Question

\[y^{\prime\prime}-3y^\prime+2y=e^x\]

Answer 1

\[y_c^{\prime\prime}-3y_c^\prime+2y_c=0\]
\[m^2-3m+2=0\]\[(m-1)(m-2)=0\]\[m=1,2\]
\[y_c=Ae^x+Be^{2x}\]

Answer 2

\[y_p^{\prime\prime}-3y_p^\prime+2y_p=e^x\]
\[y_p=Cxe^x\]\[y^\prime_p=C(1+x)e^x\]\[y^{\prime\prime}_p=C(2+x)e^x\]
\[C(2+x)e^x-3C(1+x)e^x+2Cxe^x=e^x\]\[C\left(2+x-3-3x+2x\right)e^{x}=e^x\]\[-C=1;\qquad C=-1\]
\[y_p=-xe^x\]

Answer 3

\[y=y_c+y_p\]
\[y(x)=Ae^x+Be^{2x}-xe^x\]

Answer 4

my text book has this as the solution:
\[y(x)=Ae^x+Be^{2x}-(x+1)e^x\]

Answer 5

is the book right/?
where doe the extra one come from?

Answer 6

I think you cannot take particular solution as \(e^x\) because it is in the homogeneous solution too..

Answer 7

But no, yes you can take because you are multiplying x with it..

Answer 8

i took the particular as \(y_p=Cxe^x\)

Answer 9

\[A
_1e^x-e^x=A_2e^x\]so the constant for the other e^x part gets "absorbed" into the other

Answer 10

no need to write both; they are linearly dependent

Answer 11

They are both correct.

Answer 12

Yes, I got that Uncle Rocks..

I tried by using \((Cx + D)e^x\) then I got \(C = -1 \quad and \quad D = 0\) then also..

You are right @UnkleRhaukus

Answer 13

\[y(x)=Ae^x+Be^{2x}-xe^x-e^x=(A-1)e^x+Be^x=Ce^x+Be^{2x}\]same thing, yup :)

Answer 14

\[y(x)=Ae^x+Be^{2x}-xe^x-e^x\]\[=(A-1)e^x+Be^x-xe^x\]\[=Ce^x+Be^{2x}-xe^x\]

Answer 15

Again typo in middle line...
Ha ha ha..

Answer 16

yes there is :S

Answer 17

forgot a 2

Answer 18

but I hope you can fill in the missing 2

Answer 19

it's 1:16 am here folks, gimme a break :P

Answer 20

i think i get it now, the book is just using a constant that is different to my constant by one unit,

Answer 21

yeah that's for sure the deal
I just encountered this same coefficient problem recently, so I was able to spot it pretty quickly

Answer 22

so, i'll just leave the solution how i had it
\[y(x)=Ae^x+Be^{2x}-xe^x\]
is neater than
\[y(x)=A'e^x+Be^{2x}-(x+1)e^x\]

Answer 23

Would the book be advocating?:
\[y_p=Cxe^x+e^x\]

Answer 24

i am not sure @Valpey

Answer 25

@UnkleRhaukus turning test is saying right..
You have done it correctly..

Answer 26

im just not sure how the book go to what it did, or why i took that approach

Answer 27

why it*

Answer 28

p=1,m=0 Unkle,Do you remember?

Answer 29

i do not remember

Answer 30

take a look the post that help you for all problems. Amazing individual way.

Answer 31

what do yon mean by p and m