Can Integration Techniques Solve the ODE y' = ye^(x+y)?

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Homework Statement


dy/dx=y*e^(x+y)
solve for y

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 doesn't really matter. primarily its just the integrating i want help with.
 
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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?
 


well yea its e^-t
but I am a dummy, i have to solve something this in an exam.
i still don't know how to do it.

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


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.
 


blarg! I'm sorry, i still don't get it. i think i need a bigger hint as to how to relate it and then what to do from there. I am sorry I am so slow, i really am.
 


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.
 


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.
 


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


Well, the integral you have isn't perfectly Ei(x), but Ei(-x). Can you see why?
 
  • #10


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

sorry, i don't really know how to use the latex stuff, i have never really used them before.
 
  • #11


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 <br /> Ei (-x) = \int^{-x}_{-\infty} \frac{e^t}{t} dt<br />

To get that to look like yours (remember indefinite integrals vary by additive constants) let u=-t and simplify.
 
  • #12


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
 
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