Question

what Yp should i use if i have this eqn? y''-y'-6y=1+e^-2x

Answer 1

try \(Y_p = A + Be^{-2x}\)

Answer 2

why not Ax+B ?

Answer 3

isit because 1 on the RHS doesnt have an x counter part?

Answer 4

yes but with that guess, the exponent is disappearing on left hand side

Answer 5

@experimentX

Answer 6

i used the eqn you give but i cant find the coefficients.. i used Ax+B and i got yp= -1/6+e^-2x

Answer 7

i gotta go out soon ... what method are you using?

Answer 8

undertermined coefficients

Answer 9

\(y= A + Be^{-2x}\)
\(y'= - 2Be^{-2x}\)
\(y''= 4Be^{-2x}\)
--------------------
\(y'' - y' - 6y = -6A = 1 + e^{-2x}\)


\(\implies A = \dfrac{-1}{6}\)

Answer 10

B just disappeared... :/

Answer 11

should i change the initial guess @experimentX

Answer 12

try individually ...
first for
Y_p = x
Y_p = e^(2x) or what ever ....
then use superposition theorem.

Answer 13

that you are using A is giving wrong answer.

Answer 14

it must be Cx+D

Answer 15

or try
y = Ax+B + Ce^( .... x)

good luck!! gotta go out for a while.
probably won't be back.

Answer 16

http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients#Typical_forms_of_the_particular_integral
the superposition of first and second.

Answer 17

okay... wil try thnks @experimentX :)

Answer 18

\(\dfrac{-1}{6} - \dfrac{1}{5}e^{-2x}\)

Answer 19

wolfram is giving that^^

Answer 20

so that means Ax+B + ce^-2x is correct?

Answer 21

okay okay ... this is 1/6
http://www.wolframalpha.com/input/?i=y%27%27-y%27-6y%3D1

Answer 22

the other is
http://www.wolframalpha.com/input/?i=y%27%27-y%27-6y%3De%5E%28-2x%29

Answer 23

hmm.. does this mean it cant be solved by undetermined coefficients? do method power series apply on this?

Answer 24

\(Y_p = A + Bxe^{-2x}\)

Answer 25

since the first guess \(Y_p = A + Be^{-2x}\) didnt work,
we make a new guess \(Y_p = A + Bxe^{-2x}\)

Answer 26

okay ill try

Answer 27

yeah! i got it! yay! xD

Answer 28

thanks! but, i'm not really sure how to guess what's the right Yp.. is there some kind of a clue?

Answer 29

\(y = A + Bxe^{-2x}\)
\(y' = Be^{-2x} - 2Bxe^{-2x}\)
\(y'' = -2Be^{-2x} - 2Be^{-2x} + 4Bxe^{-2x}\)
-----------------------------------------------
\(y'' - y' - 6y = 6Bxe^{-2x} - 5Be^{-2x} - 6A - 6Bxe^{-2x} = 1 + e^{-2x}\)

Answer 30

\(y'' - y' - 6y = - 5Be^{-2x} - 6A = 1 + e^{-2x}\)

\(A = \dfrac{-1}{6}\)

\(B = \dfrac{-1}{5}\)

Answer 31

yup! i got -1/6 and -1/5 too! :D

Answer 32

it is the right solution as plugging it in satisfies the right hand side, right ?

Answer 33

oh you're asking how to make initial guess for \(Y_p\) ?

Answer 34

yeah.. i'm not really sure whenever i guess..

Answer 35

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx#Second_UnderCoeff_Ex8c

Answer 36

my prof have this solution but im not really sure how he does it

Answer 37

how does he know that it precicely the solution for the complementary equation?

Answer 38

you fine wid solving the complementary solution, right ?

Answer 39

complementary means the chaaracteristic equation?

Answer 40

yup solution from characteristic equaiton

Answer 41

\(y'' + 2y' + y = xe^{-x}\)

complementary solution : \(y_c(x) = c_1e^{-x} + c_2xe^{-x}\)

Answer 42

complementary solution the solution for \(y'' + 2y' + y = 0\)

Answer 43

yeah because it has 2 distinct roots

Answer 44

ohh equal

Answer 45

\(y'' + 2y' + y = xe^{-x}\)

\(y_c(x) = c_1e^{-x} + c_2xe^{-x}\)


To find the particular solution, look at right hand side and make a guess

Answer 46

\(y'' + 2y' + y =\color{red}{ xe^{-x}}\)

Answer 47

the guess for polynomial \(\color{red}{x}\): \(Ax + B\)
the guess for exponent \(\color{red}{e^{-x}}\) : \(Ce^{-x}\)

so the overall guess is : \((Ax+B)(Ce^{-x})\)

Answer 48

but C is just an unnecessary constant, so

so the overall guess is : \((Ax+B)e^{-x}\)

Answer 49

first guess : \(Y_p = (Ax+B)e^{-x}\)

Answer 50

fine till here ?

Answer 51

yup! :D thats my guess too.. but why it isnt right?

Answer 52

it is right, and it is indeed a solution. but the bad thing is it is `part of ur complementary solution family`. it is not constrained to the `right hand side specific function`.

Answer 53

\(y_c(x) = c_1e^{-x} + c_2xe^{-x}\)

plugin \(c_1 = B, ~ c_2 = A\) above^

Answer 54

you get back ur first guess, right ?

Answer 55

so ur `first guess` is useless since its already part of complementary solution.
u need to make a guess thats not part of `complementary solution family`

Answer 56

ohh.. i see, so thats why he kept adding x then? i noticed, is it right if i say, Axe^-x is identical to xe^-x, and (Ax^2+Bx)e^-x is identical (the B part which is Bxe^-x) that's why its useless?

Answer 57

yesss
lets see why second guess is useless

Answer 58

first guess : \(Y_p = (Ax+B)e^{-x} \)
second guess : \(Y_p = (Ax^2+ Bx)e^{-x} \)

Answer 59

how is the second guess part of complementary solution itself ?

Answer 60

the Bxe^-x ? haha

Answer 61

yes, complementary solution : \(y_c(x) = c_1e^{-x} + c_2xe^{-x}\)

plugin \(c_1 =A= 0, ~~ c_2 = B\)
u get back ur second guess

Answer 62

so the second guess is bad too

Answer 63

anyway going back to our previous question.. y"-y'-6y=1+e^-3x
you put A because RHS is 1 (a constant) and you put Bxe^-3x so that it wont be the same as in the complementary solution?

Answer 64

for the original question,

complementary solution : \(y_c(x) = c_1e^{-2x} + c_2e^{3x}\)

Answer 65

diff eq : \(y''-y'-6y=1+e^{-2x}\)

our first guess : \(y_p = A + Be^{-2x}\)

Answer 66

clearly, when this is part of complementary solution family wid \(c_1 = B, ~c_2 = A = 0\)

Answer 67

its a bad guess

Answer 68

i see now.. okay, how about for this: involving trig: 9y''+y=e^-3xsin(x/3)+ xcos(x/3)
my guess would be, Ae^-3xsin(x/3)+ Be^-3xcos(x/3) what do you think?

Answer 69

\[9y"+y=e ^{-3x}\sin (\frac{ x }{ 3 }) + xcos(\frac{ x }{ 3 })\]

Answer 70

complementary solution : \(y_c = c_2 \sin (\frac{x}{3}) + c_1\cos(\frac{x}{3}) \)

Answer 71

oh it would be the same then, so means i need to add x somewhere

Answer 72

guess for sin(kt) : Asin(kt) + Bcos(kt)

Answer 73

I think u wil get 8 variables to solve

Answer 74

woow why is that?

Answer 75

\(9y"+y=e ^{-3x}\sin (\frac{ x }{ 3 }) + x\cos(\frac{ x }{ 3 }) \)


\(y_p=[Ae ^{-3x}] [B\sin (\frac{ x }{ 3 }) + C\cos(\frac{ x }{ 3 }) ] + [Dx+E][F\sin (\frac{ x }{ 3 }) + G\cos(\frac{ x }{ 3 })] \)

Answer 76

here is the particular solution from wolfram :
http://prntscr.com/3e6kl4

Answer 77

doesnt look pleasant to solve so many constants ;/

Answer 78

haha yeah xD no need to solve these just looking for Yp haha xD

Answer 79

btw im not sure how u got your guess honestly

Answer 80

i used my table...

```
function : guess
x : Ax + B
e^(kx) : Ae^(kx)
sin(kx) : Asin(kx) + Bcos(kx)
cos(kx) : Asin(kx) + Bcos(kx)
```

Answer 81

ooohh!! this useful! haha thank you! :D

Answer 82

np :)