# Diffrential equations

solve the diffrential equation:
xy' = x(e^(-y/x)) + y

y = xln|lncx|

problem:
i don't get it

can anyone explain

## Answers and Replies

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Hootenanny
Staff Emeritus
Gold Member
Try re-writing it as

$$y^\prime = e^{-y/x} + \frac{y}{x}$$

Does a substitution now spring to mind?

yeah i get that... but then i don't know what to do... do we substitute y=vx???

Hootenanny
Staff Emeritus
Gold Member
yeah i get that... but then i don't know what to do... do we substitute y=vx???
That would indeed be my next step

lnC + lnx is lnCx yeah?

Hootenanny
Staff Emeritus
Gold Member
lnC + lnx is lnCx yeah?
It is indeed. In general ln(a)+ln(b)=ln(ab).

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
Staff Emeritus
Gold Member
What do you get after the substitution?

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!

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
Staff Emeritus
Gold Member
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,

$$\frac{d}{dx}vx = x\frac{dv}{dx} + v$$

Hence, the second line of your working is actually,

$$x\frac{dv}{dx} + v = e^{-v} + v$$

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

$$\int v(x)dx \neq v\cdot x + c$$

Last edited:
yes yes yes! i see

cheers!

Hootenanny
Staff Emeritus