Chapter 8: Problem 10

Find the general solution of each of the following differential equations. \(y^{\prime}+y \tanh x=2 e^{x}\)

Short Answer

Expert verified

The general solution is \( y = 2e^x \tanh x + C \cdot \text{sech} x \).

Step by step solution

Identify the Form of the Differential Equation

The given differential equation is: \(y' + y \tanh x = 2 e^x\). This is a first-order linear differential equation in the form \(y' + P(x)y = Q(x)\), where \(P(x) = \tanh x\) and \(Q(x) = 2 e^x\).

Determine the Integrating Factor

The integrating factor \(\mu(x)\) is given by \(e^{\int P(x) \, dx}\). Here, \(P(x) = \tanh x\), so we compute:\[ \mu(x) = e^{\int \tanh x \, dx} \]Since \(\int \tanh x \, dx = \ln |\cosh x|\), we have:\[ \mu(x) = e^{\ln |\cosh x|} = \cosh x \]

Multiply the Differential Equation by the Integrating Factor

Multiply both sides of the given differential equation by the integrating factor \(\cosh x\):\[ \cosh x \cdot y' + \cosh x \cdot y \tanh x = 2 e^x \cdot \cosh x \]Since \(\cosh x \cdot \tanh x = \sinh x\), the equation simplifies to:\[ \cosh x \cdot y' + \sinh x \cdot y = 2 e^x \cosh x \]

Rewrite the Left Side as a Derivative

The left side of the equation can be written as the derivative of the product of \(y\) and the integrating factor \(\cosh x\):\[ \frac{d}{dx}(y \cosh x) = 2 e^x \cosh x \]

Integrate Both Sides

Integrate both sides with respect to \(x\):\[ \int \frac{d}{dx}(y \cosh x) \, dx = \int 2 e^x \cosh x \, dx \]The left side simplifies to \(y \cosh x\):\[ y \cosh x = \int 2 e^x \cosh x \, dx \]

Solve the Integral on the Right Side

To solve the integral on the right side, notice that:\(\int 2 e^x \cosh x \, dx\) can be solved by using integration by parts or recognizing it as a standard integral. We get:\[ \int 2 e^x \cosh x \, dx = 2 e^x \sinh x\]Thus,\[ y \cosh x = 2 e^x \sinh x + C \]

Solve for \(y\)

Finally, solve for \(y\) by dividing both sides by \( \cosh x \):\[ y = \frac{2 e^x \sinh x + C}{\cosh x} \]This can be simplified noting that \(\frac{\sinh x}{\cosh x} = \tanh x\):\[ y = 2 e^x \tanh x + C \cdot \text{sech}x \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Factor

First-order linear differential equations can often be solved using an integrating factor. The integrating factor is a function that, when multiplied by the original differential equation, transforms it into an equation that is easier to solve. For a differential equation of the form \(y' + P(x)y = Q(x)\), the integrating factor \(\mu(x)\) is given by:

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

In the specific exercise we are considering, \(P(x) = \tanh x\). The integral of \(\tanh x\) is \(\ln |\cosh x|\), so the integrating factor \(\mu(x)\) is \(e^{\ln |\cosh x|} = \cosh x\). Multiplying through by this integrating factor allows us to rewrite the differential equation in a way that simplifies the process of finding its solution.

Hyperbolic Functions

Understanding hyperbolic functions is essential for solving some differential equations. Hyperbolic functions, similar to trigonometric functions, have essential properties and are defined as follows:

\( \sinh x = \frac{e^x - e^{-x}}{2} \)
\( \cosh x = \frac{e^x + e^{-x}}{2} \)
\( \tanh x = \frac{\sinh x}{\cosh x} \)
\( \sech x = \frac{1}{\cosh x} \)
\( \csch x = \frac{1}{\sinh x} \)
\( \coth x = \frac{1}{\tanh x} \)

In our equation, the term \( \tanh x \) on the left-hand side and terms \( \cosh x \), \( \sinh x \) on both sides are hyperbolic functions. Simplifying hyperbolic expressions is crucial to reaching the general solution.

General Solution of Differential Equations

Finding the general solution to a differential equation involves several steps, with the aim of expressing the dependent variable, typically \(y\), explicitly in terms of the independent variable, typically \(x\), plus an arbitrary constant (reflecting the general nature of the solution).

For the differential equation \(y' + y \tanh x = 2 e^x\) discussed in the exercise, we first identify the integrating factor and proceed through several transformations, utilizing hyperbolic functions and integrating both sides accordingly. The simplified version of the final general solution is:

\[ y = 2 e^x \tanh x + C \cdot \sech x \]

Here, \(C\) represents the integration constant, and it is crucial in describing the family of solutions for the differential equation. Each specific value of \(C\) corresponds to a specific solution path for the given differential equation.