Question

Differential equations help

Answer 1

\[\Large \frac{d^4y}{dx^4}-y=coshx \sin hx\]

Answer 2

I guess a good start would be to put it into exponential form

Answer 3

@iambatman can you give me some help? im on top top question right now its geometry

Answer 4

ill post my attempt guys..wait

Answer 5

I converted RHS to exponential forms.
But before that,I tried to find the CF.
Here is the auxilliary equation
\[\Large (m^4-m)=0\]
\[\Large m=0,1,\omega,\omega^2\]
\[\LARGE C.F= C_1 e^x + C_2 e^{\omega x} + C_3 e^{\omega^2 x} + C_4 \]

Answer 6

after solving a bit of the omega term I get this and I think im wrong \[\Large e^\frac{-x}{2} (2 \cos \sqrt{3}x/2)\]

Answer 7

this is what I got after solving C2 and C3 terms....

Answer 8

shouldnt you be solving \(m^4-1 = 0 \) ?

Answer 9

ooh..right..:P is this solution right if the auxilliary equation was this?

Answer 10

lets see what wolfram says
http://www.wolframalpha.com/input/?i=solve+y%27%27%27%27-y%3D0

Answer 11

yeah the ans is right

Answer 12

m^2=1
m=1,-1
ce^x+c1e^-x+c2

Answer 13

and m^2=-1
m=+- i?

Answer 14

yes

Answer 15

solving PI seems difficult
(e^x-e^-x/2)(e^x+e^-x)/2

Answer 16

e^2x-e^2x/4?

Answer 17

i done seem to remember how to solve particular solution for orders > 2
need to revise

Answer 18

ohh..

Answer 19

when we have e^ax on RHS we simply replace D^2 with a^2

Answer 20

in 1/f(D)

Answer 21

when right hand side is \(e^{ax}\)

particular solution = \(\large \dfrac{e^{ax}}{f(a)}\)

are you using this ?

Answer 22

here we have \(f(D) = D^4 - 1 \)

Answer 23

\[\Large \frac{ c_1 e^x}{D^4-D} + \frac{c_2 e^{-x}}{D^4-D} +\frac{ c_4}{D^4-D} +\frac{c_5}{D^4-D} \]

Answer 24

sorry..D^4-1..im confusing it again :/

Answer 25

RHS : \(\large \dfrac{e^{2x}-e^{-2x}}{4}\)

Answer 26

damn!!
I divide it by D^4-1 and for first term D^4=2^4 and for 2nd term too the same

Answer 27

particular solution for \(\frac{e^{2x}}{4}\): \(\frac{e^{2x}/4}{2^4-1} = \frac{e^{2x}}{60} \)

Answer 28

yes :)

Answer 29

how did we get cosx and sinx in the solution then/?

Answer 30

Seems Euler's identity was used

Answer 31

it has nothing to do with particular solution

Answer 32

solving C e^i + C1 e^-i we get C 2cos(2)

Answer 33

they are part of your complementary solution

Answer 34

yes i know that...

Answer 35

consider this characteristic root : 0 + 1i

Answer 36

a solution for \(y^4-y = 0\) is \(e^{(0+1i)x}\)

yes ?

Answer 37

yes

Answer 38

i got it...:P

Answer 39

split that into real and imaginary parts

Answer 40

#blundermistake!!

Answer 41

Yeah, it's pretty easy from there

Answer 42

its done!

Answer 43

I have another quick question.
How do I find P.I when I have 2^x on RHS?

Answer 44

okie there is a theorem about what you need to do for solutions when you get complex roots

it will be there in your textbook

Answer 45

yep,I was getting confused for not proceeding stepwise

Answer 46

not sure, i think this method works only when the right hand side functions are convertable to form e^(ax) or polynomial

Answer 47

yep,there are various cases when we have things on RHS but I cant find for 2^x

Answer 48

2^x = e^(ln2 * x)

Answer 49

that should do i think ^

Answer 50

bingo!! :O

Answer 51

thanks everyone :)

Answer 52

Thanks ganeshie xD