Question

help!!!

solve the differential eqn: x''-2x'+x=e^(2t)+t

given x(0)=0 and x'(0)=4

Answer 1

you can use variation of parameters or undetermined coefficients
which do you prefer?

Answer 2

either way you should first get the complimentary solution, i.e. solve\[x''-2x'+x=0\]

Answer 3

undetermined co-efs....(f-1)(f-1)=0

x=Ae^x+Bxe^x

Answer 4

ok, the we have the particular given by\[y_p=-y_1\int\frac{y_2g(t)}Wdt+y_2\int\frac{y_1g(t)}Wdt\]where g(t)=e^(2t)+t, and W is the Wronskian. Do you have the Wronskian yet?

Answer 5

then*

Answer 6

noooo, we haven't done that yet... wat does W indicate?

Answer 7

\[W(y_1,y_2)=\left|\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right|\]

Answer 8

(O__o)... oh my!!!

Answer 9

erm, if you don't know at all what I'm saying then I think you need to review variation of parameters
the above is a determinant of the matrix. do you remember determinants for 2x2 matrices?

Answer 10

Oh I'm sorry you wanted undetermined coefficients huh...

Answer 11

ow yes... that cheap stuff... yes... in a 2x2 matrix with [a b,c d]... becomes ad-bc

Answer 12

my mistake I was doing variation of parameters
so for undetermined coefficients, what is your guess for the particular going to be?

Answer 13

i tried 4 guesses... stil got contradictions... :(

i tried x= Ae^(2t)+Ax+t

Answer 14

you got x and t mixed together, is that what you tried to use?

Answer 15

yes :(

Answer 16

yp=Ae^(2t)+Bt+C

Answer 17

the guess for the e^(2t) part is Ae^(2t)
the guess for the t is Bt+C
add them together

Answer 18

0=A+B+C for x

4=2A+B for x'

Answer 19

did you plug in the initial conditions or something?

Answer 20

wait... after substituition


i'm stuck here 1-B+Ae^(2t)+C=e^(2t)+t

Answer 21

I don't see how you got 1-B

Answer 22

x''=4Ae^(2t)
x'=2Ae^(2t)+B
x=Ae^(2t)+Bt+C

Answer 23

after substitution

yp=Ae^(2t)+Bt+C
y'=2Ae^(2t)+B
y''=4Ae^(2t)+1

Answer 24

B is a constant

Answer 25

when u differentiate B, becomes 0 or 1?

Answer 26

lone constants differentiate to zero

Answer 27

owwww, ok, let me try again

Answer 28

i got A=1 B=1 C=1

Answer 29

C is not 1, but the others are

Answer 30

When i equated coeffic.... i got (-B+C)=0

Then -B=-C... C and B hav the same values

Answer 31

I think you dropped a 2 from the -x' part, look again

Answer 32

still getting the same answer... 4Ae^(2t)-4Ae^(2t)-2B+Ae^(2t)+Bt+C=e^(2t)+t

Answer 33

that's my substitution

Answer 34

this is right

Answer 35

so it simplifies to what?

Answer 36

Bt=t
-2B+B+C=0
Ae^(2t)=e^(2t)

Answer 37

what is the Bt=t thing?

Answer 38

Ae^(2t)+Bt-2B+C=e^(2t)+t
is what I was going for

Answer 39

-2B?

Answer 40

set the coefficients equal
what is the coefficient of the e^(2t) term on the left?

Answer 41

it's A

Answer 42

and on the right?

Answer 43

1

Answer 44

right.so A=1
now what is the coefficent of t on the left and rigth? what is B then ?

Answer 45

B=1

Answer 46

good
and what is the constant term on each side?

Answer 47

-2B+C=0

Answer 48

so what is C ?

Answer 49

C=2

Answer 50

yes

Answer 51

wow... so my soln becomes yp=e^(2t)+t+2

Answer 52

that's the particular, yep

Answer 53

y=Ae^(x)+bxe^(x)+e^(2t)+t+2??

Answer 54

yes, now you can apply the initial conditions

Answer 55

and keep your x's and t's straight...

Answer 56

x(t)=c1e^(t)+c2te^(t)+e^(2t)+t+2

Answer 57

let me try it out

Answer 58

x(0)=0 i leave the t as it is in my substitution?

Answer 59

x=0 and t=0, so you have to plug in 0 for t

Answer 60

i'm getting A=-3... B=0

Answer 61

A=-3 yes, B=0 no

Answer 62

you can't find B without the other condition. Did you apply iot yet?

Answer 63

it*

Answer 64

yes... x'(0)=4

x'=Ae^(2t)+Bte^(2t)+e^(2t)

Answer 65

that is not right for x'

Answer 66

and those first two e^(2t)'s should be e^t's

Answer 67

ow yes... i'm kinda confused :(

Answer 68

\[x(t)=Ae^{t}+Bte^{t}+e^{2t}+t+2\]\[x'(t)=Ae^t+B(t+1)e^t+2e^{2t}+1\]

Answer 69

u really fast... u on matlab or mathematica?

i got B=6

Answer 70

no computer help, I just know these things and already solved it
it is not B=6, maybe you got that from my typo though

Answer 71

not 6? :(

Answer 72

\[x'(0)=A+B+2+1=4\]

Answer 73

B=4?? :) yes?

Answer 74

yes :)
and that's basically the end of the problem

Answer 75

not really :)

Answer 76

add the complimentary and the particular
the end

Answer 77

are u tired of helping me? :(... can u atleast check my last step... they asked me to find x(1)

Answer 78

well, what is x(t) ?

Answer 79

x(1)= 13.107

Answer 80

final answer :)

Answer 81

yep, that's right :)

Answer 82

thank u... its midnyt here in south africa... got a test at 2pm... thank u so much... wil post another question if i do get stuck sumwher

Answer 83

You're welcome. I'll be happy to help if I'm available. See ya!

Answer 84

:)