Question

how to find the \(y_p\) of this using undetermined coefficients

\[(D^2 - 1)y = e^x +1\]

Answer 1

have you found \(y_c\) already?

Answer 2

Have you found \(y_c\) ???

Answer 3

\[(D^2 - 1)y_c = 0\]\[m^2-1=0\]\[m=\pm1\]\[y_c=Ae^x+Be^{-x}\]

Answer 4

^that's what i got for yc too

Answer 5

notice how the non-homogenous term \(e^x\) on the right hand side of the original DE is a scalar multiple of one of the complementary terms \(Ae^x\),
this corresponds to RESONANCE , so you multiply what you might first think for \(y_p\) by x
\[y_p=(A+Be^x)x\]

Answer 6

huh?

Answer 7

The particular solution will be the sum of the particular solutions.
\[y''-y=e^x\] and
\[y''-y=1\]

Answer 8

and that means?

Answer 9

go slow =_=

Answer 10

im not following

Answer 11

\[y_p=y_1+y_2\]\[y_{p_1}=A\]\[y_{p_2}=Bxe^x\]

Answer 12

...i dont think that was any slower o.O

Answer 13

what bit are you stuck on?

Answer 14

just imagine i dont know how to use undetermined coefficients

Answer 15

i dont have much practice yet so im not yet good at it

Answer 16

ok so you have a table right/?

Answer 17

table?

Answer 18

maybe showing my solution will help...im not sure if this is right
\[y_p = Ae^x + B\]
\[Dy_p = Ae^x\]
\[D^2 y_p = ae^x\]
\[(D^2 - 1)y = e^x + 1\]
\[Ae^x - (Ae^x + B) = e^x + 1\]
\[-B = e^x + 1\]
\[B = 1 - ^x\]
is this right? if yes what's next?

Answer 19

that should be \[B = 1 - e^x\]

Answer 20

yeah , that y_p wont work because you already have a e^x term in the complementary solution

Answer 21

so what's the right way?

Answer 22

\[Ax e^x + B?\]

Answer 23

yeah i guess that could work, (i havent tried it)

Answer 24

That should work

Answer 25

so what's next?

Answer 26

\[y_p=Axe^x+B\]\[y^\prime_p=\]\[y^{\prime\prime}_p=\]

Answer 27

\[Dy_p = A(xe^x + e^x)\]
\[D^2y_p = A(xe^x + 2e^x)\]
\[(D^2 - 1)y = e^x +1\]
\[Axe^x + 2Ae^x - Axe^x - B = e^x+1\]
is that right?

Answer 28

looks good

Answer 29

so what's next?

Answer 30

now find A,B

Answer 31

how?

Answer 32

gaussian method?

Answer 33

ok
\[Axe^x + 2Ae^x - Axe^x - B = e^x+1\]
\[2Ae^x - B = e^x+1\]

Answer 34

A = 1/2
B = -1?

Answer 35

simply equate co efficients ., yeap

Answer 36

so then....sub into yp?

Answer 37

so you now have A and B, plug them into y_p, yep

Answer 38

\[\frac{xe^x}{2} +1\]

Answer 39

and finally
\[y(x)=y_c+y_p\]

Answer 40

oh wait...it needs yc

Answer 41

-1 not +1

Answer 42

\[C-1 e^x + c_2e^{-x} + \frac{xe^x}{2} - 1\]

Answer 43

ugh darn latex

Answer 44

\[C_1 e^x + C_2 e^{-x} + \frac{xe^x}{2} - 1\]

Answer 45

you know how to copy paste right?

Answer 46

with the right click-show math as-tex commands

Answer 47

yeah..just too lazy

Answer 48

anyhoo,

\[\Huge\color{red} \checkmark\]

Answer 49

nice thanks

Answer 50

more/?

Answer 51

haha lol yes please