(sin x)y''' − 3xy'' + 2y = tan x

Home • Forums • Blogs • Glossary • Recent

About • Contact us • Disclaimer • Privacy Policy • Hosted by WebFaction • Powered by Drupal

About • Contact us • Disclaimer • Privacy Policy • Hosted by WebFaction • Powered by Drupal

Forum posts (unless otherwise specified) licensed under a Creative Commons Licence.

All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.

All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.

You are interested in listing the interval of validities of the third-order differential equation $(\sin x)y'''-3xy''+2y = \tan x$.That explains why you didn't specify initial condition like $y(0) = 4$.

Here's how to get it...

Modifying the differential equation to look like this:

$$y'''+p(x)y''+q(x)y'+r(x)y = g(x)$$

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 $n\pi $ interval.

$$\sin x = 0 \implies x = \color{red}{n}\pi\,, n \in \mathbb Z$$

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

$$\cos x =0 \implies x =(2n+1)\color{blue}{\frac{\pi}{2}}\,, n\in \mathbb Z$$

Therefore, the intervals of the differential equation $(\sin x)y'''-3xy''+2y = \tan x$ would be...

$$\mathbb R \setminus \{n\pi \mid \, n \in \mathbb Z\} \setminus \{(2n+1)\frac{\pi}{2} \mid \, n \in \mathbb Z\}$$

It refers to the fact that the solution set is $\mathbb R$ minus than the values $2nπ$ and $(2n+1)\frac{π}{2}$. It is the difference of sets. It an has interval ranging from $-\infty$ to $\infty$, excluding $2nπ$ and $(2n+1)\frac{π}{2}.$