MHB Cassi's question at Yahoo Answers regarding a first order linear ODE

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The discussion focuses on solving the first-order linear ordinary differential equation (ODE) given by y' + y cot(x) = 2cos(x) on the interval (0, π). By multiplying through by sin(x) and applying the product rule, the solution is derived as y(x) = sin(x) + C csc(x). It is established that for any non-zero constant C, the solution becomes undefined at x = kπ, where k is an integer. Therefore, the only solution that remains valid across the entire real line is y(x) = sin(x). This analysis confirms that there is exactly one solution to the initial value problem defined on (-∞, +∞).
MarkFL
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Here is the question:

Initial value problem?


Find all the solutions of y'+ycotx=2cosx on the interval (o, pi). Prove that exactly one of these is a solution on (-infinity, +infinity)

I have posted a link there to this thread so the OP can view my work.
 
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Hello Cassi,

We are given the ODE:

$$y'+y\cot(x)=2\cos(x)$$

If we multiply through by $\sin(x)\ne0$ (which we can do given the interval) we obtain:

$$\sin(x)\frac{dy}{dx}+\frac{d}{dx}(\sin(x))y=2\sin(x)\frac{d}{dx}(\sin(x))$$

Observing that we have the product rule on the right, we may integrate as:

$$\int\,d\left(\sin(x)y \right)=2\int \sin(x)\,d(\sin(x))$$

$$\sin(x)y=\sin^2(x)+C$$

And thus, dividing through by $\sin(x)$, we obtain:

$$y(x)=\sin(x)+C\csc(x)$$

Now, for any choice of $C\ne0$, we encounter division by zero for $x=k\pi$ where $k\in\mathbb{Z}$, and so the only solution defined over all the reals is:

$$y(x)=\sin(x)$$
 
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