Question

Find all solutions of \(y^{\prime}+y \cot x=2 \cos x\) on the interval \((0, \pi)\). Prove that exactly one of these is also a solution on \((-\infty, \infty)\)

Short answer

The general solution to the ODE is given by \(y(x) = \frac{\sin^2(x) + C_1}{\sin(x)}\). There is exactly one solution that is valid on the entire real line, as the solution is undefined at the multiples of \(\pi\) except for \(x = \frac{\pi}{2}\).

Step-by-step solution

Identify the Integrating Factor

The integrating factor (I.F.) is given by the formula: \(I.F. = e^{\int P(x) dx}\). In our case, \(P(x) = \cot(x)\). So, we have: \(I.F. = e^{\int \cot(x) dx}\). Now, we'll find the integrating factor by computing the integral.

Compute the Integral

Integrate \(\cot(x)\) with respect to x: \(e^{\int \cot(x) dx} = e^{\int \frac{\cos(x)}{\sin(x)} dx} = e^{\ln(\sin(x))}\). Now, we can simplify the expression: \(I.F. = e^{\ln(\sin(x))} = \sin(x)\).

Multiply the Integrating Factor

Now, we'll multiply the integrating factor to both sides of the ODE: \(y^{\prime}\sin(x) + y\cot(x)\sin(x) = 2\cos(x)\sin(x)\). This simplifies to: \(y^{\prime}\sin(x) + y\cos(x) = 2\cos(x)\sin(x)\).

Integrate Both Sides

Integrate both sides with respect to x: \(\int\left(y^{\prime}\sin(x) + y\cos(x)\right) dx = \int 2\cos(x)\sin(x) dx\). Observe that the left side is an exact derivative. We can rewrite it as \(\int(d(y\sin(x))) = \int 2\cos(x)\sin(x) dx\). So, we have: \(y\sin(x) = \int 2\cos(x)\sin(x) dx + C\).

Solve for y(x)

Now, we need to solve for y(x). First, we'll compute the integral: \(\int 2\cos(x)\sin(x) dx = \sin^2(x) + C_1\). Next, rewrite the equation for y(x): \(y(x) = \frac{\sin^2(x) + C_1}{\sin(x)}\).

Check for Solution Validity

Let's analyze the solution for the whole real line: If \(\sin(x) = 0\), the solution is undefined. Since we're interested in solutions on \((-\infty, \infty)\), we should check for values at the multiples of \(\pi\). For those values, the left side of the ODE is zero (because \(\sin(x) = 0\)), and the right side is \(\cos(x)\). For the right side to be zero as well, \(\cos(x)\) needs to be 0 for that particular \(x\), otherwise, the solution won't be valid for the whole real line. Only one such value of \(x\) exists at \(x = \frac{\pi}{2}\). Thus, the general solution to the ODE is: \(y(x) = \frac{\sin^2(x) + C_1}{\sin(x)}\), with exactly one solution being also valid on \((-\infty, \infty)\).