
#1
Sep1111, 04:58 AM

P: 108

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! 



#2
Sep1111, 06:19 AM

HW Helper
P: 2,148

That is not a solution.
It is a separable differential equation. 



#3
Sep1111, 01:57 PM

P: 108

thanks very much!
I separated leading to lny=lnx+C so then you just find the substitution c/x? (I misread the text: l is /) 



#4
Sep1111, 02:29 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

Kreyszig ODE[tex]e^{lny}= e^{lnx+ C}[/tex] [itex]lny= lny^{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^{lnx}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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