Flash Flash Revolution - View Single Post - [differential equations] stuck on a problem

**iCeCuBEz v2** · 01-14-2013, 10:45 PM

Hey this problem seems fairly easy but I keep seeming to mess it up... I'm probably just missing something really obvious.

'Verify by substitution that each given function is a solution of the given differential equation.' (just involves basic derivatives really)

y'' - 2y' + 2y = 0

y1 = (e^x)(cosx)
y2 = (e^x)(sinx)

All this involves is product rule but for some reason I keep messing it up

Also, I've been having difficulty with solving systems of linear equations with elementary row operations so if someone could do an example of one for me that would be great. I'll give this system of linear equations.

x1 - 6x2 = 5
x2 - 4x3 + x4 = 0
-x1 + 6x2 + x3 + 5x4 = 3
-x2 +5x3 + 4x4 = 0

note: all of the numbers after the x's are supposed to be subscripts.

I'll give 10,000 credits to the first person who responds with a correct answer for each problem.

01-14-2013, 10:45 PM	#1
iCeCuBEz v2 XFD Join Date: Mar 2008 Location: Connecticut Age: 33 Posts: 4,924	[differential equations] stuck on a problem Hey this problem seems fairly easy but I keep seeming to mess it up... I'm probably just missing something really obvious. 'Verify by substitution that each given function is a solution of the given differential equation.' (just involves basic derivatives really) y'' - 2y' + 2y = 0 y1 = (e^x)(cosx) y2 = (e^x)(sinx) All this involves is product rule but for some reason I keep messing it up Also, I've been having difficulty with solving systems of linear equations with elementary row operations so if someone could do an example of one for me that would be great. I'll give this system of linear equations. x1 - 6x2 = 5 x2 - 4x3 + x4 = 0 -x1 + 6x2 + x3 + 5x4 = 3 -x2 +5x3 + 4x4 = 0 note: all of the numbers after the x's are supposed to be subscripts. I'll give 10,000 credits to the first person who responds with a correct answer for each problem. Last edited by iCeCuBEz v2; 01-14-2013 at 11:25 PM..