First order linear - substitution - why take the +ve square root?

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January 29th 2010, 08:56 AM
Stonehambey

First order linear - substitution - why take the +ve square root?

Hi guys,

So the equation looks like this,

$\frac{dy}{dx} + \left(\frac{1}{2}\tan x \right)y = -(2\sec x)y^3,\,\,\,-\frac{\pi}{2}<x<\frac{\pi}{2}$

and we're required to use the substitution $z=y^{-2}$

Well I do it all and get to

$z = \frac{4x+c}{\cos x}\implies y^2 = \frac{\cos x}{4x+c}$

The answer in the back of the book takes the positive square root of this, and I can't see why. It would be easy to shrug it off, since I've done the hard part, but I really would like to know why the positive square root is taken, it's probably something simple I've missed (Wink)

Thanks

Stonehambey
January 29th 2010, 06:41 PM
Calculus26

Note y = 0 is an equiilbrium solution so y is always positive or always negative if y(0) isn't zero.

Without an initial condition I can't see why either to necessarily take the pos sqrt.

eg if y(0) =-1 then y = -cos(x)/(4x+c)

-1 = -1/(c) so c = 1

y = -cos(x)/(4x+1) would be a solution
January 30th 2010, 02:54 AM
Stonehambey

I'm unsure what an equilibrium solution is, will go and look that up...

That aside, am I right in thinking that the best way to write the general solution to this equation (given that there were no initial conditions) is

$y^2 = \frac{\cos x}{4x+c}$ ?
January 30th 2010, 04:35 AM
shawsend

3 Attachment(s)

I think the solution is:

$y^2=\frac{\cos(x)}{4x+c}$

where $c$ is dependent on the initial conditions. The first plot below is the solutionn for $c=2$ , the second is for $c=10$ and the third is the complete solution space $y^2=\frac{\cos(x)}{4x+z}$ (with help from the Mathematica forum). The particular solutions then are cross-sections of the solution-space. Note in the first plot, the solution is only defined for $(-1/2,\pi/2)$ in the region $(-\pi/2,\pi/2)$ but when $c=10$ we have a closed contour representing the solution in $(-\pi/2,\pi/2)$ .
January 30th 2010, 11:08 AM
Calculus26

an equiilibrium solution is a solution of the form y = c

In this case y = 0 is just such a solution.

The imprtance bein on an interval where the conditions of existence and uniqueness are met solution curves don't cross.

that is why if the IC is pos we take pos sqrt and if the IC is negative we take neg sqrt

I agree http://www.mathhelpforum.com/math-he...b73bd133-1.gif would be the proper way to express the solution.