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Differential equation

Date: Wed, 14 Dec 1994 15:24:40 -0500 (EST)
From: Lizzy Hechenbleikner

I am trying to solve the differential equation:

 dy/dx=2y(root of (y^2-1))
        at x=0 and y=2

you got that? it's 2y times the square root of y squared minus 1


Date: Thu, 15 Dec 1994 12:13:56 -0500 (EST)
From: Dr. Ken
Subject: Re: differential equation

Hello there!

There are two ways I know of to do your problem, and they basically 
boil down to the same way.  I'll show you both of them.

1)  We have dy/dx = 2y Sqrt{y^2 - 1}.  We want to simplify this some 
by doing a substitution.  A general rule when doing a substitution is 
to find the nastiest part of the integral and get rid of it when you do 
your substitution.  In this case, the Sqrt{y^2 - 1} looks pretty nasty.  
In fact, it looks like we're going to have to do a trig substitution.  So
let's do it.

Let y = Sec[u], because we're going to take advantage of the fact that 
Sec^2 - 1 = Tan^2.  So we have dy = Sec[u]*Tan[u]*du, and our 
problem  becomes

----------------  =   2*Sec[u]*Sqrt{Tan^2[u]}  =  2*Sec[u]*Tan[u].

We can cancel the Secant and the Tangent, and we're left with 
du/dx = 2.  Much easier than what we started with.  So u = 2x + c, 
for some constant c.

Now we plug back in to find everything in terms of x and y.  Since
y=Sec[u], we have y = Sec[2x + c].  Then we use the values you gave 
for the intitial conditions to find c:  When x=0, y=2.  So this means
2=Sec[c].  What angle has a Secant of 2, i.e. what angle has a Cosine 
of 1/2?  Pi/3 or -Pi/3.  So our answer is y = Sec[2x +- Pi/3].  Fantastic!

2)  We can rewrite the equation like this:

----------------  =  dx      and then take the integral of both sides.  
2y*Sqrt{y^2 - 1}

When you do this, you'll have to make basically the same substitution 
you made in the previous way (y = Sec[u]).  So it's really a matter of 
choice which method you use.  One way lets you deal with the 
differential equation longer, the other way puts everything in terms of 
integrals and then makes you solve integrals.  So I'll leave it up to you 
to actually solve this integral.  I hope you get the same answer I got!

Thanks for the question!

-Ken "Dr." Math
Associated Topics:
College Calculus

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