Question

Hyperbolic functions..

Answer 1

\(\large y= c_1 \sinh (kx) + c_2 \cosh (kx) \)

Answer 2

take the derivative, wat do u get ?

Answer 3

use below :

(sinhx)' = coshx
(coshx)' = sinhx

Answer 4

okay, the derivative of sinh is cosh and derivative of cosh is -sinh..

Answer 5

derivative of cosh is just sinh

Answer 6

there is no negative sign

Answer 7

\(\large y= c_1 \sinh (kx) + c_2 \cosh (kx)\)

take derivative


\(\large y'= c_1 \cosh (kx)(k) + c_2 \sinh (kx)(k)\)

\(\large ~~~= c_1k \cosh (kx) + c_2k \sinh (kx)\)

Answer 8

take derivative again

Answer 9

you wud get :


\(\large y'= c_1k \cosh (kx) + c_2k \sinh (kx)\)

\(\large y''= c_1k \sinh (kx)(k) + c_2k \cosh (kx)(k)\)

\(\large ~~~~= c_1k^2 \sinh (kx) + c_2k^2 \cosh (kx)\)

\(\large ~~~~= k^2\left(c_1\sinh (kx) + c_2 \cosh (kx)\right)\)

Answer 10

the thing inside parenthesis should look familiar ?b

Answer 11

well, i thought the derivative of cos is -sin

Answer 12

im sorry but imma little bit confused.. but.. how do i take the derivative after this? im sorry

Answer 13

derivative of \(\cos(x)\) is \(-\sin (x)\)

derivative of \(\cosh(x)\) is \(\sinh (x)\)

Answer 14

yes.. they're correct and what's after this? =)

Answer 15

What's \(\large k^2 y =\color{red}?\)

Answer 16

stare at last line

Answer 17

\(\large ~~~~= k^2\left(c_1\sinh (kx) + c_2 \cosh (kx)\right)\)

Answer 18

the thing inside parenthesis is exactly same as ur given function \(y\) right ?

Answer 19

we're done actually :)

Answer 20

just replace the entire thing inside parenthesis wid "y"

Answer 21

\(\large y'' ~~~~= k^2\left(c_1\sinh (kx) + c_2 \cosh (kx)\right)\)

\(\large y'' ~~~~= k^2\left(y\right)\)

Answer 22

slap that right hand side stuff to left side

Answer 23

\(\large y'' - k^2\left(y\right) = 0\)

Answer 24

Thank you so much!!!!!!!!!!!! =)

Answer 25

let me knw if smthng doesnt make sense :)

Answer 26

u wlc ! :)