Simplifying differential equations (go easy on me, I know I'm a noob)

**Kaitosan** · April 22nd 2009, 03:34 PM

How do you simplify (abs(1-y)) = 9e^[(-1/2)(x^2)]?

The answer is y = 1 + 9e^[(-1/2)(x^2)] but for some reason I keep getting y= 1 - 9e^[(-1/2)(x^2)].

I know it has something to with the absolute value sign so I know I'm missing something here. I'd appreciate any help regarding operations with the absolute value sign, such as how to dissolve it properly or something like that. I'm sure I've learned it somewhere a while ago but I'm sure I forgot it or something and I need a refresh.

Thanks in advance.

**skeeter** · April 22nd 2009, 04:08 PM

Originally Posted by Kaitosan

How do you simplify (abs(1-y)) = 9e^[(-1/2)(x^2)]?

The answer is y = 1 + 9e^[(-1/2)(x^2)] but for some reason I keep getting y= 1 - 9e^[(-1/2)(x^2)].

I know it has something to with the absolute value sign so I know I'm missing something here. I'd appreciate any help regarding operations with the absolute value sign, such as how to dissolve it properly or something like that. I'm sure I've learned it somewhere a while ago but I'm sure I forgot it or something and I need a refresh.

Thanks in advance.

what was the original DE and its initial condition?

**Kaitosan** · April 22nd 2009, 06:44 PM

Originally Posted by skeeter

what was the original DE and its initial condition?

dy/dx = x(1-y)

with f(0) = 10

But that information isn't necessary for anyone to complete the last step of the problem, the part of which I'm stuck because I basically forgot the algebraic operations necessary to solve expressions with absolute value symbols.

**skeeter** · April 23rd 2009, 04:36 AM

Originally Posted by Kaitosan

dy/dx = x(1-y)

with f(0) = 10

But that information isn't necessary for anyone to complete the last step of the problem, the part of which I'm stuck because I basically forgot the algebraic operations necessary to solve expressions with absolute value symbols.

yes, it is necessary to see the initial condition.

$<br /> |1-y| = 9e^{-\frac{x^2}{2}}<br />$

remember how absolute value works?

$|1-y| = 1-y$ if $(1-y) > 0$

$|1-y| = -(1-y)$ if $(1-y) < 0$

since $y = 10$ , $(1-y) < 0$ ...

$<br /> -(1-y) = 9e^{-\frac{x^2}{2}}<br />$

$-1 + y = 9e^{-\frac{x^2}{2}}$

$y = 1 + 9e^{-\frac{x^2}{2}}$

**Kaitosan** · April 23rd 2009, 04:44 AM

Ah, so if an absolute value expression turns out to be positive, you remove the symbol; otherwise, "negatify" the expression so it remains positive. Got it.

**Kaitosan** · April 23rd 2009, 02:26 PM

Hey, I've another question. What would you say is the range and the domain of that function? It looks like the function can't be over y=1. Is that right? Also, suppose that an expression like (abs(x-3)) is thrown in along with (abs(1-y)), does that mean that x must be more than or equal to three while y must be less than or equal to 1?

Also, do the initial conditions affect the range and domain in any way?

I'd love some clarification if you can offer it. Thanks!

**skeeter** · April 23rd 2009, 03:09 PM

Originally Posted by Kaitosan

Hey, I've another question. What would you say is the range and the domain of that function? It looks like the function can't be over y=1. Is that right? Also, suppose that an expression like (abs(x-3)) is thrown in along with (abs(1-y)), does that mean that x must be more than or equal to three while y must be less than or equal to 1?

Also, do the initial conditions affect the range and domain in any way?

I'd love some clarification if you can offer it. Thanks!

range of $y = 1 + 9e^{-\frac{x^2}{2}}$ is $1 < y \leq 10$

I don't follow your 2nd question ... how is $|x-3|$ related to $|1-y|$ ?

yes ... initial conditions can affect both domain and range of a function.

consider the DE $\frac{dy}{dx} = \frac{x}{y}$ , $y \neq 0$

find the solution with initial point (1,2), then find the solution with initial point (3,-1) ... you'll find that domain and range for both solutions are very different.

**Kaitosan** · April 23rd 2009, 07:34 PM

One last question! Are the "initial conditions" the very first coordinates that are defined for a function?

**mollymcf2009** · April 23rd 2009, 08:03 PM

Originally Posted by Kaitosan

How do you simplify (abs(1-y)) = 9e^[(-1/2)(x^2)]?

The answer is y = 1 + 9e^[(-1/2)(x^2)] but for some reason I keep getting y= 1 - 9e^[(-1/2)(x^2)].

I know it has something to with the absolute value sign so I know I'm missing something here. I'd appreciate any help regarding operations with the absolute value sign, such as how to dissolve it properly or something like that. I'm sure I've learned it somewhere a while ago but I'm sure I forgot it or something and I need a refresh.

Thanks in advance.

The absolute value of a number is its distance from zero. It is asking you how FAR something is from something else, NOT in what direction it is. In other words, if you have:

$|y-1| = 2$

Then it is asking:

$y-1 = 2$ and $y-1 = -2$

So for your problem:

$|1-y| = 9e^{-\frac{1}{2}x^2}$

AKA

$-9e^{-\frac{1}{2}x^2} = 1 - y$

and

$9e^{-\frac{1}{2}x^2} = 1 - y$

So, the distance is always positive because you don't care where it is you just care how far away it is!
Hope that helps you!

**Calculus26** · April 23rd 2009, 10:46 PM

don't forget your equiilibriium solutions here y =1 is an equillibrium soln

By uniqeness solution curves don't cross so if your initial condition starts

with y(0) > 1 then y is always greater than 1

If y( 0 ) < 1 then y stays less than 1 and you would use |1-y| =1-y

and your "mistaken" solution would now be the correct one

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