# Diffrential equations

1. Apr 5, 2008

### sara_87

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

y = xln|lncx|

problem:
i don't get it

can anyone explain

2. Apr 5, 2008

### Hootenanny

Staff Emeritus
Try re-writing it as

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

Does a substitution now spring to mind?

3. Apr 5, 2008

### sara_87

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

4. Apr 5, 2008

### Hootenanny

Staff Emeritus
That would indeed be my next step

5. Apr 5, 2008

### sara_87

lnC + lnx is lnCx yeah?

6. Apr 5, 2008

### Hootenanny

Staff Emeritus
It is indeed. In general ln(a)+ln(b)=ln(ab).

7. Apr 5, 2008

### sara_87

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

8. Apr 5, 2008

### Hootenanny

Staff Emeritus
What do you get after the substitution?

9. Apr 5, 2008

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!

10. Apr 5, 2008

### sara_87

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?

11. Apr 5, 2008

### Hootenanny

Staff Emeritus
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: Apr 5, 2008
12. Apr 5, 2008

### sara_87

yes yes yes! i see

cheers!

13. Apr 5, 2008

### Hootenanny

Staff Emeritus
No problem