Question

Proof.

Answer 1

\[\frac{ d^{n} }{ dx^{n} }(e^{ax}\sin(bx)) = r^{n}*e^{ax}Sin(bx+n \theta)\]

Answer 2

I'll show you what I have so far

Answer 3

my best guess is induction

Answer 4

I thought we just apply product rule like this:

\[h(x) = f(x)*g(x) \]

\[h'(x) = g(x)f'(x) + f(x)g'(x)\]

\[\frac{ d^{n} }{ dx^{n} }(e^{ax}\sin(bx)) = r^{n}*e^{ax}Sin(bx+n \theta)\]


\[\frac{ d }{ dx }(e^{ax}\sin(bx)) = \sin(bx)'e^{ax}+e'^{ax}\sin(bx)\]


\[\cos(bx)*b*e^{ax}+\sin(bx)*e^{ax}*a\]

but how do i proceed?

\[e^{ax}*(\cos(bx)*b+\sin(bx)*a)\]

Answer 5

Ugh... they also said

\[r^{2} = a^{2}+b^{2} \]

\[\theta = \tan^{-1}(b/a)\]

how do I substitute this into x?

Answer 6

oh now i have to think
it is one of those trig identities

Answer 7

i always forget this one, the one where like \[\cos(2x)+\sin(2x)=\sqrt2\sin(2x+\frac{\pi}{4})\]

Answer 8

you know this one?

Answer 9

I'm not familiar with this identity

Answer 10

it comes from manipulating the double addition angle formula, i used to have nice notes on it, let me see if i can find it

Answer 11

Thank you

Answer 12

i can't seem to find them, but they are in any pre calc book
the fact is this \[a\sin(x)+b\cos(x)=r\sin(x+\alpha)\] where \(r=\sqrt{a^2+b^2}\) and \(\tan(\alpha)=\frac{b}{a}\)

Answer 13

Thank you I'm going to look this over

Answer 14

i doubt you have to prove that for this proof

Answer 15

here is a video of the proof, let me keep looking
https://www.youtube.com/watch?v=G5LEOplnBxg

Answer 16

wait that looks a-lot like what I had for the derivative.

\[a\sin(x)+b\cos(x)=r\sin(x+\alpha)\]

\[e^{ax}*(\cos(bx)*b+\sin(bx)*a)\]

\[a\sin(x)+b\cos(x)=r\sin(x+\alpha)\]

\[e^{ax}*r*\sin(x+\alpha)\]

Answer 17

Going to look at the video you posted.

Answer 18

actually if you google asin(x)+bcos(x) you will get lots
can''t find my nice one though, hope i didn't get rid of it

Answer 19

Cool thanks !