# [SOLVED]2.2.3 de with tan x

#### karush

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

ok just want to see if this first step is

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

#### MarkFL

Staff member
$$\displaystyle \mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x)$$

#### karush

##### Well-known member
$$\displaystyle \mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x)$$
something fishy????

#### MarkFL

Staff member
$$\displaystyle -\ln(\cos(x))=\ln\left((\cos(x))^{-1}\right)=\ln(\sec(x))$$

#### karush

##### Well-known member
$$\displaystyle \mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x)$$
$\textit{So then distribute$u(x)$}$
$$(\sec x)(y^\prime+(\tan x)y)=(\sec x) \sin {2x}$$
$\textit{hence}$
$$(y \, sec(x))^\prime=(\sec x) \sin {2x}$$
$\textit{then}$
$\displaystyle y \, sec(x) =\int (\sec x) \sin {2x} \, dx=-2\cos(x)+c_1$

$$(c-2x \cos x+2\sin x)\cos x$$
W|A returned this
$$y(x)=c_1\cos(x)-2\cos^2(x)$$

Last edited:

#### MarkFL

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

$$(c-2x \cos x+2\sin x)\cos x$$
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

Staff member
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
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