Differential equationDate: 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 THANKS A WHOLE LOT!!!! 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 Sec[u]*Tan[u]*du ---------------- = 2*Sec[u]*Sqrt{Tan^2[u]} = 2*Sec[u]*Tan[u]. dx 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: du ---------------- = 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/