1. ## differential equations

I have begun studying differential equations and I think I understand problems involving two variables of x and y, but the first problems they have listed involve e

dy/dx = e^x+y

Ok, I would think that you would do something like this:

dy/dx -e^x = dy/dx e^y

but really am unsure...

2. ## it is a first-order linear DE

it is a first-order linear DE

3. Originally Posted by confusedagain
I have begun studying differential equations and I think I understand problems involving two variables of x and y, but the first problems they have listed involve e

dy/dx = e^x+y

Ok, I would think that you would do something like this:

dy/dx -e^x = dy/dx e^y

but really am unsure...
This is an inhomogeneous linear 1-st order ODE. Its general solution is
the sum of the general solution to the homogeneous equation:

$
\frac{dy}{dx}-y=0
$

and a particular integral of:

$
\frac{dy}{dx}-y=e^x
$

We solve the homogeneous equation by using a trial solution $y=e^{\lambda x}$, which when substituted into the equation gives us the
indicial equation:

$
\lambda - 1=0
$

So the general solution of the homogeneous equation is $y=Ae^{x}$.

Now as the right hand side of the inhomogeneous equation is in fact a solution of the homogeneous equation tradition dictates that we try $y=xe^x$ out as a particular solution. In this case:

$
\frac{d}{dx}\left(xe^x\right)=e^x + xe^x
$

so $y=xe^x$ is indeed a particular integral as expected, so the general solution to the original ODE is:

$
y=Ae^x + xe^x=x^x(A+x)
$

RonL

4. Originally Posted by curvature
it is a first-order linear DE
Please try to keep the width of attacked images to less than 801 pixels,
so that the image can be displayed without the need to scroll on smaller
screens.

RonL

5. Originally Posted by CaptainBlack
Please try to keep the width of attacked images to less than 801 pixels,
so that the image can be displayed without the need to scroll on smaller
screens.

RonL
Thanks for your advice. However I don't know how to control the width. Please tell me how.

curvature

6. Originally Posted by curvature
Thanks for your advice. However I don't know how to control the width. Please tell me how.

curvature
Well you can start by not reattaching a larger image after I have resized one for you.

Copy in image into an image processing tool, any photo editor will do, and
use the resize tool to set the size to something smaller, then save.

RonL

7. Originally Posted by CaptainBlack
Well you can start by not reattaching a larger image after I have resized one for you.

Copy in image into an image processing tool, any photo editor will do, and
use the resize tool to set the size to something smaller, then save.

RonL
The images are directly saved as JPEG format from PPT slides. The images are a little bit wider than expected. However the main page can be seen without scrolling the screen.

8. ## the solution

this image is 800by600

9. You could just learn the LaTeX, which isn't all that hard. For example, click on the equation below:
$y = e^{-\int P(x) dx} \left ( \int Q(x) e^{\int P(x) dx} dx + C \right )$

-Dan

10. Originally Posted by confusedagain
dy/dx = e^x+y
I've got a doubt

Is $\frac{dy}{dx}=e^{x+y}$ or $\frac{dy}{dx}=e^x+y$ ??

In the first case, would be an equation with separable variables, and the second case like they said before, a linear ODE.

11. Anyway, I'll solve $
\frac{dy}{dx}=e^x+y
$

$\frac{dy}{dx}=e^x+y\iff{y'-y=e^x}$

Let's find an integrating factor defined by $\mu(x)=\exp\left(-\int~dx\right)=e^{-x}$

Now multiplying the equation by this factor yields

\begin{aligned}
e^{-x}y'-e^{-x}y~&=~1\\
(e^{-x}y)'~&=~1\\
e^{-x}y~&=~\int~dx\\
e^{-x}y~&=~x+c\\
y~&=~xe^x+ce^x
\end{aligned}

12. ## clarification of problem

dy/dx = e^(x+y)

13. Well, in that case, we have that $\frac{dy}{dx}=e^{x+y}\iff\frac{dy}{dx}=e^x\cdot{e^ y}\implies\frac1{e^y}~dy=e^x~dx$

Can you take it from there?