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

• MHB
• karush
In summary, William Boyce and Richard DiPrima state that sec(x)y'+\tan(x)\sex(x)y=\sec(x)\sin(2x). Sec(x)y'+\tan(x)\sex(x)y=-2\cos(x)+c_1 can be written as y(x)=-2\cos^2(x)+c_1\cos(x) and can be integrated to obtain y(x)=-2\cos^2(x)+c_1.
karush
Gold Member
MHB
$\tiny{2.2.3}$
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 $\tiny{\color{blue}{From \, Text \, Book: \,Elementary \, Differential \, Equations \, and \, Boundary \, Value \, Problems \, by: \, William \, Boyce \, and \, Richard \, C. \, DiPrima \, Rensselaer \, Polytechnic \, Institute, \, 1969}}$

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

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

something fishy?

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

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

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

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

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

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
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.

## 1. What is the given differential equation?

The given differential equation is -2.2.3 de y'+(\tan x)y=\sin {2x}; -\pi < x < \pi/2.

## 2. What is the order of the differential equation?

The order of the differential equation is first order.

## 3. What is the domain of the variable x?

The domain of the variable x is -\pi < x < \pi/2.

## 4. What is the coefficient of the y' term?

The coefficient of the y' term is -2.2.3.

## 5. What is the general solution of the given differential equation?

The general solution of the given differential equation is y(x) = \frac{1}{\cos^2 x + C}, where C is an arbitrary constant.

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