Question

is there a concise solution?

Answer 1

hm

Answer 2

to this:

\(y'' - \omega^2 ~ y = C \sinh \omega x\)

\(y_c\ = y(\pm \omega)\), which is a real PITA.

the fuller eqn is : \(y'' - \omega^2 ~ y = B \sin \omega x + C \sinh \omega x\)

that first part comes out real easy using a complex approach, ....

Answer 3

\(\sinh x = \dfrac{e^x-e^{-x}}{2}\)

superposition works smoothly right ?

Answer 4

It is pretty easy if we use this result :

A particular solution of \(P(D) = e^{ax}\) is \(\dfrac{e^{ax}}{P(a)}\) when \(P(a) \ne 0\)

A particular solution of \(P(D) = e^{ax}\) is \(\dfrac{e^{ax}}{P'(a)}\) when \(P(a) = 0\) and \(P(a)\ne 0\)

Answer 5

where \(P(D) = Ay''+By'+Cy\)

Answer 6

i tried \(y_p = A \cosh(\omega x) + B \sinh(\omega x)\). that crashes badly.

the computer takes it your way, ganesh.

the characteristic equation has \(\pm \omega\) in the \(y_c\) solution.

i've got dreadful bandwith, so apologies for crossing.

i'm listening now....😀

Answer 7

superposition principle says that a particular solution to the eqn :
\(P(D) = f(t)+g(t)\)

can be obtained by adding particular solution of each of below eqns
\(P(D) = f(t)\)
\(P(D) = g(t)\)

Answer 8

yep

Answer 9

fixed a mistake here :

It is pretty easy if we use this result :

A particular solution of \(P(D) = e^{ax}\) is \(\dfrac{e^{ax}}{P(a)}\) when \(P(a) \ne 0\)

A particular solution of \(P(D) = e^{ax}\) is \(\dfrac{\color{red}{x}e^{ax}}{P'(a)}\) when \(P(a) = 0\) and \(P(a)\ne 0\)

Answer 10

\(y'' - \omega^2 ~ y = C \sinh \omega x =\frac{C}{2}e^{\omega x} -\frac{C}{2}e^{-\omega x}\tag{1}\)



Here \(P(D) = D^2 - \omega^2\)
Notice that \(P(\pm \omega) = 0\).

A particular solution to \((1)\) is :
\[y_p = \dfrac{xC/2e^{\omega x}}{P'(\omega)} + \dfrac{-xC/2e^{-x}}{P'(-\omega)} =\dfrac{xC/2e^x}{2\omega} + \dfrac{-xC/2e^{-x}}{-2\omega} \]

Answer 11

yeah that looks correct finally

Answer 12

does that simplify to \(y_p = \dfrac{C}{\omega}x\cosh \omega x\)

Answer 13

yes