Trig derivative questions.

Sep 2009
37
1
\(\displaystyle I\ have\ an\ equation\ and\ solution.\ I'm\ unclear\ on\ a\ couple\ of\ things.\)

\(\displaystyle An\ analysis\ of\ the\ motion\ of\ a\ particle\ resulted\ in\ the\ differential\ equation\)

\(\displaystyle \frac{d^2y}{dt^2}\ +\ b^2y\ =\ 0\)

\(\displaystyle Show\ that\ y=a\ sin\ bx\ is\ a\ solution\ of\ this\ equation.\)

\(\displaystyle Solution\)

\(\displaystyle y=a\ sin\ bx\)

\(\displaystyle \frac{dy}{dt}\ =\ ab\ cos\ bt\ Why\ has\ x\ become\ t\ and\ why\ is\ there\ another\ b\ in\ the\ equation?\)
\(\displaystyle What\ is\ the\ rule\ for\ finding\ the\ derivative?\)

\(\displaystyle \frac{d^2y}{dt^2}\ =\ -ab^2\ sin\ bt\)

\(\displaystyle \frac{d^2y}{dt^2}\ =\ -b^2(a\ sin\ bt)\)

\(\displaystyle \frac{d^2y}{dt^2}\ =\ -b^2y \)

\(\displaystyle \frac{d^2y}{dt^2}+b^2y=-b^2y+b^2y=0\)

\(\displaystyle y=a\ sin\ bt\ is\ the\ solution\ of \ the\ given\ differential\ equation.\)
 
May 2009
959
362
The particular solution should be \(\displaystyle y = a \sin bt \).

Then using the chain rule, \(\displaystyle \frac{dy}{dt} = a \cos bt \ \frac{d}{dt} \ (bt) = ab \cos bt \)

and \(\displaystyle \frac{d^{2}y}{dt^{2}} = -ab \sin bt \ \frac{d}{dt} (bt) = - ab^{2} \sin bt \)
 
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dwsmith

MHF Hall of Honor
Mar 2010
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The particular solution should be \(\displaystyle y = a \sin bt \).

Then using the chain rule, \(\displaystyle \frac{dy}{dt} = a \cos bt \ \frac{d}{dt} \ (bt) = ab \cos bt \)

and \(\displaystyle \frac{d^{2}y}{dt^{2}} = -ab \sin bt \ \frac{d}{dt} (bt) = - ab^{2} \sin bt \)
Why isn't the solution \(\displaystyle y=C_1cos(bt)+C_2sin(bt)\)?
 
May 2009
959
362
Why isn't the solution \(\displaystyle y=C_1cos(bt)+C_2sin(bt)\)?
That's the general solution. What's given is a particular solution (when \(\displaystyle C_{1}=0 \) and \(\displaystyle C_{2} = a \) ).
 
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