Question

y'- 3y = x.e^(2x) + 6
with undetermined coef method
any help ???

Answer 1

first solve the homogenious equation

y'-3y=0

Answer 2

this can be put in as

m-3=0

Answer 3

m=3 k=1

Answer 4

\[y_c=c_1e^{3x}\]

Answer 5

Yeah so true

Answer 6

@mukushla .. can you finish this. I gotta go do something

Answer 7

@mukushla is says to do this by undetermined coefficients
I think the correct guess for the particular is \[Y_p=(Ax+B)e^{2x}+C\]

Answer 8

@mukushla if u do not want to do it , do not do it ok?

Answer 9

@TuringTest you are right but i could not understand too much, can u explain more please ?

Answer 10

@SkykhanFalcon ok

Answer 11

plug that guess into the original equation

Answer 12

\[Y_p=(Ax+B)e^{2x}+C\]\[Y_p'=2(Ax+B)e^{2x}+Ae^{2x}\]\[Y_p''=4(Ax+b)e^{2x}+4Ae^{2x}\]

Answer 13

that's correct turing

Answer 14

\[Y_p''-3Y_p=xe^{2x}+6\]\[4Axe^{2x}+4Be^{2x}+4Ae^{2x}-3Axe^{2x}-3Be^{2x}-3C=xe^{2x}+6\]thanks for double-checking :)

Answer 15

@TuringTest i love u dude :D thank u so much

Answer 16

so true and so helpful

Answer 17

\[Axe^{2x}+Be^{2x}-3C=xe^{2x}+6\]from which we can infer\[A=1\]\[B=0\]\[-3C=6\implies C=-2\]so\[Y_p=xe^{2x}-2\]I hope....

Answer 18

@TuringTest can i ask a question

Answer 19

you do not understand the

\[(Ax+b)e^x+c\]?

your particular solution will be in the form of your f(x) on the right side

since you have xe^x, you have a polynomial of power one and e^x therefore the base for that would be

\[(Ax+B)e^x\]

lastly you have 6 which is a c so your particular will be in the form

\[(Ax+B)e^x+c\] this is your particular solution wwith undetermined coefficients. since it's a solution , you can sub it into the equation like turing did and solve as such

Answer 20

oh crap it's y' not y''

Answer 21

I messed up by going to far :S

Answer 22

lol o well haha

Answer 23

@TuringTest how can you know Yp= (Ax+B)e^2x +C ???

Answer 24

I just explained it Skykhan however i can give you a table that shows some of the ways...

Answer 25

oo Thank u

Answer 26

\[Y_p'-3Y_p=xe^{2x}\]\[2(Ax+B)e^{2x}+Ae^{2x}-3Axe^{2x}-3Be^{2x}-3C=xe^{2x}+6\]\[-Axe^{2x}+(A-B)e^{2x}-3C=xe^{2x}+6\]hope I did it right that time

Answer 27

you may also find these resources useful

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

http://integral-table.com/downloads/ODE-Summary.pdf