Question

Find all solutions of \(y^{\prime} \sin x+y \cos x=1\) on the interval \((0, \pi)\). Prove that exactly one of these solutions has a finite limit as \(\mathrm{x} \rightarrow 0\), and another has a finite limit as \(\mathrm{x} \rightarrow \pi\).

Short answer

#Short Answer# The general solution of the given first-order linear differential equation on the interval \((0, \pi)\) is \(y(x) = \frac{x+C}{\sin x}\), where \(C\) is the constant of integration. There is exactly one solution with a finite limit as \(x \rightarrow 0\), which is \(y(x) = \frac{-x^2}{\sin x}\). Similarly, there is exactly one solution with a finite limit as \(x \rightarrow \pi\), which is \(y(x) = \frac{\pi\sin x - x^2}{\sin x}\).

Step-by-step solution

Rewrite the given equation and identify the integrating factor

First, let's rewrite the given differential equation as a standard first-order linear differential equation: $$\frac{dy}{dx}+\frac{\cos x}{\sin x} y= \frac{1}{\sin x}.$$Now our integrating factor will be the exponential of the integral of the coefficient of the dependent variable, \(y\):$$\mu(x) = e^{\int\frac{\cos x}{\sin x} dx}.$$

Calculate the integrating factor

We can now compute the integrating factor:$$\mu(x) = e^{\int\frac{\cos x}{\sin x} dx}=e^{\ln|\sin x|} = |\sin x|.$$For the interval \((0, \pi)\), the sine function is positive, so in this case, the integrating factor is simply $$\mu(x)=\sin x.$$

Multiply the equation by the integrating factor

Now we will multiply the original equation by the integrating factor, \(\sin x\):$$\sin x (\frac{dy}{dx}+\frac{\cos x}{\sin x} y) = \sin x \frac{1}{\sin x}.$$ This simplifies to:$$y^{\prime} \sin^2 x + y \cos x = 1.$$

Integrate both sides

The left-hand side of the equation is now the derivative of the product \(y\sin x\). To find the general solution, we integrate both sides of the equation:$$\int (y^{\prime} \sin^2 x + y \cos x) dx = \int 1 dx.$$ This gives us: $$y\sin x = x + C,$$ where \(C\) is the constant of integration.

Obtain the general solution

Now we will solve for \(y\) to obtain the general solution:$$y(x) = \frac{x+C}{\sin x}$$

Analyze the limits

Now, we will analyze the behavior of the general solution as \(x \rightarrow 0\) and \(x \rightarrow \pi\). As \(x \rightarrow 0\), the sine function tends to 0, so \(y(x) \rightarrow \infty\), unless \(C=-x\). If \(C=-x\), then $$y(x) = \frac{-x^2}{\sin x},$$ and as \(x \rightarrow 0\), we have \(y(x) \rightarrow 0\). Thus, exactly one solution has a finite limit as \(x \rightarrow 0\). As \(x \rightarrow \pi\), the sine function tends to 0 again, and \(y(x) \rightarrow \infty\). But if \(C=\pi-x\), then $$y(x) = \frac{\pi\sin x-x^2}{\sin x},$$ and as \(x \rightarrow \pi\), we have \(y(x) \rightarrow 0\). Thus, exactly one solution has a finite limit as \(x \rightarrow \pi\).

Key concepts

Integrating Factor

When tackling first-order linear differential equations, the method involving an integrating factor is particularly useful. This technique simplifies the process of finding the general solution by transforming the differential equation into a form that can be integrated directly.

An integrating factor, denoted by \(\textbf{µ}(x)\), is a function that, when multiplied by the original differential equation, allows for the terms to be combined into a single derivative of a product of functions. It's derived by taking the exponential of the integral of the coefficient of \(y\), which in our exercise is \(\frac{\cos(x)}{\sin(x)}\).

For the problem at hand, \(\textbf{µ}(x)\) simplifies to \(|\sin(x)|\), and on the specified interval \((0, \pi)\), the sine function is invariably positive, which makes the absolute value unnecessary, thus reducing it to \(\sin(x)\).

Multiplying the entire differential equation by this integrating factor leads to the simplification of the original problem into an integrable form. This allows us to find a function whose derivative with respect to \(x\) gives us the left side of the original equation. In essence, the integrating factor is the key to 'unlocking' the problem and laying the path toward the general solution.

General Solution of Differential Equations

The 'general solution' of a differential equation represents a family of solutions containing an arbitrary constant, which can be determined by applying initial conditions or special constraints. To obtain the general solution, it is necessary to integrate the manipulated form of the differential equation once the integrating factor has been applied.

In our exercise example, after applying the integrating factor \(\sin(x)\), the general solution is expressed as \(y(x) = \frac{x + C}{\sin(x)}\), where \(C\) is the constant of integration. The term 'general' is emphasized here because this solution encompasses all possible solutions to the given differential equation within the interval \((0, \pi)\).

The arbitrary constant \(C\) plays a crucial role as it allows for the customization of the solution to meet specific conditions. When the value of this constant is chosen based on certain criteria -- for instance, boundary conditions or initial values -- the general solution becomes a 'particular solution' tailored to a specific situation.

Limits of Solutions

Understanding the 'limits of solutions' helps to predict the behavior of differential equation solutions as the independent variable approaches a certain value. This concept is particularly crucial when dealing with boundaries or singularities within the domain of interest.

In the context of our exercise, we investigate the behavior of the general solution \(y(x) = \frac{x + C}{\sin(x)}\) as \(x\) approaches 0 and \(\pi\). Due to the nature of the sine function nearing values where it becomes zero, the solutions can tend towards infinity, unless the constant of integration \(C\) is adjusted to counteract this effect.

For a finite limit to exist as \(x\) approaches 0, we require \(C\) to be set to \(-x\), because the numerator and denominator will both approach zero, allowing for the application of L'Hôpital's rule or a similar analysis. Conversely, as \(x\) nears \(\pi\), for the limit to remain finite, \(C\) must be \(\pi - x\). The exploration of these limits is essential to not only satisfy the conditions of the problem but also to gain deeper insights into the behavior of solutions in the vicinity of points of interest within the given interval.