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Mathematics 10 Online
OpenStudy (anonymous):

Find the value(s) of w for which y = cos(wt) is a solution to the differential equation below ((d^2y)/(dt^2))+9y=0

OpenStudy (phi):

Did you plug in cos(wt) for y into your differential equation ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

i get y"=-9(cos(wt)

OpenStudy (anonymous):

$$y=\cos wt\\y'=-w\sin wt\\y''=-w^2\cos wt\\\text{so }\frac{d^2y}{dt^2}+9y=0\implies-w^2\cos wt+9\cos wt=0\\(-w^2+9)(\cos wt)=0\\\text{so }9=w^2\implies w=\pm3$$

OpenStudy (phi):

oldrin showed you the details. you also have to take the 2nd derivative of y you get (9-w^2)=0 or cos wt =0 the first gives you w= ±3

OpenStudy (anonymous):

thanks guys

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