Register to reply

Kreyszig ODE

by mathnerd15
Tags: kreyszig
Share this thread:
mathnerd15
#1
Sep11-11, 04:58 AM
P: 110
Hi! I think I have to ask this since I'm having health problems-

from Kreyszig, for xy'=-y how do you verify the solution y=h(x)=clnx by differentiating
y'=h'(x)=-clnx^2? I don't see how you get the x^2 term
also for ODEs the solution is on an open interval a<x<b but how does it include special cases of the intervals -inf<x<b, a<x<inf, -inf<x<inf; wouldn't the open interval a<x<b exclude -inf<x<b?
thanks very much!
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
lurflurf
#2
Sep11-11, 06:19 AM
HW Helper
P: 2,264
That is not a solution.
It is a separable differential equation.
mathnerd15
#3
Sep11-11, 01:57 PM
P: 110
thanks very much!
I separated leading to-
-ln|y|=ln|x|+C
so then you just find the substitution c/x? (I misread the text: l is /)

HallsofIvy
#4
Sep11-11, 02:29 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,569
Kreyszig ODE

Quote Quote by mathnerd15 View Post
thanks very much!
I separated leading to-
-ln|y|=ln|x|+C
so then you just find the substitution c/x? (I misread the text: l is /)
Well, you don't just "find" a substitution. You always solve an equation of the form f(y)= F(x) for y by taking [itex]f^{-1}[/itex] of both sides. The inverse function to ln(x) is, of course, [itex]e^x[/itex] so take the exponential of both sides:
[tex]e^{-ln|y|}= e^{ln|x|+ C}[/tex]

[itex]-ln|y|= ln|y^{-1}|[/itex] so the left side is [itex]y^{-1}= 1/y[/itex]. Because [itex]e^{a+ b}= e^ae^b[/itex] the right side is [itex]e^{ln|x|}e^C= C' |x|[/itex] where C'= e^C.
That is, [itex]1/|y|= C'|x|[/itex] or [itex]|y|= C'/|x|[/itex]. We can then "absorb" the absolute values into C' by allowing it to be positive or negative.


Register to reply