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I was interested in getting the interval of validities of the third-order differential equation $(\sin x)y'''-3xy''+2y = \tan x$. I just want to list those intervals. That explains why I don't have an initial condition like $y(0) = 4$.

My work

Modifying the differential equation to look like this:

$$y'''+p(x)y''+q(x)y'+r(x)y = g(t)$$

It becomes...

$$(\sin x)y'''-3xy''+2y = \tan x$$ $$\frac{(\sin x)y'''-3xy''+2y = \tan x}{\sin x}$$ $$y'''-\frac{3x}{\sin x} y'' + \frac{2}{\sin x}y = \frac{\tan x}{\sin x}$$ $$y'''-\frac{3x}{\sin x} y'' + \frac{2}{\sin x}y = \frac{1}{\cos x}$$

To determine the intervals, we need to see where the $p(x)$ , $r(x)$, and $g(x)$ to be continuous. We must avoid those points that render $p(x)$ , $r(x)$, and $g(x)$ to be discontinuous. Those points we must avoid are:

Setting $\sin x$ to be zero makes $p(x)$ and $r(x)$ to be discontinuous. Sine function goes to zero every $2\pi n$ interval.

Setting $\cos x$ to be zero makes $g(x)$ to be discontinuous. Cosine function goes to zero every $(2n+1)\pi$ intervals.

Therefore, the intervals of the differential equation $(\sin x)y'''-3xy''+2y = \tan x$ would be $\infty < x < 2\pi n$, $2\pi n < x < (2n+1)\pi$, and $(2n+1)\pi < x < \infty$. These are where the $p(x)$ , $r(x)$, and $g(x)$ to be continuous.

Did I get these intervals right?

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  • $\begingroup$ No, your solutions for $\sin x = 0$ and $\cos x = 0$ are not correct. $\endgroup$ Commented Dec 29, 2017 at 8:54
  • $\begingroup$ the intervals where sine and cosine doesnt have a zero are of the kind $(k\pi/2,\pi/2+k\pi/2)$ for all $k\in\Bbb Z$. $\endgroup$
    – Masacroso
    Commented Dec 29, 2017 at 8:58

1 Answer 1

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Note that $$\sin x = 0 \implies x = \color{red}{n}\pi\,, n \in \mathbb Z$$ and $$\cos x =0 \implies x =(2n+1)\color{blue}{\frac{\pi}{2}}\,, n\in \mathbb Z$$ So, the required intervals should be $$\mathbb R \setminus \{n\pi \mid \, n \in \mathbb Z\} \setminus \{(2n+1)\frac{\pi}{2} \mid \, n \in \mathbb Z\}$$

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  • $\begingroup$ oh..you mean....$\infty < x < n\pi $, $ n\pi < x < (2n+1)\frac{\pi}{2}$, and $(2n+1)\frac{\pi}{2} < x < \infty$? $\endgroup$ Commented Dec 29, 2017 at 9:01
  • $\begingroup$ @PalautotKa No, I mean the real number line excluding the points of the form $2n\pi$ and $(2n+1)\pi/2$. Understand like this. $x$ is a function of $n$ that when you put an integer value outputs in terms of pi. You cannot then define $\infty < x < n\pi$. Understood? This doesn’t make sense, no? $\endgroup$
    – user371838
    Commented Dec 29, 2017 at 9:07
  • $\begingroup$ A little.....The expression $\mathbb R \setminus \{n\pi \mid \, n \in \mathbb Z\} \setminus \{(2n+1)\frac{\pi}{2} \mid \, n \in \mathbb Z\}$....what is the translation into words? $\endgroup$ Commented Dec 29, 2017 at 10:36
  • $\begingroup$ @PalautotKa It refers to the fact that the solution set is $\mathbb R$ minus than the values $2n\pi$ and $(2n+1)\frac{\pi}{2}$. It is the difference of sets. $\endgroup$
    – user371838
    Commented Dec 29, 2017 at 10:38
  • $\begingroup$ You mean it has interval ranging from $-\infty $ to $\infty$, excluding $2n\pi$ and $(2n+1)\frac{\pi}{2}$? $\endgroup$ Commented Dec 29, 2017 at 10:45

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