Question

y''-2y'+5y=5x^2-4x+2; y(0)=0 y'(0)=2

Answer 1

got the complimentary solution yet?

Answer 2

no =(

Answer 3

that's just the solution to the homogeneous equation\[y''-2y'+5y=0\]can you do that?

Answer 4

I can, I just don't know what to do with y(0)=0 e.c. given

Answer 5

we are going to use that at the end to find the constants c1 and c2

do you have to use a particular method to find the particular solution?
variation of parameters, undetermined coefficients, etc.?

Answer 6

yeah, I need the particular solution

Answer 7

I repeat:

do you have to use a particular method to find the particular solution?
variation of parameters, undetermined coefficients, etc.?

Answer 8

I don't...

Answer 9

Well I guess undetermined coefficients is what I jump to

do you have any familiarity with that technique?

Answer 10

I do

Answer 11

ok, so you wanna try it on paper while I do the same?

Answer 12

ok

Answer 13

ok I got my answer for the particular solution, let me know what you get so we can compare

Answer 14

-x^2+6/5x-26\25

Answer 15

er... no not quite
what was your guess for the particular solution?

Answer 16

well, f(x)=5x^2-4x+2=Ax^2+Bx+C
F'(x)=2Ax+B
F''(x)=2A
2A-2Ax-B-5Ax^2-5Bx-5C=5x^2-4x+2
A=-1
B=6/5
C=-26/25

Answer 17

you just messed up on the system is all...

Answer 18

wait, is it +5y or -5y ???

Answer 19

...in the original problem I mean...

Answer 20

also you forgot to multiply y' by 2...

Answer 21

Oh... now i got it

Answer 22

now A=1, B=-1,6, C=-0, 64 am I right?

Answer 23

are you sure your original question is

y''-2y'+5y=5x^2-4x+2

???

(all the +'s and -'s correct?)

Answer 24

and how do you have multiple answers for B and C o-0 ???

Answer 25

yes, they are correct

Answer 26

in that case you are still wrong, but closer
A=1 is correct

but how do you have multiple answers for the other constants?

Answer 27

B=0!!!

Answer 28

yeah ;)

Answer 29

I've just found the mistake

Answer 30

hooray!
so your particular is...?

Answer 31

it is x^2

Answer 32

yep :)
now the full solution is the linear combinations of the and complimentary solutions plus the particular \[y=y_c+y_p\]and what is that in your case?

Answer 33

y=e^x(C1cos2x+C2sin2x)+x^2

Answer 34

good, this is our \(general\) solution because it still has C1 and C2 in it

to find the exact solution we need to apply the initial conditions

we already know what y(x) is, so we can get an expression for y(0).
what about y'(x) ?
what then is y'(0) ?

Answer 35

y(0)=C1=> C1=0

Answer 36

yep :)

Answer 37

and C2=?

Answer 38

y'(x)= e^x(C2sin2x)+e^x*C2cosx*2+2x

Answer 39

I think you had a slight typo on the cosine argument
y'(x)= e^x(C2sin2x)+e^x*C2cos2x*2+2x

Answer 40

but that doesn't change anything anyway :P

Answer 41

so what is C2 ???

Answer 42

C2 is 1

Answer 43

right, so the final answer to out IVP is...?

Answer 44

our*

Answer 45

y=e^x*sin2x+x^2

Answer 46

yes indeed :)
and in case you don't believe me, ask the wolf

http://www.wolframalpha.com/input/?i=solve+y%27%27-2y%27%2B5y%3D5x%5E2-4x%2B2+%2Cy%280%29%3D0%2C+y%27%280%29%3D2

(note that the wolf does not have the best reputation when it comes to solving DE's)

Answer 47

Thank u so much!!!! Now I've made it out

Answer 48

you're welcome :) see ya!

Answer 49

I have 3 more equations to solve if you don't mind helping me