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[SOLVED] 2.2.3 de with tan x

karush

Well-known member
Jan 31, 2012
2,928
$\tiny{2.2.3}$
$\textsf{find the solution:}$
$$y^\prime+(\tan x)y=\sin {2x} \quad -\pi < x < \pi/2$$
$\textit{find u(x)}$
$$u(x)=\exp\int \tan x \, dx = -e^{\ln(\cos x)}=-\cos x$$


ok just want to see if this first step is :cool:


$\tiny{\color{blue}{From \, Text \, Book: \,Elementary \, Differential \, Equations \, and \, Boundary \, Value \, Problems \,
by: \, William \, Boyce \, and \, Richard \, C. \, DiPrima \,
Rensselaer \, Polytechnic \, Institute, \, 1969}}$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x)\)
 

karush

Well-known member
Jan 31, 2012
2,928

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle -\ln(\cos(x))=\ln\left((\cos(x))^{-1}\right)=\ln(\sec(x))\)
 

karush

Well-known member
Jan 31, 2012
2,928
\(\displaystyle \mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x)\)
$\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:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\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:

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

This can be written as:

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

Integrate:

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

Multiply through by $\cos(x)$:

\(\displaystyle y(x)=-2\cos^2(x)+c_1\cos(x)\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

karush

Well-known member
Jan 31, 2012
2,928
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
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.