Ordinary differential equations question

[Obnoxious]

In the equation: [math]\frac{d^2x}{dt^2} + K_1\frac{dx}{dt} + K_0x = 0[/math]

 

Can someone please explain to me how I'm suppose to be able to use the trig answer [math]x=A\sin(rt + b)[/math] or [math]x=A\cos(rt + b)[/math] to get to the characteristic equation?

 

Everytime I try to mush the equations around I get stuck at this junction:

[math]-r^2x + K_1Ar\cos(rt + b) + K_0x = 0[/math]

or

[math]-r^2x - K_1Ar\sin(rt + b) + K_0x = 0[/math]

I seriously need some help on this as I'm learning this stuff on my own, and therefore have no professor to turn to with questions.

[Dave]

The idea is that given [math]x=A\sin(rt + b)[/math], you can work out [math]x'(t)[/math] and [math]x''(t)[/math], then substitute these back into the equation to get some trigonometric problem to solve for t. You shouldn't have any x's in it at all.

[Obnoxious]

Okay, so basically, I have to actually work out the expression and solve for r?

[math]-Ar^2\sin(rt+b) + K_1Ar\cos(rt+b) + K_2A\sin(rt + b) = 0[/math]

as in expand the thing with trig identities and mush out an answer? There's no easy way?

[Tom Mattson]

There's no easy way?

 

Well there's certainly an easier way. The basis [imath]\{\sin(x)' date='\cos(x)\}[/imath'] spans the exact same vector space as the basis [imath]\{exp(ix),exp(-ix)\}[/imath]. Convert your trig functions to complex exponentials and have a ball.