Question

Solve using variation of parameters
\[y''-2y'+y=e^{2x}\]

Answer 1

complimentary....

Answer 2

meaning?

Answer 3

\[y_c=c_1e^{-x}+c_2xe^{-x}\]

Answer 4

right...

Answer 5

now we need the wronskian, do you know about that?

Answer 6

nope

Answer 7

it is the determinant of a matrix

Answer 8

for\[y_c=c_1y_1+c_2y_2\]the Wronskian is\[W(y_1,y_2)=\left|\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right|\]

Answer 9

that's a determinant in case you forgot, so find that determinant

Answer 10

\[\left|\begin{matrix}e^{-x}&xe^{-x}\\-e^{-x}&-e^{-x}(x-1)\end{matrix}\right|\]
=\[e^{-x}(e^{-x}(x-1))-xe^{-x}e^{-x}\]

Answer 11

completely wrong?

Answer 12

oh it's plus

Answer 13

yep you caught it :)

Answer 14

\[e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]

Answer 15

that's it?

Answer 16

you dropped a minus

Answer 17

where?

Answer 18

ohhhh

Answer 19

\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]

Answer 20

right

Answer 21

simplify and be happy :)

Answer 22

does it equal\[ e^{-x}\]?
\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}\]
\[{-x}(e^{-x}(x-1))+xe^{-x}=1\]

Answer 23

\[(e^{-x}(x-1))+xe^{-x}=1\]

Answer 24

typo

Answer 25

no, why did you set it equal to e^-x, we can't set it equal to anything yet...

Answer 26

oh my bad haha

Answer 27

so how should I simplify it then?

Answer 28

\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}\]now try again while I try to eat ;)

Answer 29

-(x-1)=1-x
1-x=1-x

Answer 30

no no no ...wait a second

Answer 31

now i get 2-x=2-x

Answer 32

nope still wrong

Answer 33

x=1

Answer 34

stop laughing

Answer 35

Let's try that again
\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}\]

Answer 36

they're just equal to each other

Answer 37

\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]

Answer 38

\[-e^{-2x}(x-1)+xe^{-2x}\]

Answer 39

\[e^{-2x}((1-x)+x)\]

Answer 40

\[e^{-2x}\]

Answer 41

yes?

Answer 42

yes, sorry, I was afk

Answer 43

LOL ok

Answer 44

do you know the formula for the particular solution for variation of parameters?

Answer 45

something like that?

Answer 46

almost unidentifiable to me in that form, but correct I'm sure
I have the more direct formula memorized
http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx
(the green box with the formula that says "variation of parameters)

Answer 47

is the first one a y''?
\[y''+q(t)y'+r(t)y=g(t)\]

Answer 48

yeah, it looks screwy I know

Answer 49

do we know what q(t) and r(t) are? are they the coefficients of y' and y? Probably not it looks like

Answer 50

yes and in this case they are the coefficients, which in this case are constants
the formula is just pointing out that they may not be

Answer 51

oh I see, soo...
\[y''+q(t)y'+r(t)y=g(t)\]
\[y''-2y'+y=e^{2x}\]
so now I integrate

Answer 52

right, using the formula...
r(t) and r(t) don't even come into the formula for the particular solution in variation of parameters

Answer 53

q(t) and r(t)...
they determine the complimentary only, which as you can see *is* part of the formula for the particular

Answer 54

\[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{-x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx\]

Answer 55

the first integrand is wrong I think

Answer 56

I think you put the wrong g(x)

Answer 57

\[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx\]

Answer 58

oh wait, wait, the denominator in each integrand is the wronskian, which we found to be e^(-2x)
I think you are confusing notation with the parentheses

Answer 59

Oh lol so the denominator is just e^{-2x}

Answer 60

right

Answer 61

notice that makes our integrals quite tolerable :)

Answer 62

\[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}}dx\]
That's a lot easier to integrate

Answer 63

oh yeah :) especially after a little simplification

Answer 64

\[-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]

Answer 65

so what was the point of doing variation of parameters

Answer 66

well now you have a particular solution

Answer 67

y(x)=yc+yp
you got you now so you're good to go

Answer 68

you got yp*

Answer 69

I did not check your integral btw

Answer 70

that's ok I did wolfram

Answer 71

right-o, I'd do the same right now :)

Answer 72

oh I see
\[c_1e^{-x}+c_2xe^{-x}-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]

Answer 73

yep, and of course if this is an IVP we now proceed to find c1 and c2
otherwise we're done having found the general solution

Answer 74

good, wow, that was a lot
THanks Turing!

Answer 75

Always a pleasure Sofiya :D

Answer 76

Looking back over this, I don't understand how you guys got a negative double root to the characteristic equation. I got m_1 = m_2 = +1. Plugging into the normal formula for repeated roots, getting e^x and xe^x, not e^-x. Am I missing something? @ganeshie8

Answer 77

@jim_thompson5910 @agent0smith

Answer 78

@Zarkon @terenzreignz Is anybody else familiar with ODE's seeing this, or am I missing something?

Answer 79

@zepdrix Anybody?

Answer 80

@Mendicant_Bias For the homogeneous solution? Yes the characteristic equation appears to be giving +1 as a repeated root, as you indicated

Answer 81

So yah, that was a mistake early in the thread :C