Integrating factor of the differential equation 2coshxcosydx= sinhxsinydy

To solve the differential equation given by \(2 \cosh x \cos y \, dx = \sinh x \sin y \, dy\), we first need to rearrange it into a more manageable form. This equation can be expressed as a first-order differential equation, which we can analyze for potential integrating factors.

Rearranging the Equation

We can rewrite the equation in the standard form:

\( \frac{dy}{dx} = \frac{2 \cosh x \cos y}{\sinh x \sin y} \)

This form allows us to identify the functions involved more clearly. The left-hand side is simply \(dy/dx\), while the right-hand side is a function of both \(x\) and \(y\).

Identifying the Integrating Factor

To find an integrating factor, we look for a function \( \mu(x, y) \) such that multiplying the entire equation by \( \mu \) makes it exact. An equation is exact if it can be expressed as the derivative of a single function. The general form of an exact equation is:

\( M(x, y) \, dx + N(x, y) \, dy = 0 \)

Here, \( M(x, y) = -\sinh x \sin y \) and \( N(x, y) = 2 \cosh x \cos y \).

Checking for Exactness

To check if the equation is exact, we compute the partial derivatives:

• \( \frac{\partial M}{\partial y} = -\sinh x \cos y \)
• \( \frac{\partial N}{\partial x} = 2 \sinh x \cos y \)

Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is not exact. Therefore, we need to find an integrating factor.

Finding the Integrating Factor

For equations of the form \( M(x, y) \) and \( N(x, y) \), a common approach is to look for an integrating factor that is a function of either \( x \) or \( y \) alone. In this case, we can try \( \mu(y) = \sin y \) or \( \mu(x) = \cosh x \). Let's test \( \mu(y) = \sin y \):

Multiplying the entire equation by \( \sin y \) gives:

\( -\sinh x \sin^2 y \, dx + 2 \cosh x \cos y \sin y \, dy = 0 \)

Now, we check for exactness again:

• \( \frac{\partial}{\partial y}(-\sinh x \sin^2 y) = -\sinh x (2 \sin y \cos y) \)
• \( \frac{\partial}{\partial x}(2 \cosh x \cos y \sin y) = 2 \sinh x \cos y \sin y \)

Now, we see that \( -\sinh x (2 \sin y \cos y) \) and \( 2 \sinh x \cos y \sin y \) are equal, confirming that the modified equation is now exact.

Solving the Exact Equation

To solve the exact equation, we can integrate \( M \) with respect to \( x \) and \( N \) with respect to \( y \). The solution will be of the form:

\( F(x, y) = C \), where \( C \) is a constant.

Integrating \( M \) gives:

\( F(x, y) = -\sinh x \sin^2 y + g(y) \)

Next, we differentiate \( F \) with respect to \( y \) and set it equal to \( N \):

\( \frac{\partial F}{\partial y} = -\sinh x (2 \sin y \cos y) + g'(y) = 2 \cosh x \cos y \sin y \)

From this, we can solve for \( g'(y) \) and integrate to find \( g(y) \). Finally, we combine everything to find the general solution of the original differential equation.

Final Thoughts

By following these steps, we can effectively find the integrating factor and solve the differential equation. This method not only helps in solving this particular equation but also provides a systematic approach to tackle similar problems in differential equations.