
#19
Apr1110, 09:38 PM

P: 21

error in which part?? yes i got the answer of 2k³4k²k/(2k²+1)² using....(u'vuv')/v² am i right?? 



#20
Apr1110, 10:00 PM

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P: 5,004





#21
Apr1110, 10:04 PM

P: 21

i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right] [/tex] owk i think i got it, is it correct....i(iu~(k)k²u~(k)) that will give me u~(k)+ik²u~(k)>u~(k)[1+ik²]?? owk how do i separate u~(k) so that it don't cancel each other(left and right side)?? 



#22
Apr1110, 10:17 PM

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P: 5,004

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}\right]=\frac{d}{dk}\left[k\tilde{u}\right]=\left(\tilde{u}\frac{dk}{dk}+k\frac{d\tilde{u}}{dk}\right)=\tilde{u}k\frac{d\tilde{u}}{dk}[/tex] This is a basic application of the product rule. 



#23
Apr1110, 10:32 PM

P: 21

so now the equation will be u~(k)[2k²+2]=k du~/dk so u~(k)= [k/(2k²+2) du~/dk Is it correct if i diffrentiate towards k on the right side??? 



#24
Apr1110, 10:40 PM

HW Helper
P: 5,004

[tex]\frac{d\tilde{u}}{\tilde{u}}=2\frac{k^21}{k}dk[/tex] 



#25
Apr1110, 11:12 PM

P: 21

u~(k)=e^[(k²)(2 ln k)] ....u~(k)=e^(k²)/e^(2 ln k) then i transform it using the table?? right? 



#26
Apr1110, 11:31 PM

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#27
Apr1110, 11:52 PM

P: 21

should i just let it be in terms of convolution plus the homogenous equation right? that should be my final general solution. 



#28
Apr1210, 09:23 AM

HW Helper
P: 5,004

Right, although in this case, I think your second solution comes from solving [itex]u''(x)=0[/itex] and [itex]xu'u=0[/tex]. which isn't really "the homogeneous equation".




#29
Apr1210, 09:57 PM

P: 21

anyway thanks a lot for your help! i really appreciate it.. 


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