Obnoxious Posted March 24, 2006 Share Posted March 24, 2006 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. Link to comment Share on other sites More sharing options...

Dave Posted March 24, 2006 Share Posted March 24, 2006 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. Link to comment Share on other sites More sharing options...

Obnoxious Posted March 25, 2006 Author Share Posted March 25, 2006 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? Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 25, 2006 Share Posted March 25, 2006 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. Link to comment Share on other sites More sharing options...

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