Solve DE y' = \frac{y+y^2}{x+x^2} - Separation of Variables

[beetle2]

Homework Statement

$y' = \frac{y+y^2}{x+x^2}$

Homework Equations

separation of variables

The Attempt at a Solution

I start with
$y' = \frac{y+y^2}{x+x^2}$
which is
$\frac{dy}{dx} = \frac{y+y^2}{x+x^2}$
next step is

$dy = \frac{y+y^2}{x+x^2}dx$

than I divide both sides by $y+y^2$

so gives

$\frac{dy}{y+y^2} = \frac{1}{x+x^2}dx$

so then I integrate both sides.

$\int\frac{dy}{y+y^2} = \int\frac{1}{x+x^2}dx$

which gives

$ln\right[\frac{\mid y\mid}{\mid y+1\mid}\left]$=$ln\right[\frac{\mid x\mid}{\mid x+1\mid}\left]$

Is this right so far?

 

[vela]

Don't forget the arbitrary constant.

 

[beetle2]

Sorry,

$ln\left[\frac{\mid y\mid}{\mid y+1\mid}+ C \right]$ = $ln\left[\frac{\mid x\mid}{\mid x+1\mid}+C \right]$


I now multiply both sides by $\mid y+1\mid$

$ln\left[\frac{\mid y+1\mid \mid y\mid}{\mid y+1\mid} \right]$ = $ln\left[\frac{\mid y+1\mid \mid x\mid}{\mid x+1\mid}+ C \mid y+1\mid \right]$

 

[beetle2]

Divide by |y+1| Equals

$ln\left y \right]$ = $ln\left[\frac{\mid x\mid}{\mid x+1\mid}\right]$

 

[beetle2]

Do I then take the inverse log to make

$y = \frac{\mid x\mid}{\mid x+1\mid} + C$

 

[gomunkul51]

How did you knew straight away that the integral of 1/(x+x^2) is ln(|x/(x+1)|) ?
:)

 

[vela]

Not quite. You can combine the two arbitrary constants into one to get

$$\log \left|\frac{y}{y+1}\right| = \log \left|\frac{x}{x+1}\right| + c$$

Now take the inverse log.

 

[Char. Limit]

No, not yet...

You need a log(A) = log(B), but you have a log(A)=log(B)+C. So, say C=log(D), then say that log(B)+log(D)=log(BD), then your equation can be inverse logged. But not before.

 

[vela]

No, not yet...

You need a log(A) = log(B), but you have a log(A)=log(B)+C. So, say C=log(D), then say that log(B)+log(D)=log(BD), then your equation can be inverse logged. But not before.

Sure it can.

log A = log B + c
→ e^{log A} = e^{log B + c} = e^{log B}e^c
→ A = Be^c

 

[Char. Limit]

Sure it can.

log A = log B + c
→ e^{log A} = e^{log B + c} = e^{log B}e^c
→ A = Be^c

Good point. I just feel it's easier for me if I do it my way.

 

[beetle2]

log A = log B + c
→ elog A = elog B + c = elog Bec
→ A = Bec




So I've got

$$\log \left|\frac{y}{y+1}\right| = \log \left|\frac{x}{x+1}\right| + c$$

So i take the inverse log of both sides

$$e^{\log \left|\frac{y}{y+1}\right|} = e^\log \left|\frac{x}{x+1}\right| + c}$$
$$e^{\log \left|\frac{y}{y+1}\right|} = e^\log \left|\frac{x}{x+1}\right|} e^c$$

in turn gives


$$\left|\frac{y}{y+1}\right|} = \left|\frac{x}{x+1}\right|}e^c$$



How do I get rid of the $$y+1$$ in the dinominator of LHS?

 

[Char. Limit]

First, make some restrictions on x so you can lose the absolute values, then multiply both sides by y+1. You can solve for y from there.

 

[beetle2]

muitiply both sides by $$y+1$$ for $$x \neq 0$$

gives

$$y= \frac{x(y+1)e^c}{x+1}$$

I keep rearranging but I can't seem to get the (y+1) out of the RHS it just seems to be changing sides

 

[vela]

First, to get rid of the absolute values, recall that |a|=|b| means a=±b.

Second, assume x≠0 and y≠0 for the moment. If you take the reciprocal of both sides, you get

$$\left|1+\frac{1}{y}\right| = e^{-c}\left|1+\frac{1}{x}\right|$$

Can you see where to go from there?

 

[beetle2]

Using that |a|=|b| means a=±b.

$$\left|1+\frac{1}{y}\right| = e^{-c}\left|1+\frac{1}{x}\right|$$


becomes



$$\frac{-(y+1)e^{-c}}{y}) = \frac{-(y-1)e^{-c}}{y})$$

so


$$(-e^{-c}+\frac{-1}{y}) = (-e^{-c}-\frac{-1}{y})$$

is that right so far

 

[vela]

Nope. Check your algebra.

 

[beetle2]

Using that |a|=|b| means a=±b.

$$\left|1+\frac{1}{y}\right| = e^{-c}\left|1+\frac{1}{x}\right|$$


becomes



$$\frac{-(y+1)e^{-c}}{y} = \frac{-(y-1)e^{-c}}{y}$$

so


$$\frac{-(y+1)e^{-C)}{y}) = \frac{-(y-1)e^{-C)}{y})$$

is that right so far

 

[beetle2]

I mean,

$$\frac{-(y+1)e^{-C)}{y} = \frac{-(y-1)e^{-C}{y}$$

 

[beetle2]

$$\frac{-(y+1)e^{-C}}{y} = \frac{-(y-1)e^{-C}}{y}$$

 

[beetle2]

i s that better

$\frac{-(y+1)e^{-C}}{y} = \frac{-(y-1)e^{-C}}{y}$

 

[vela]

I'm not sure what you're doing. The exponential is always positive so you can pull it inside the absolute value to get

$$\left|1+\frac{1}{y}\right| = \left| e^{-c}\left(1+\frac{1}{x}\right)\right|$$

So you have |a|=|b| where

$$a= 1+\frac{1}{y}$$

$$b = e^{-c}\left(1+\frac{1}{x}\right)$$

Try taking it from there.

 

[beetle2]

How this look?

$$\left|1+\frac{1}{y}\right| = e^{-c}-(1+\frac{1}{x}) = e^{-c}-1-\frac{1}{x}$$


or


$$1+\frac{1}{y} = e^{-c} 1+\frac{1}{x}$$

 

[beetle2]

What I was trying to do was something I remembered when working with absolute values on both sides of the equation. which was that there could be two solutions depending on + or -.

So I thought there might be more then 1 solution.

So using what I've got.

$$\left|1+\frac{1}{y}\right| = \left| e^{-c}\left(1+\frac{1}{x}\right)\right|$$
take 1 from both sides.

$$\frac{1}{y}\ = e^{-c}\left(1+\frac{1}{x}\right)-1$$

Take reciprocal of both sides

$$\frac{y}{1}\ = \frac{1}{e^{-c}\left(1+\frac{1}{x}\right)-1}$$


$$y = \frac{-xe^{c}}{x(e^c-1)-1}$$