Separable Differential Equation

[lilshai]

ok, so I have this first order differential equation for which I am trying to find the general solution for. It is 2xy' = y + 2. I am trying to reduce it to a separable equation (one variable on one side, the other on the other side) by introducing a change of variable, namely v = y/x which is also vx = y. By the product rule, I get y' = xv' + v which I can use to substitute into the original O.D.E.

I did some simplification and I end up with xv' + 1/2*v = 1/x. Is there a way to separate this equation into x's on one side and v's on the other??

Thanks.

 

[Daniel_Feldman]

I'm not quite sure here. Are you required to introduce a change of variable? If not, then here's what I would do:

Rewrite y' as dy/dx.

2xdy/dx = y + 2.

2xdy=(y+2)dx

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

Antidifferentiate both sides

2ln(y+2)=ln(abs(x))+C


Note: abs=absolute value.

Then I'd solve for y from there.


Hope this helps,

-Daniel-

 

[soroban]

Hello, lilshai!

Daniel is absolutely correct . . . it is already separable!

. . . . .dy
. . 2x --- . = . y + 2
. . . . .dx

. . .2 dy . . . . dx
. . ------- . = . ---
. . y + 2 . . . . .x

Integrate: .2 ln|y + 2| . = .ln|x| + c

. . . . . . . . . . .ln|y + 2| . = . ½ln|x| + c . = . ln|x^½| + c
. . . . . . . . . . . . . . . . . . . . . . . ._ . . . . . . . . . . . . . . _
. . . . . . . . . . .ln|y + 2| . = . ln√x| + ln(C) . = . ln|C√x}
. . . . . . . . . . . . . . . . . . . . . . . _
. . . . . . . . . . . . . y + 2 . = . C√x
. . . . . . . . . . . . . . . . . . . . . . . _
. . . . . . . . . . . . . . . . y . = . C√x - 2

 

[lilshai]

Thanks guys! I think I made it more complicated than I had to...I tend to do that sometimes. Just one question, on the step after this one:

2 ln|y + 2| = ln|x| + c

don't you also have to divide the c by 2 as well?

You did get the correct answer though.

 

[Gene]

Since c is an arbitrary constant this c is twice as big but it is still arbitrary. If it were something with a fixed value then, yes, you would have to divide by 2.
--------------------
Gene

 

[lilshai]

So we never modify the constant? let's say we had an equation such as x/3 = 4x + c and I wanted to get rid of the 1/3 on the left hand side. If I multiplied both sides by 3 would my resulting equation be x = 12x + c as opposed to x = 12x + 3c?

 

[tkhunny]

I used to try to track that sort of thing. Every time I modified the constant I substituted for it with a new constant. D, E, F...

It never made a difference.
It never made anything easier or more clear.
No one ever cared.
A few folks wondered why I was going to that trouble.

Notice how you already skipped over one such consideration:

2 ln|y + 2| = ln|x| + c ??

Shouldn't that be:

2 ln|y + 2| + c₁ = ln|x| + c ₂

Don't worry about it.

 

[lilshai]

wait a minute...am I missing something here?

Where did the natural log on the C come from?

ln|y + 2| = ln√x| + ln(C) = ln|C√x}

I still don't understand why we don't divide the constant by 2 yet we still take the natural log of the constant. I thought we were leaving it alone.

I understand everything until this step:

ln|y + 2| = ½ln|x| + c

After that :?:

 

[tkhunny]

Notice how it changed from 'c' to 'C' when the logarithm was introduced. The difficulty is the nature of the constant.

Adding a constant like this

log(a) = log(b) + c

Is equivalent to multiplying by a constant like this

a = e^log(b)+c = e^log(b)*e^c = b*e^c = C*b

You do have to consider the nature of the constant and its relationship to the variable of interest. Dividing by 2 does not change it's relationship with the variable. Adding some does not change its relationship. Other operations, like the transformation from logarithms to exponentials, do change it's relationship.

 

[soroban]

Hello, lilshai!

Let me give it a try . . .

Recall what integration is . . . we're "anti-differentiating".

Some wiseguy has a function, f(x), and he differentiates it and gets: 6x²
. . He gives you the derivative and challenges to find f(x).
. . You anti-differentiate it and get: 2x³.
. . And he says, "Wrong! .It was f(x) = 2x³ + 73 . . . ha-ha!"
Not funny, right? .He could have had ANY constant tacked on.

That's why we write: .∫ 6x² dx .= .2x³ + C . . . where C is an arbitary constant.
. . That is, "The integral of six x-squared is two x-cubed plus some constant."

Suppose we decide to divide your answer by two.
. . We get: .x³ + C/2

But since C is some arbitrary constant, when we divide it by 2, we get another arbitrary constant.

We didn't know the value of C, so we certainly don't know the value of C/2.
. . Why not call it "C"? . . . just an arbitrary constant.

And that's why something like: .3y .= .3x² + 6x + C .becomes: .y .= .x² + 2x + C

 

[lilshai]

oh ok, that makes sense now! Now I understand why certain books use K instead of C when writing the general solution! I also get the part about the constant, good explanation. Thanks tkhunny and soroban. :wink:

 

[lilshai]

Suppose we decide to divide your answer by two.
. . We get: .x³ + C/2

But since C is some arbitrary constant, when we divide it by 2, we get another arbitrary constant.

We didn't know the value of C, so we certainly don't know the value of C/2.
. . Why not call it "C"? . . . just an arbitrary constant.

And that's why something like: .3y .= .3x² + 6x + C .becomes: .y .= .x² + 2x + C

ok, I have a question about the constant in relation to integration again. I was looking at this other problem where I am solving xdx - y²dy = 0. Solving it as a normal separable equation by integration, we have x²/2-y³/3 = c. Solving for y, we get y = (3/2*x²+k)^1/3 where k = -3c. This would imply that the constant was modified when multiplying both sides of the equation by 3, correct? This was in a math book.

 

[Gene]

Let's say you solved for initial conditions and got c=-3. Then you multiplied by 3, solved for initial conditions and got k=9
WHAT DOES IT MATTER?
If you apply initial conditions at the first equation it becomes x2/2-y3/3 = -3 and it is no longer an arbirtary constant and when you multiply by 3 you get y = (3/2*x2+9)^(1/3)
If you don't apply initial conditions till the second equation it becomes y = (3/2*x2+9)^(1/3)
I say the answer is (3/2*x2+9)^(1/3). You pick out where I applied the initial conditions.