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Diffrential equations

  • Thread starter sara_87
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763
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solve the diffrential equation:
xy' = x(e^(-y/x)) + y


answer:
y = xln|lncx|


problem:
i don't get it

can anyone explain
 

Answers and Replies

Hootenanny
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Try re-writing it as

[tex]y^\prime = e^{-y/x} + \frac{y}{x}[/tex]

Does a substitution now spring to mind?
 
763
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yeah i get that... but then i don't know what to do... do we substitute y=vx???
 
Hootenanny
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763
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lnC + lnx is lnCx yeah?
 
Hootenanny
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763
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ok
no matter what i do i'm not getting that answer! and its not supposed to a hard question!!

i get this:

y = x lnCx
 
Hootenanny
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What do you get after the substitution?
 
2,981
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ok
no matter what i do i'm not getting that answer! and its not supposed to a hard question!!

i get this:

y = x lnCx
What are the last few steps leading up to your answer? Without seeing your work, it is difficult to say where your error is. Though I am inclined to say that you are probably not taking the natural log of the entire right hand side. That is usually where my error is.

Edit: I notice Hootenany has got you covered. You are in good hands. Back to my DE!
 
763
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i didn't want to write it out at first because 1. i can't write it the fancy form and 2. i'm affraid i made a very stupid embarrasing mistake, but ok this is what i did:

y' = (e^(-y/x)) + y/x

then y=vx and so v=y/x


d(vx)/dx = (e^(-v)) + v

then i integrate:

vx = x(e^(-v)) + vx + c

then i substituted y back in

is this right so far?
 
Hootenanny
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You need to be careful here since v is not a constant, instead it is a function of x, i.e. v=v(x). Therefore, by the product rule,

[tex]\frac{d}{dx}vx = x\frac{dv}{dx} + v[/tex]

Hence, the second line of your working is actually,

[tex]x\frac{dv}{dx} + v = e^{-v} + v[/tex]

Edit: Just to add a little more detail, since v is a function of x, in general,

[tex]\int v(x)dx \neq v\cdot x + c[/tex]
 
Last edited:
763
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yes yes yes! i see

cheers!
 
Hootenanny
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