Trig derivative questions.

**Splint** · May 18th 2010, 08:53 PM

$I\ have\ an\ equation\ and\ solution.\ I'm\ unclear\ on\ a\ couple\ of\ things.$

$An\ analysis\ of\ the\ motion\ of\ a\ particle\ resulted\ in\ the\ differential\ equation$

$\frac{d^2y}{dt^2}\ +\ b^2y\ =\ 0$

$Show\ that\ y=a\ sin\ bx\ is\ a\ solution\ of\ this\ equation.$

$Solution$

$y=a\ sin\ bx$

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

$\frac{d^2y}{dt^2}\ =\ -ab^2\ sin\ bt$

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

$\frac{d^2y}{dt^2}\ =\ -b^2y$

$\frac{d^2y}{dt^2}+b^2y=-b^2y+b^2y=0$

$y=a\ sin\ bt\ is\ the\ solution\ of \ the\ given\ differential\ equation.$

**Random Variable** · May 18th 2010, 09:06 PM

The particular solution should be $y = a \sin bt$ .

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

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

**dwsmith** · May 18th 2010, 09:11 PM

Originally Posted by Random Variable

The particular solution should be $y = a \sin bt$ .

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

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

Why isn't the solution $y=C_1cos(bt)+C_2sin(bt)$ ?

**Random Variable** · May 18th 2010, 09:18 PM

Originally Posted by dwsmith

Why isn't the solution $y=C_1cos(bt)+C_2sin(bt)$ ?

That's the general solution. What's given is a particular solution (when $C_{1}=0$ and $C_{2} = a$ ).

**dwsmith** · May 18th 2010, 09:29 PM

Originally Posted by Random Variable

That's the general solution. What's given is a particular solution (when $C_{1}=0$ and $C_{2} = a$ ).

Ok

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