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Kreyszig ODE

by mathnerd15
Tags: kreyszig
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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!
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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
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Thanks
PF Gold
P: 39,682
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.


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