# Ordinary differential equations question

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In the equation: $\frac{d^2x}{dt^2} + K_1\frac{dx}{dt} + K_0x = 0$

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

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

$-r^2x + K_1Ar\cos(rt + b) + K_0x = 0$

or

$-r^2x - K_1Ar\sin(rt + b) + K_0x = 0$

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.

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The idea is that given $x=A\sin(rt + b)$, you can work out $x'(t)$ and $x''(t)$, 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.

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

$-Ar^2\sin(rt+b) + K_1Ar\cos(rt+b) + K_2A\sin(rt + b) = 0$

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

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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.

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