Question

$$ \csc y d x+\sec x d y=0 $$

Short answer

The general solution to the given first-order differential equation \(\csc y \, dx + \sec x \, dy = 0\) is obtained by separating the variables and integrating both sides. The final general solution is \(y = \arccos (\sin x - C)\), where C is the integration constant.

Step-by-step solution

Separation of Variables

To separate the variables, divide both sides by \(\csc y \cdot \sec x\), so we get $$ \frac{dx}{\sec x} = -\frac{dy}{\csc y} $$ Since \(\sec x = \frac{1}{\cos x}\) and \(\csc y = \frac{1}{\sin y}\), we can rewrite the equation as $$ \frac{dx}{\frac{1}{\cos x}} = -\frac{dy}{\frac{1}{\sin y}} $$ This simplifies to $$ \cos x \, dx = - \sin y \, dy $$

Integrate Both Sides

Now we will integrate both sides with respect to their variables. For the left side, we have $$ \int \cos x \, dx $$ And for the right side, we have $$ \int - \sin y \, dy $$ Integrating both sides yields $$ \sin x = \cos y + C $$ where C is the integration constant.

Solve for y

Now we will solve for y, which will give us the general solution to the differential equation. To do this, we'll take the inverse cosine of both sides and get $$ y = \arccos (\sin x - C) $$ This is the general solution to the given differential equation.

Key concepts

Separation of Variables

The technique of separation of variables is a fundamental method used to solve ordinary differential equations (ODEs) where all terms involving one variable are moved to one side of the equation and all terms involving another variable to the opposite side. This method relies on the ability to express the ODE in a form where the two variables can be separated by a simple algebraic manipulation.

For example, consider the ODE \[\begin{equation} \frac{dx}{\frac{1}{\text{cos } x}} = -\frac{dy}{\frac{1}{\text{sin } y}} \br\end{equation}\]\ where we already have an expression in which the differentials of each variable are on opposite sides. Once we reach this point, we can integrate each side separately with respect to its corresponding variable. It's essential that we express the ODE in the separated form accurately, as any mistakes here can lead to incorrect results. This technique is particularly powerful because it simplifies the problem into integrals that are often easier to evaluate.

Integration

Once we have separated our equation into terms of one variable each, we proceed to the integration step. Integrating is a way of finding the function whose derivative is given. It is a process intimately linked to differentiation, the reverse of taking a derivative, and is a crucial tool when solving differential equations. In the given exercise, we are faced with two integrals: \[\begin{equation}\text{left side: } \text{cos } x \text{right side: } -\text{sin } y \br\end{equation}\]\ Using fundamental integration rules, we find the antiderivatives of these functions. The trigonometric integrals are especially common when dealing with differential equations and usually lead to inverse trigonometric functions or logarithmic expressions as solutions. Note that when integrating, we always add a constant of integration, often denoted by 'C', because the derivative of a constant is zero.

Inverse Trigonometric Functions

After integration, we often encounter inverse trigonometric functions. These functions are the inverses of the trigonometric functions and are used to determine the measure of the angle when the value of the trigonometric function is known. To finalize the solution to an ODE, we typically isolate the dependent variable in terms of the independent variable and these constants. When we do this with trigonometric equations, we end up using inverse trigonometric functions, such as arcsin, arccos, and arctan.

For instance, in our exercise, once we've integrated to find \[\begin{equation}\text{sin } x = \text{cos } y + C \br\end{equation}\]\, we proceed to solve for y. To isolate y, we use the inverse cosine function, denoted by \text{arccos}\, which gives us \[\begin{equation} y = \text{arccos } (\text{sin } x - C) \br\end{equation}\]\ Th function \text{arccos}\ is defined as returning the angle whose cosine is the given number. When working with inverse trigonometric functions, it's vital to consider the domain and range, as they have restrictions due to the periodic nature of trigonometric functions.