:

\[
\mu(x) = e^{\int P(x) dx}
\]

Let's compute the integrating factor.

\[
\mu(x) = e^{\int \tanh(x) dx}
\]

Since \( \int \tanh(x) dx = \ln(\cosh(x)) \), we have:

\[
\mu(x) = e^{\ln(\cosh(x))} = \cosh(x)
\]

### Step 2: Multiply the ODE by the Integrating Factor
Multiplying both sides of the differential equation by \( \mu(x) = \cosh(x) \):

\[
\cosh(x) \frac{dy}{dx} + y \cosh(x) \tanh(x) = \cosh(x) \text{sech}^3(x)
\]

The left-hand side simplifies to:

\[
\frac{d}{dx} \left( y \cosh(x) \right)
\]

The right-hand side simplifies to:

\[
\cosh(x) \cdot \text{sech}^3(x) = \text{sech}^2(x)
\]

So, the equation becomes:

\[
\frac{d}{dx} \left( y \cosh(x) \right) = \text{sech}^2(x)
\]

### Step 3: Integrate Both Sides
Now, integrate both sides with respect to \( x \):

\[
y \cosh(x) = \int \text{sech}^2(x) dx
\]

Since \( \int \text{sech}^2(x) dx = \tanh(x) \), we get:

\[
y \cosh(x) = \tanh(x) + C
\]

where \( C \) is the constant of integration.

### Step 4: Apply the Initial Condition
We are given that \( y(0) = 0 \). At \( x = 0 \), we have:

\[
0 \cdot \cosh(0) = \tanh(0) + C
\]

Since \( \tanh(0) = 0 \) and \( \cosh(0) = 1 \), this simplifies to:

\[
0 = C
\]

Thus, \( C = 0 \), and the solution becomes:

\[
y \cosh(x) = \tanh(x)
\]

Therefore, the solution to the differential equation is:

\[
y(x) = \frac{\tanh(x)}{\cosh(x)} = \text{sech}(x)
\]

### Step 5: Find \( y(\ln(2)) \)
Now we need to compute \( y(\ln(2)) \):

\[
y(\ln(2)) = \text{sech}(\ln(2))
\]

We can use the identity \( \cosh(\ln(2)) = \frac{5}{4} \). Thus:

\[
y(\ln(2)) = \frac{1}{\cosh(\ln(2))} = \frac{1}{\frac{5}{4}} = \frac{4}{5}
\]

### Final Answer:
\[
y(\ln(2)) = \frac{4}{5}
\]

Would you like further details or have any questions?

Here are some related questions you might find useful:
1. What is the general method to solve a first-order linear ODE?
2. How does the integrating factor help solve linear differential equations?
3. What are the hyperbolic trigonometric identities used in solving this ODE?
4. How does the initial condition affect the solution to an ODE?
5. How do you compute the values of hyperbolic functions at logarithmic points?

**Tip:** When solving first-order linear ODEs, always check if multiplying by the integrating factor simplifies the equation into an easily integrable form.

Consider the differential equation \(\frac{dy}{dx} + y \tanh(x) = \text{sech}^3(x)\). If \(y(0) = 0\), then find \(y(\ln(2))\)?

Explore how to solve a first-order linear differential equation involving hyperbolic functions. Learn how to find the solution using the integrating factor method and apply an initial condition to compute the value of the function at a specific point.

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