3
$\begingroup$

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?

  • 0
    Note that you can use \sin, \cos for the $\sin,\cos$ functions. see the difference between $sin(x)$ and $\sin(x)$2012-11-03
  • 0
    Unless $\beta(x)$ is a special function, I see that since $sin^2(x)+cos^2(x)=1$ which makes $\beta(x)=cos(x)$ is a solution.2012-11-03
  • 1
    @EmmadKareem That was my first thought, but it gives $-\sin^2(x)+\cos^2(x)$, notice the "negative sign" (pun!)2012-11-03
  • 0
    @Logan, thanks, I need thicker glasses and a new brain :)2012-11-03

1 Answers 1

7

$\sin(x) \beta'(x) + \cos(x) \beta(x) = \sin(x) \beta'(x) + (\sin(x))' \beta(x) = 1$. 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 $$\beta(x) \sin(x) = \int \csc(y) \sin(y) dy = x + c$$

  • 0
    You can also finish it by observing that if $\lim_{x \to 0} \dfrac{x+c}{\sin(x)}$ exists then, $\lim_{x \to 0} \dfrac{x+c}{\sin(x)}\cdot \sin (x)=0$.2012-11-03
  • 0
    Thanks Marvis , But what about my solution ,Is it wrong?2012-11-03
  • 0
    @Nabil Did you check if it satisfies the equation? I think it doesn't...2012-11-03
  • 0
    Yes ,I tried but I get a complicated expression . I think that my solution is wrong .2012-11-03
  • 0
    But $cos(x)$ is a solution that's finite as $x$ tends to $0$ . So your solution is not the most general right?2012-11-03
  • 0
    @Nabil I don't know how you get your solution. If you are using integrating factor, then I have worked it out in the edit.2012-11-03
  • 0
    @Nabil I don't see how $\cos(x)$ is a solution. Note that $$\sin(x) \cos'(x) + \cos(x) \cos(x) = \color{blue}{-} \sin^2(x) + \cos^2(x)$$2012-11-03
  • 0
    I have made a mistake .I didn't see $sin(x)$ in front of $\beta(x)$ That's why my solution is wrong.2012-11-03
  • 0
    Yes , I used the integrating factor method .2012-11-03