-2.2.3 de y'+(\tan x)y=\sin {2x}; -\pi < x < \pi/2

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SUMMARY

The discussion focuses on solving the first-order linear differential equation \(y' + (\tan x)y = \sin{2x}\) for the interval \(-\pi < x < \frac{\pi}{2}\). The integrating factor is determined to be \(\mu(x) = \sec(x)\), leading to the equation \((y \sec(x))' = \sec(x) \sin{2x}\). Upon integrating, the solution is derived as \(y(x) = -2\cos^2(x) + c_1\cos(x)\), which is confirmed by the use of the W|A tool. The discussion also touches on user experience issues with forum navigation.

PREREQUISITES
  • Understanding of first-order linear differential equations
  • Familiarity with integrating factors in differential equations
  • Knowledge of trigonometric identities and their derivatives
  • Experience with mathematical software tools like Wolfram Alpha (W|A)
NEXT STEPS
  • Study the method of integrating factors in differential equations
  • Learn about trigonometric integrals, specifically \(\int \sec(x) \sin(2x) \, dx\)
  • Explore the application of differential equations in real-world scenarios
  • Investigate user interface design improvements for online forums
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Students and educators in mathematics, particularly those focusing on differential equations, as well as developers and designers interested in enhancing user experience on educational forums.

karush
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$\tiny{2.2.3}$
1000
$\textsf{find the solution:}$
$$y^\prime+(\tan x)y=\sin {2x} \quad -\pi < x < \pi/2$$
$\textit{find u(x)}$
$$u(x)=\exp\int \tan x \, dx = -e^{\ln(\cos x)}=-\cos x$$ok just want to see if this first step is :cool:$\tiny{\color{blue}{From \, Text \, Book: \,Elementary \, Differential \, Equations \, and \, Boundary \, Value \, Problems \,
by: \, William \, Boyce \, and \, Richard \, C. \, DiPrima \,
Rensselaer \, Polytechnic \, Institute, \, 1969}}$
 
Last edited:
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$$\mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x)$$
 
MarkFL said:
$$\mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x)$$

something fishy?

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$$-\ln(\cos(x))=\ln\left((\cos(x))^{-1}\right)=\ln(\sec(x))$$
 
MarkFL said:
$$\mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x)$$

$\textit{So then distribute $u(x)$}$
$$(\sec x)(y^\prime+(\tan x)y)=(\sec x) \sin {2x}$$
$\textit{hence}$
$$(y \, sec(x))^\prime=(\sec x) \sin {2x}$$
$\textit{then}$
$\displaystyle y \, sec(x)
=\int (\sec x) \sin {2x} \, dx=-2\cos(x)+c_1$

ok don't see this heading toward the book answer
$$(c-2x \cos x+2\sin x)\cos x$$
W|A returned this
$$y(x)=c_1\cos(x)-2\cos^2(x)$$
 
Last edited:
karush said:
$\textit{So then distribute $u(x)$}$
$$(\sec x)(y^\prime+(\tan x)y)=(\sec x) \sin {2x}$$
$\textit{hence}$
$$(y \, sec(x))^\prime=(\sec x) \sin {2x}$$
$\textit{then}$
$\displaystyle y \, sec(x)
=\int (\sec x) \sin {2x} \, dx=-2\cos(x)+c_1$

ok don't see this heading toward the book answer
$$(c-2x \cos x+2\sin x)\cos x$$
W|A returned this
$$y(x)=c_1\cos(x)-2\cos^2(x)$$

When you multiply by $\mu(x)$, you get:

$$\sec(x)y'+\tan(x)\sex(x)y=\sec(x)\sin(2x)$$

This can be written as:

$$\frac{d}{dx}\left(\sec(x)y\right)=2\sin(x)$$

Integrate:

$$\sec(x)y=-2\cos(x)+c_1$$

Multiply through by $\cos(x)$:

$$y(x)=-2\cos^2(x)+c_1\cos(x)$$
 
Are you using Tapatalk by any chance? I have code in place to let me know when posts have been edited, so I don't miss added content (it's better to use a new post for new content), but Tapatalk lazily bypasses all my custom code and so I'm thinking that's why I didn't know you had added to your last post. Also, I restored one of the posts, because when it was deleted, then one of my posts didn't make much sense.
 
ok well I really don't like the long threads
when most of the posts are just a few lines and vast majority is irrelevant space
it would be better have a minimize or close out feature rather than delete the posts.
doing endless scrolling just to see the process gets old fast.

with that however i have tried other forums
but this is by far the most user freindly

I thot stack exchange was just a get lost fast jungle
yahoo very hard to read
most have no live view latex
 
I can't imagine trying to use the internet on a telephone. I'm sorry scrolling on a telephone is such a chore...they should fix that.
 

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