# Integration helpWatch

#1
How do I integrate e^-(1/x)?
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4 years ago
#2
(Original post by Vorsah)
How do I integrate e^-(1/x)?
You can't!

(i.e. it's one of many functions that doesn't have an elementary integral )
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#3
(Original post by davros)
You can't!

(i.e. it's one of many functions that doesn't have an elementary integral )
In that case, can you find my mistake in solving this differential equation because I cant?

The integrating factor is xe^-(1/x)
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4 years ago
#4
Although I have not looked at this in detail, I do not think that the above function is integrable, at least in terms of elementary functions ...
Integration is NOT a guaranteed process.
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4 years ago
#5
(Original post by Vorsah)
In that case, can you find my mistake in solving this differential equation because I cant?

The integrating factor is xe^-(1/x)
Assuming you've done the manipulation OK, you don't need to find the integral you quoted in your first post.
You need to find the integral of (x^-2)(e^-1/x) which looks like recognition / substitution to me
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4 years ago
#6
your integral is not the one you quote but but a recognisable one ...
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#7
(Original post by davros)
Assuming you've done the manipulation OK, you don't need to find the integral you quoted in your first post.
You need to find the integral of (x^-2)(e^-1/x) which looks like recognition / substitution to me
Why can't you do it by parts?
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4 years ago
#8
(Original post by Vorsah)
Why can't you do it by parts?
Well how are you going to do it by parts? As I pointed out first of all, you can't integrate e^-1/x on its own
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#9
(Original post by davros)
Well how are you going to do it by parts? As I pointed out first of all, you can't integrate e^-1/x on its own
Thanks, can you check my final answer?

The power is -1/x btw
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4 years ago
#10
give me a minute to solve the ODE
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4 years ago
#11
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4 years ago
#12
(Original post by Vorsah)
Thanks, can you check my final answer?

The power is -1/x btw
Well you haven't used the boundary condition y(1) = 0 so you can find the value of the constant

Also, if you have a power of -1/x in the denominator of a fraction, you can rewrite more simply as the positive power +1/x.
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#13
(Original post by davros)
Well you haven't used the boundary condition y(1) = 0 so you can find the value of the constant

Also, if you have a power of -1/x in the denominator of a fraction, you can rewrite more simply as the positive power +1/x.
thank you
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#14
(Original post by davros)
Well you haven't used the boundary condition y(1) = 0 so you can find the value of the constant

Also, if you have a power of -1/x in the denominator of a fraction, you can rewrite more simply as the positive power +1/x.
when should you use the modulus sign when you have ln in the differential equation?

Im solving an equation and the conditions are y=3 when x=0 to find the constant.

and how do you get rid of them in the end?
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4 years ago
#15
(Original post by Vorsah)
when should you use the modulus sign when you have ln in the differential equation?

Im solving an equation and the conditions are y=3 when x=0 to find the constant.

and how do you get rid of them in the end?
You have use the modulus sign when you don't know in advance what values are going to be input into the logarithm.

You can only remove the modulus when you know for certain that the expression inside the logarithm is positive - this might be something you're told in the question, or it might be clear if the equation is modelling a physical quantity like time that can never be negative.
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