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
)Original post by davros
You can't!
(i.e. it's one of many functions that doesn't have an elementary integral )
(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)
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)
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
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
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?
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
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
Original post by Vorsah
Thanks, can you check my final answer?
The power is -1/x btw
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.
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.
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
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.
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?
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?
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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