[differential equations] stuck on a problem

[iCeCuBEz v2]

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.

[foilman8805]

Dusted off some cobwebs in my brain for this one as I'm about 4 years removed from university differential equations, but I have a solution for your first problem. I can work on your second problem in a bit, but the matrix math is a little more tricky than derivatives so it'll take me some time (and again, it's been awhile).

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

So let's just start by finding the first and second derivatives.

y = e^x*cosx

First derivative is as you said - just a simple product rule. Derivative of e^x is just e^x. Derivative of cosx is -sinx. Product rule states (f*g)' = f'*g + f*g'.

Therefore, y' = e^x*cosx - e^x*sinx. Agree?

Second derivative will follow the same convention, you just have to do it twice. Derivative of -e^x is -e^x. Derivative of sinx is cosx. Following the product rule on each term in y', I get (before simplification):

y'' = (e^x*cosx - e^x*sinx) + (-e^x*sinx - e^x*cosx)

Which simplifies to:

y'' = -2e^x*sinx

Putting it all together in y'' - 2y' + 2y:

-2e^x*sinx - 2*(e^x*cosx - e^x*sinx) + 2*(e^x*cosx)

Multiply yo' shit out and you get:

-2e^x*sinx - 2e^x*cosx + 2e^x*sinx + 2e^x*cosx = 0

All of your terms cancel, therefore 0 = 0.

[foilman8805]

Working on the second problem.

[dag12]

The second one doesn't have a solution. Did you type it correctly?
For reference:
http://www.wolframalpha.com/input/?i...%2C4%2C0%5D%5D

Note the last row of the rref form has 0,0,0,0,1 as the last row.

[foilman8805]

Ninja'd. I just started working it out and I came to the same conclusion.

[qqwref]

For the linear equation one, first we set it up as a matrix equation of the form A*x=b:

[  1 -6  0  0 ]   [x1]   [5]
[  0  1 -4  1 ]   [x2]   [0]
[ -1  6  1  5 ] * [x3] = [3]
[  0 -1  5  4 ]   [x4]   [0]

To get a solution, the procedure is basically
- Find a row with the first variable, then delete some multiple of that row from every other row with that variable in it, so the first variable isn't in any other rows. Make sure to delete from both A and b, but not x.
- Find a row with the second variable and without the first variable, then delete some multiple of that row from every other row, so the second variable isn't in any other rows.
- Continue like that (always finding a row with the next variable, but with none of the variables you already did). Ideally you will end up with only one non-0 number in A in each row. However, if you are able to make one row of A completely zeroed, your equation has either zero solutions or multiple solutions.

So for this example, let's first eliminate x1 in all rows except the first, by adding the first row to the third row:

[  1 -6  0  0 ]   [x1]   [5]
[  0  1 -4  1 ]   [x2]   [0]
[  0  0  1  5 ] * [x3] = [8]
[  0 -1  5  4 ]   [x4]   [0]

Now let's take the second row, which has x2 but not x1. We add 6 times the second row to the first row, and add the second row to the fourth row:

[  1  0 -24 6 ]   [x1]   [5]
[  0  1 -4  1 ]   [x2]   [0]
[  0  0  1  5 ] * [x3] = [8]
[  0  0  1  5 ]   [x4]   [0]

Now let's take the third row, which has x3 but not x1 or x5. Add 4 times the third row to the second row, and 24 times the third row to the first row. Subtract the third row from the fourth row.

[  1  0  0 126]   [x1]   [197]
[  0  1  0 21 ]   [x2]   [ 32]
[  0  0  1  5 ] * [x3] = [  8]
[  0  0  0  0 ]   [x4]   [ -8]

Okay, so now our equations are x1 + 126*x4 = 197, x2 + 21*x4 = 32, x3+5*x4 = 8, and 0 = -8, assuming I did the arithmetic right. That last equation (0 = -8) is obviously impossible so this set of equations has no solutions! Too bad.



HOWEVER, if the numbers had been a little different, we might have ended up with something like this:

[  1  0  0  0 ]   [x1]   [5]
[  0  1  0  0 ]   [x2]   [6]
[  0  0  1  0 ] * [x3] = [3]
[  0  0  0  1 ]   [x4]   [7]

in which case there is one unique solution, which is just x = b, or in other words, x1 = 5, x2 = 6, x3 = 3, x4 = 7.

[dag12]

I'm assuming you made up the problem or something just to see how to go through solving a system of linear equations. The particular problem you posed has no solution, but qqwerf's post regarding gaussian elimination should be helpful regardless. Nice post - very detailed, haha (unlike my own post)!

[iCeCuBEz v2]

nope I typed the problem correctly I just picked one out of the book and it said determine if the system is consistent so it is ok that there is no solution. I just wanted to see the row operations.

thank you qqwref and foilman :3

I understand all three of the basic row operations but for some reason the answers I get don't match the back of the book so I'm obviously doing something wrong I just don't know what.

[qqwref]

Maybe you're making mistakes in the arithmetic? I find that it's often really hard to keep track of stuff like -5 - 2*-1, so it could be that you're doing all the hard stuff right. And you can always check stuff with wolframalpha, like dag12 mentioned (you can even put in only 2-3 rows and see what it does with those).

[iCeCuBEz v2]

yeah probably

[iCeCuBEz v2]

97 on my quiz in linear algebra what up XD

[j-rodd123]

u go girl!!!!

[foilman8805]

good shit, bro