A Purely Mathematical

[Tomoyo]

dy/dx = (y+1)/x

Find the solution to said differential equation of the form y = f(x) which passes through the point (-1, 1) and state the domain.

I and several of my classmates arrived at one solution which all books and my teacher say is incorrect, but was not able to explain.

Go. You guys are smart. Theoretically, at least.

My work:

Spoiler :

dy/(y+1) = dx/x
Ln|y + 1| = Ln|x| + c
Ln|(1) + 1| = Ln|(-1)| + c
Ln(2) = c
Ln|y + 1| = Ln|x| + Ln(2)
|y + 1| = 2|x|
y + 1 = 2|x| or y + 1 = -2|x|
1 + 1 = 2|-1| or 1 + 1 = -2|-1|
2 = 2 or 2 = -2
Because the second equation does not hold for the point (-1, 1), it's wrong.

Answer is y = 2|x| - 1. And the domain is all real numbers. Right?

 

[Perfection]

seems ligit

 

[Ecofarm]

EDIT: Bah, I was wrong. Good luck guys.

 

[Catharsis]

I tried two methods and got y = -1 as the answer both times. That's obviously not right.

EDIT: Wait, I think I forgot the constant of integration the second time.

 

[Atticus]

Answer is y = 2|x| - 1. And the domain is all real numbers. Right?[/spoiler]

0 can't be in the domain, because your function has no derivative at that point. But otherwise it seems right. What was the solution your teacher told you?

 

[Catharsis]

Separating variables seems to give y = 2|x| - 1, as you said. Integrating factor got me y = -2x - 1, but I didn't use any modulus functions so that's probably wrong.

 

[Truronian]

That's correct, though as Atticus says 0 is not in the domain.

One nice shortcut you could have used is:

ln|x| + c = ln(A|x|)

(c,A constants)

 

[Perfection]

Actually, I think the correct answer is just y=2x-1

sub that into the equation and you get

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

What I think is going on is that the general solution is kx-1, and with all your weird absolute value crap you spliced two solutions together (via an absolute value), in these cases the relation doesn't hold at the vertices.

 

[Truronian]

EDIT: Ignore me. I can't read.

 

[Mise]

Actually, I think the correct answer is just y=2x-1

sub that into the equation and you get

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

What I think is going on is that the general solution is kx-1, and with all your weird absolute value crap you spliced two solutions together (via an absolute value), in these cases the relation doesn't hold at the vertices.

y=2x-1 doesn't pass through (-1,1)

 

[Tomoyo]

I thought that the first derivative not existing at x = 0 didn't matter in the case of the solution as y = 2|x| - 1 has a corner at x = 0, but I'm certainly no expert at that. (i.e. even though the derivative does not exist, the function does)

My teacher (not very vehemently) said that the domain had to be x < 0. In fact, this question was on the AP Calculus test several years ago, so we had access to the rubric, which says the same thing.

 

[Mise]

Ahh, yeah, it has to be x<0, because (-1,1) only holds for x<0:

y=2|x|-1
x = (y+1)/2 ... or ... -(y+1)/2

Only the second holds for (-1,1), so x is always the opposite sign to y; where y is positive, x is negative, and vice versa. We know y is always positive, so x has to be always negative.

 

[Truronian]

Ahh, yeah, it has to be x<0, because (-1,1) only holds for x<0:

y=2|x|-1
x = (y+1)/2 ... or ... -(y+1)/2

Only the second holds for (-1,1), so x is always the opposite sign to y; where y is positive, x is negative, and vice versa. We know y is always positive, so x has to be always negative.

I think the issue is if you include the positive x's there are an infinity of solutions of the form:

y =

-2x-1 for x<0
kx-1 for x>0, k&#8800;0

(Which is effectively what Perfection was saying).

 

[Mise]

Huh, I didn't actually read the whole of Perfection's post...

 

[Perfection]

oops sign error

 

[Supermath]

I don't think there's enough information to determine the solution for x > 0. You can get one of two functions.

|y + 1| = 2|x|
y + 1 = 2x or y + 1 = -2x
1 + 1 = (2)(-1) or 1 + 1 = -(2)(-1)
2 = -2 or 2 = 2

y = -2x - 1

Since there are two possible solutions at x > 0, the domain has to be (-inf, 0).

Of course, I could be wrong; I never actually paid attention in class.

 

[nihilistic]

At this following step you failed to consider all the possibilities:

|y + 1| = 2|x|

That basically meant that the left and right sides evaluates to the same absolute value, not that the right side is necessarily positive because you chose y as the dependant variable. y+1 at y=1 being positive and 2x at x=-1 being negative leaves 2 possibilities:
y = 2|x| - 1 and y = -2x - 1

Differentiate both of them and you'll see that only y = -2x - 1 works.

You'll also see that y = 2|x| - 1 not only has a smaller domain that fits the differential equation, but that it is unfortunately undifferentiable at that point that you were to consider: (-1,1).

Also note that integration problems are usually absurdly easy to verify: just differentiate your answer.

 

[Mise]

Maple agrees with y=-2x-1.

 

[Kal'thzar]

And here I was wondering If I couldn't do maths, of course I was drunk when I did it but still....

I got y=-2x-1


ln(y+1) = lnx + c
ln(y+1) = lnx + lnA
ln(y+1) = ln(Ax)
y+1 = Ax
y = Ax -1
(-1,1)
1= -A -1
A = -2
y=-2x-1