Question

what is the corresponding linear DE to the particular solution \[y = 3x^2 - \cosh 2x\]

how to do this?

Answer 1

take derivative and remember you have eliminate the constant here it is 3..

Answer 2

basically i have no idea what root corresponds to cosh or any hyperbolic function

Answer 3

why take the derivtive?

Answer 4

Don't you want to find DE??

Answer 5

DE is differential equation..

You are given with solution..
And you have to find corresponding DE.

Answer 6

why derivative O.o

Answer 7

Do you know the form of DE ??

Answer 8

im given a particular solution...i need to find the DE

Answer 9

so wha?

Answer 10

That is what I am saying..

Take the derivative first time..

Answer 11

so whenever im given a solution and i need a diff eq i take the derivative?

Answer 12

Yes.. you will..

Answer 13

hmm \[y' = 6x - 2\sin h 2x\]
wat now?

Answer 14

you have to eliminate all the constants..

Answer 15

Or may be @UnkleRhaukus will help you better..

ha ha ha..

Answer 16

what do you mean??

Answer 17

Call Uncle Rocks..

Answer 18

lol..you start it and then you let others finish :p that's not fair

Answer 19

Sorry I don't want to give you false or wrong knowledge..

Answer 20

Take derivative one more time:

\[y'' = 6 - 4\cosh(2x)\]

Answer 21

By taking derivative one more time 6 will be eliminated:

\[y'' = -8\sinh(2x)\]

Answer 22

...and i still dont know the roots that correspond to hyperbolic functions :(

Answer 23

May Be this is your answer:

\[y'' + 8\sinh(2x) = 0\]

Answer 24

actually it's (D^4 - 4D^2); y=0

dunno what that means

Answer 25

You are given with:

\(y = 3x^2 - \cosh 2x\)

Can you take it derivative four times ???

Do it hurry up..

Answer 26

Wait let me try first..

Answer 27

No I am not getting this actually..

Answer 28

lol =))) hard sint it

Answer 29

Meaning?

Answer 30

hard isnt it* lol

Answer 31

No it is not hard..

I studied 4 years before about DE..

My concepts gone far away from me..

Uncle will help you now..

Answer 32

i hope so

Answer 33

\[\cosh x=\frac{e^x+​e^{​-x}}{​2}\]

Answer 34

oh yeahhh

Answer 35

so cosh2x is....?

Answer 36

Replace x with 2x..

Answer 37

mmhmm

Answer 38

from there i derive?

Answer 39

break \(y_p \) up into \(y_{p_1}+y_{p_2}+y_{p_3}\)

Answer 40

that means what again?

Answer 41

\[y_p = Ax^2 - B\cosh 2x\]\[y_{p_1}=Ax^2\]\[y_{p_2}=Be^{x}\]\[y_{p_3}=Ce^{-x}\]

Answer 42

woah what? how?

Answer 43

those constant are not ment to be the same

Answer 44

but why? how? wwhat?

Answer 45

take the derivatives of y_p_1 and find A

Answer 46

why do you have Be^x and Ce^-x

Answer 47

they came form the cosh x

Answer 48

but it's cosh2x

Answer 49

A B and C can take any values they are constants..

You have changed cosh(2x) into exponential terms..

Answer 50

well stick some 2's on

Answer 51

eh?

Answer 52

in the exponents

Answer 53

Be^2x and Ce^-2x

Answer 54

what do you mean stick some 2's on?

Answer 55

oh

Answer 56

like that
.

Answer 57

so where did 1/2 go off to?

Answer 58

it got absorbed in the constant like waterineyes was saying

Answer 59

ohh..okay so what's next?

Answer 60

\[y^\prime_{p_1}=\]

Answer 61

2A x