Question

the m' for a homogeneous linear DE that has \(y = 3e^x \sin^2 x\) as particular solution is/are what?

what does this question even mean?

Answer 1

my main problem here is i dont know what m' means

Answer 2

@UnkleRhaukus please help T_T

Answer 3

\[y_c=Ae^{mx}\]

Answer 4

but where is the DE?

Answer 5

And just to clarify, m' is the first derivative of m, m'' would be the second derivative, m''' would be the third and so on...

Answer 6

that's what i dont get... i know m is the root...so what does m' mean o.O i take the derivative of a root?

Answer 7

say you had a DE , like \[y''-y'-2y=0\]
\[y=Ae^{mx};\qquad y'=Ame^{mx};\qquad y''=Am^2e^{mx}\]
\[Am^2e^{mx}-Ame^{mx}-2Ae^{mx}=0\]\[A(m^2-m-2)e^{mx}=0\]\[m^2-m-2=0\]\[(m-2)(m+1)=0\]\[m=2,-1\]
\[y=Be^{2x}+Ce^{-x}\]

Answer 8

okay...im not getting you....

Answer 9

the m's are the roots of the homogenous equation

Answer 10

i think the roots in this problem is

\[m = 1 \pm i\]
right?

Answer 11

hmm how come it's sin^2 o.O

Answer 12

are you ment to use reduction of order because you now know a particular solution.

Answer 13

i dont know what you're talking bout...im not good with fancy terminologies

Answer 14

Do you know which satisfy the equation are called solutions ??

Answer 15

hmm?

Answer 16

Finding the particular solution mean that we are finding the solution to this DE which will satisfy this DE..

Answer 17

perhaps a demo? im not good with words...im dyslexic with words heh

Answer 18

Can I attach a file??

Answer 19

uhmm sure?

Answer 20

Firstly I should give you demo right??

Answer 21

hmm nevermind i got unkle's demo already

Answer 22

perhaps just a guidance how to solve this now

Answer 23

This is simple one not related to m ..

Answer 24

Uncle can explain you far more better than me..

Answer 25

aww @UnkleRhaukus come back T_T

Answer 26

hmm

Answer 27

this is easy :D

Answer 28

really? so how?

Answer 29

yeh man it is !

Answer 30

so how?

Answer 31

man take 3 aside ok

Answer 32

im not really interested whether t's hard or not..what i need to know is how

Answer 33

it's*

Answer 34

yeah sure...then next?

Answer 35

then open sin^2x

Answer 36

open?

Answer 37

i mean in terms of double angle of cos

Answer 38

(1-cos2x)/2

Answer 39

oh...still cant see it though

Answer 40

now take 1/2 as common ok?

Answer 41

\[\frac 32 e^x(1 - \cos (2x) )?\]

Answer 42

yup

Answer 43

so a = 1 and b = 2?

\[m = 1\pm 2i?\]

Answer 44

i have a different apporach from here

Answer 45

oh cool..i see it now..thanks

Answer 46

lgb can you do or should i help

Answer 47

Go through this @lgbasallote

Answer 48

\[\frac{3e^x}{2} - \frac{3e^x \cos 2x}{2}\]
so m = 1 and m = \(1 \pm 2i\)