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Old May 2, 2012 | 01:08 PM
  #20651  
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Ingen som var förbi Skokloster igår?

Btw någon som kan hjälpa mig med den här?
Visa att y=x^2*ln x är en lösning till differentialekvationen x*(dy/dx)-2y=x^2
 

Last edited by Blackout; May 2, 2012 at 01:37 PM.
Old May 2, 2012 | 01:11 PM
  #20652  
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Originally Posted by Axman
Jox banga ur, han vågade inte träffa Axman, :-)
Vafan säger du?! Den lilla räkan ....
 
Old May 2, 2012 | 01:58 PM
  #20653  
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Originally Posted by Blackout
Ingen som var förbi Skokloster igår?

Btw någon som kan hjälpa mig med den här?
Visa att y=x^2*ln x är en lösning till differentialekvationen x*(dy/dx)-2y=x^2
Jo, men:

y=x^2*ln(x) blir då y deriveras m.a.p. x:
dy/dx = 2x*ln(x) +x^2/x = x(2ln(x) +1)

Sätts detta in i: y=x/2(dy/dx - x) (omskriven form av x*dy/dx - 2y = x^2)
Får vi nu:

y = x/2 * (2ln(x) + x - x) = x^2*ln(x)

Räcker det som bevis?
 
Old May 2, 2012 | 02:02 PM
  #20654  
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vafan!
 
Old May 2, 2012 | 02:09 PM
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Originally Posted by Blackout
Ingen som var förbi Skokloster igår?
Jag tänkte åka dit men jag var så sinnesjukt bakfull. Det var riktigt illa. Är fortfarande inte helt återställd.

sen bjuder jag på den här helt random.

Kate Upton's Bikini Cat Daddy - YouTube
 
Old May 2, 2012 | 02:10 PM
  #20656  
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Haha, hård Valborg Adde?
 
Old May 2, 2012 | 02:19 PM
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Originally Posted by strebo
Haha, hård Valborg Adde?
först utgång på dagen innan valborg. hård utgång. vaknade vid 10 på valborg och innan frukost så börjar jag dricka skumpa. Jag gör aldrig sånt men jag hade LOVAT att vara nere på stan innan 11 så jag hade fan inget val, det var helt olidligt annars. Men efter en skumpa var man på topp igen och körde på hela natten. Men fy helvete igår... usch!! man är för gammal för sånt här.
 
Old May 2, 2012 | 02:23 PM
  #20658  
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Haha, fan vad gött! Jag hade min mest olidliga bakfylla någonsin i vuxen ålder i Söndags .... och då hade jag "bara" varit på middag (som ballade ur) i lördags. När jag däremot tänkte boka en enkel till Burma och bli mr.magooooo på Valborg så var jag som ny igår ändå. Förstår mig inte på det där alls
 
Old May 2, 2012 | 02:55 PM
  #20659  
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Originally Posted by Bronto
Jo, men:

y=x^2*ln(x) blir då y deriveras m.a.p. x:
dy/dx = 2x*ln(x) +x^2/x = x(2ln(x) +1)

Sätts detta in i: y=x/2(dy/dx - x) (omskriven form av x*dy/dx - 2y = x^2)
Får vi nu:

y = x/2 * (2ln(x) + x - x) = x^2*ln(x)

Räcker det som bevis?
Hum, satt och pillade lite själv innan jag såg ditt svar och jag fick till något i stil med det där;
y=x^2*ln x
y deriverad med produktregeln borde bli något i stil med dy/dx=2x*ln x+x
insatt i x*dy/dx-2y=x^2 blir det
x(2x*ln x+x)-2(x^2*ln x) = 2x^2*ln x+x^2 - 2x^2*ln x = x^2
Men jag kan va helt ute och cykla, det har hänt förr
 
Old May 2, 2012 | 03:49 PM
  #20660  
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Originally Posted by strebo
Haha, fan vad gött! Jag hade min mest olidliga bakfylla någonsin i vuxen ålder i Söndags .... och då hade jag "bara" varit på middag (som ballade ur) i lördags. När jag däremot tänkte boka en enkel till Burma och bli mr.magooooo på Valborg så var jag som ny igår ändå. Förstår mig inte på det där alls
haha låter som vi hade liknande helg båda två.
 



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