How to solve 2nd order d.e ? Is this the right start?

1. Jul 9, 2008

laura_a

1. The problem statement, all variables and given/known data
find the general solution of
y' = (y + y^2)/(x + x^2)

3. The attempt at a solution

I've tried a number of ways the first most obvious way I figured was to multiply the x+x^2 over so I did that but then when I expand I end up with a y' in both of the terms and I can't find any form of expression who's d/dx has two y' in it, because I believe it's impossible? Anyway, what I really want help with is, is there a trick to these, because this is my first look at them for many years and I just can't pick up what to do with them in order to put them in a position where you can intergrate and then solve...

My other idea for a solution was

y'(1+x) = [y(1+y)]/x

But then when I expand I still get y' on both of the LHS terms ... Is this a separable equation? I am studying this via correspondance so I've only learnt as far as I've read tonight and I'm totally lost... any help will be greatly appreciated

***Extra

I've just tried one method from my text book but it seems totally different to the first method I used of direct integration

ANyways I've got

(x+x^2)y' - (y + y^2) = 0
[(x^2)/2 + (x^3)/3]y' - [(y^2)/2 + (y^3)/3]y' = 0
y'(x^2/2 + x^3/3 - y^2/2 - y^3/3) = 0 (not sure if I can just do this or not, but it's worth a try)
so I ended up with a nicer looking eqn which was

1/6(3x^2 + 2x^3 - 3y^2 - 2y^3) = c

Clearly I have no idea what I'm doing, but I thought that was worth a shot... I still don't understand how it seems in the text book you can just use that method instead of the other method where you have to set it up in terms of something that looks like the differential of something else and then integrate it...?

Last edited: Jul 9, 2008
2. Jul 9, 2008

vipulsilwal

y' = (y + y^2)/(x + x^2)

dy/(y + y^2)=dx/(x + x^2)
dy/y(y+1)=dx/x(x+1)

lets first see the right hand side
dy*(1/y-1/y+1)
dy/y-dy/(y+1)
integrating,
ln(y)-ln(y+1)+c1
ln(y/y+1)+c1

similarly on right hand side
ln(x/x+1)+c2
equating we get
ln(y(x+1)/x(y+1))=c
y(x+1)/x(y+1)=e^c

e^c as constant k
y(x+1)/x(y+1)=k

3. Jul 9, 2008

tiny-tim

Hints are best!

Hi vipulsilwal!

Very nice.

But the OP asked for "the right start", and you've given the whole answer.

Hint: If you'd stopped where I've quoted to, it would just have been a hint.

4. Jul 9, 2008

HallsofIvy

Staff Emeritus
And that is NOT a "second order" differential equation.

5. Aug 7, 2008

laura_a

I've been working on this question for ages now

And I have to find the solution for which y(2)=1
This is the last line of working out
$$\frac{y(x+1)}{x(y+1)} =k$$

The problem is, I've got an expression with x's and y's in it. In all of the examples in my text book you always end up with y= something with x's and k's in it. But I can't solve that for y, I've tried and end up with a big mess. I assumed I need to use that fact that y(2)=1 and perhaps get simulataneous equations going? One of the problems I really have is that I don't know what y is so I can't find y(2)=1

Should I go back to

$$ln \frac{y}{y+1} + C_1 = ln \frac{x}{x+1} + C_2$$

I was thinking maybe I should split them up before taking the e of both sides

Can/SHould I do

$$ln(y) - ln(y+1) + C_1 = ln(x) - ln(x+1) +C_2$$

?

Thanks

Last edited: Aug 7, 2008
6. Aug 7, 2008

tiny-tim

Hi laura!

Trick: just simplify the y bit, to get one y on its own:

(y+1)/y = 1 + 1/y.

So 1 + 1/y = (x + 1)/kx,

so k = … ? and y = … ?

7. Aug 8, 2008

laura_a

ahh... thanks! Got it now