Question

y''' - y' = 25cos(2x)
@TuringTest so what about that we have more than a λ. how can i use these

Answer 1

i'm not entirely sure what you're asking in this case lambda usually comes with eigenvalues and functions

Answer 2

for both the complimentary and particular you will have to throw in some extra x's I think. How many I'm not so sure.

Answer 3

if you mean what if we have more than one solution for your coplementary

Answer 4

in last question you say m-3 = 0

Answer 5

when you particular = one of your complementary solutions , then you must multiply by x's until you get ne that is not

Answer 6

in that m(m-1)(m+1)=0

Answer 7

yes treat the primes as exponents so now you have

\[m^3-m=0\]

\[m(m^2-1)=0\]
m=0 of multiplicity , m=1 k=1, m=-1 k =1

Answer 8

writing this when you have m=0 it's just a constant and then adding x per multiplicity and for +1 you have two rational values your complementary will be

\[y_c=c_1+c_2e^x+c_3e^{-x}\]

Answer 9

so in this case nothing changes, because the guess for the particular is already linearly independent
so nothing changes

Answer 10

...for finding the particular I mean

Answer 11

ok i get that

Answer 12

wait i could not understand the relationship between k and m

Answer 13

alright so now for the particular you have a solution that will take the form of 25cos(x)

since when taking the derivative of cos you change from sin to cos, the possible solution is

Acos(x)+Bsin(x)

Answer 14

k=multiplicity m is a root

Answer 15

i.e. how many times that particular root m occurs is a number k called the multiplicity of that root

Answer 16

say if you had
\[m^3=0\]

it'd be m=0 k=3

Answer 17

oh i get that too :D thanks

Answer 18

so now that you know that a possible particular solution is in the form Acos(x)+Bsin(x)... just take the derivatives like turing did last time and plug them into the equation. Then solve

Answer 19

@Outkast3r09 when we choose particular solution as Acos(x) + B sin (x)
the equ is 0 = 25 cos x
:/

Answer 20

0 = 25cos (x)
i am sorry

Answer 21

ok i did :D

Answer 22

Let me make sure hold one

Answer 23

it was 25cos(2x) i calculated as 25cosx :S

Answer 24

ahh yes that's my bad also your particular will be in the form Acos(2x)+Bsin(2x)

Answer 25

@Outkast3r09 but in the end i found 6Asin(2x)-6Bcos(2x) = 25 cos(2x)
but answer should be -5sinxcosx = -2.5sin 2x

Answer 26

did i make a mistake somewhere

Answer 27

\[y_p=Acos(2x)+Bsin(2x)\]
\[y_p'=-2sin(2x)+2cos(2x)\]
\[y_p''=-4cos(2x)+-4sin(2x)\]
\[y_p'''=6sin(2x)-6cos(2x)\]

Answer 28

http://www.wolframalpha.com/input/?i=y%27%27%27+-+y%27+%3D+25cos%282x%29
it says so

Answer 29

\[6asin(2x)-6bcos(2x)-2asin(2x)+2bcos(2x)=25cos(2x)\]

Answer 30

i forgot about the a's and b's but you can see how they got there correct?

Answer 31

wait is not it wrong ?

Answer 32

yp′′′=8sin(2x)−8cos(2x)

Answer 33

ahh yes lol sorry i'm doing this all through the computer -.-

\[8asin(2x)-8bcos(2x)-(-2asin(2x)+2bcos(2x)=25cos(2x)\]
\[8asin(2x)-8bcos(2x)+2asin(2x)-2bcos(2x)=25cos(2x)\]
\[10asin(2x)-10bcos(2x)=25cos(2x)\]

-10b=25
10a=0

25/-10=b=-2.5

Answer 34

10a=0 ....a=0

Answer 35

so your particular is

0cos(2x)-2.5cos(2x)

they just use the double angle theorem

Answer 36

oh dudddeee :( i forgot the minus

Answer 37

-2.5cos(2x)=-2.5(2sin(x)cos(x)=-5sin(x)cos(x)

Answer 38

thank u so much :/ i am sorry i made u tired