# Solving ODE y' = ye^(x+y)

1. Nov 4, 2009

### brandy

1. The problem statement, all variables and given/known data
dy/dx=y*e^(x+y)
solve for y

3. The attempt at a solution
move the y's over to the other side:
dy/(y*e^y)=e^x.dx
then you integrate and this is where i get screwed up.
i use inverse product rule where u=y and v'=e^y, and i also tried u=e^y and v'=y. but this didnt get me anywhere
so i did it again for the integral on the right hand side for each and it still didnt get me anywhere.
HELP!
also, i doubt if id be able to put it back into y= forms so yea could you help with that too? if not it doesnt really matter. primarily its just the integrating i want help with.

2. Nov 4, 2009

### Gib Z

Re: dy/dx=y*e^(x+y)

It's not solvable in terms of elementary functions, but can be done with a special Function. You may have seen the the more common problem $$\int \frac{e^x}{x} dx$$. It is defined to be equal to the Exponential Integral function. It's most common definition is $$Ei (x) = \int^x_{-\infty} \frac{e^t}{t} dt$$.

Can you see how the integral you have is a certain case of that function?

3. Nov 4, 2009

### brandy

Re: dy/dx=y*e^(x+y)

well yea its e^-t
but im a dummy, i have to solve something this in an exam.
i still dont know how to do it.

whats with the Ei(x) part? and how does this give you the integral

4. Nov 4, 2009

### Gib Z

Re: dy/dx=y*e^(x+y)

Basically the function you want to integrate, as well as the one I showed you with the e^x on the numerator, both can't actually be done by any normal methods! It has no simple answer, but since it comes up so often we just denote it with a shorter symbol because its useful! Think about it, using the Fundamental theorem of calculus, differentiate Ei (x), what do you get? =]

So now, express the Integral you have in terms of the Ei function.

5. Nov 5, 2009

### brandy

Re: dy/dx=y*e^(x+y)

blarg! i'm sorry, i still dont get it. i think i need a bigger hint as to how to relate it and then what to do from there. im sorry im so slow, i really am.

6. Nov 5, 2009

### Gib Z

Re: dy/dx=y*e^(x+y)

Its ok, many people seem perturbed by the idea of just "naming" a solution. The short story is, the integral you have to work out Can't be done in any normal way. So if you got that one in a test, you should just leave it because its not doable. They will probably make sure in the actual exam they give you possible ones.

7. Nov 5, 2009

### brandy

Re: dy/dx=y*e^(x+y)

oh wait, so if you differentiate Ei(x) then you get ex/x
but what about when its e-x/x
i'm not sure i understand how to relate it.

but you'd just write that the integral of ex/x=Ei(x)??? or = that thing you wrote. and would you keep it with respect to t or x.

8. Nov 5, 2009

### brandy

Re: dy/dx=y*e^(x+y)

and we will be getting that exact question in the test, i am 110% sure of this. so :S

9. Nov 5, 2009

### Gib Z

Re: dy/dx=y*e^(x+y)

Well, the integral you have isn't perfectly Ei(x), but Ei(-x). Can you see why?

10. Nov 5, 2009

### brandy

Re: dy/dx=y*e^(x+y)

yea i can see why.
so would it be
$$\int e^(-y) /y * dx$$
= Ei (-y) = $$\int$$ $$^{-y}_{-\infty}$$ et /t *dt]

sorry, i dont really know how to use the latex stuff, i have never really used them before.

11. Nov 5, 2009

### Gib Z

Re: dy/dx=y*e^(x+y)

Well not exactly =[ We're getting variables a bit mixed up here. Look at the definition of Ei i gave, x is a variable in the bounds, not the integrand.

So $$Ei (-x) = \int^{-x}_{-\infty} \frac{e^t}{t} dt$$

To get that to look like yours (remember indefinite integrals vary by additive constants) let u=-t and simplify.

12. Nov 5, 2009

### brandy

Re: dy/dx=y*e^(x+y)

huh? i didnt put it in the integrand. how can you simplfy that?
and would it become e^-u /-u becaues its e^t/t or because i have a negative x its e^-t /t and therefore it becomes e^u / -u

i'm so sorry, i apologise for my inept brain uut you are really helping me so thankyou. verily much