Question

hey everyone :)

Answer 1

hi there

Answer 2

how are u? u from Boondocks? :D

Answer 3

i am well, i am not from boondocks, how are you?, do you have a mathematics question?

Answer 4

Nice reply @UnkleRhaukus :)

Answer 5

\[\boxed{\color{grey}{\ddot\smile}}\]

Answer 6

\[\Huge{\boxed{\color{purple}{\ddot{\smile}}}}\]

Answer 7

yes.... general solution to the this differential equation!!! :D

y''+4y'+4y=e^(-2x)

:)

Answer 8

\[y''+4y'+4y=e^{-2x }\]
so step one is to find the complementary homogeneous solution
\[y_c''+4y_c'+4y_c=0\]

Answer 9

yes... like solve for x on x^2+4x+4=0

(x+2)(x+2)=0

Answer 10

then Yc= Qe^(-2t) + Rxe^(-2t)

Answer 11

so now... i have to find the particular solution...:)

Answer 12

right

Answer 13

are you going to use the method of undetermined coefficients?

Answer 14

yes.... that's the one i used to get my Yc

Answer 15

you already have \(y_c\), you need \(y_p\)

Answer 16

we might first look to try\[y_p=Ce^{-2x}\], but this wont work , do you know why it wont work?

Answer 17

we'll have to try
\[y_p=Cxe^{-2x}\]

Answer 18

huh?

Answer 19

i'm getting zero on RHS... no "C" value

Answer 20

uncle? u there?

Answer 21

i thought you already found y_c

Answer 22

\[y_p''+4y_p'+4y_p=e^{-2x }\]

Answer 23

wait!!!!! i'm done with the homo part... now i want to solve for e^(-2x) and complete my solution

Answer 24

{{y[x] -> E^(-2 x) C[1] + E^(-2 x) x C[2] + e^x/(2 + Log[e])^2}}

Wolfram Mathematica 8 gave me that<==

Answer 25

i got half of the question right

Answer 26

\[y_p=Cxe^{-2x}\]
\[y'_p=?\]\[y''_p=?\]

Answer 27

once you have these, plug into
\[y_p''+4y_p'+4y_p=e^{-2x }\]
and solve fo the constant \(C\)

Answer 28

yes... when i used that it gave me a zero

Answer 29

how?

Answer 30

what did you get for the derivatives of y_p

Answer 31

subst back to LHS... tried to solve

Answer 32

\[y′_p=?\]

Answer 33

use the product rule

Answer 34

try it!

Answer 35

\[y'_p=(Cxe^{-2x})'=(Cx)'e^{-2x}+Cx(e^{-2x})'=Ce^{-2x}-2Cxe^{-2x}\]

Answer 36

woooow!!! thanks

Answer 37

can you now find \(y_p''\) and substitute these to find \(C\)

then you will have \(y_p\)


your gereral solution is
\[y=y_c+y_p\]