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I'm trying to solve $\sin(x)\frac{d}{dx}\beta \left ( x \right )+\cos(x)\beta (x)=1$ What I get is : $\beta (x)=\beta \left ( \alpha \right )e^{\sin(\alpha )-\sin(x)}+e^{-\sin(x)}\int_{\alpha }^{x}e^{\sin(t)}dt$ But I think that this solution is incorrect .The textbook says that there's exactly one solution that has a finite limit as $x$ tends to $0$ . But all the solutions I get have a finite limit . So what's the correct solution?

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    @Logan, thanks, I need thicker glasses and a new brain :)2012-11-03

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$\sin(x) \beta'(x) + \cos(x) \beta(x) = \sin(x) \beta'(x) + (\sin(x))' \beta(x) = 1$. Hence, $(\sin(x) \beta(x))' = 1 \implies \sin(x) \beta(x) = x + c \implies \beta(x) = \dfrac{x+c}{\sin(x)}$ Now as $x \to 0$, we want $\lim_{x \to 0} \beta(x)$ to exist i.e. $\lim_{x \to 0} \dfrac{x+c}{\sin(x)}$ exists i.e. $\lim_{x \to 0} \dfrac{x+c}{\sin(x)} = 1 + \lim_{x \to 0} \dfrac{c}{\sin(x)} \,\,\,\,\,\,\, \text{exists}.$ Hence, $c=0$ and therefore $\beta(x) = \dfrac{x}{\sin(x)}$.

EDIT

To proceed through your way, we have that $\beta'(x) + \cot(x) \beta(x) = \csc(x)$ Hence, the integrating factor $I(x) = \exp \left(\displaystyle \int M(y) dy\right) = \exp(\log(\sin(x))) = \sin(x)$. Hence, the solution is $\beta(x) \sin(x) = \int \csc(y) \sin(y) dy = x + c$

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    Yes , I used the integrating factor method .2012-11-03