# Trig derivative questions.

• May 18th 2010, 08:53 PM
Splint
Trig derivative questions.
$I\ have\ an\ equation\ and\ solution.\ I'm\ unclear\ on\ a\ couple\ of\ things.$

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

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

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

$Solution$

$y=a\ sin\ bx$

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

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

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

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

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

$y=a\ sin\ bt\ is\ the\ solution\ of \ the\ given\ differential\ equation.$
• May 18th 2010, 09:06 PM
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 $\frac{d^{2}y}{dt^{2}} = -ab \sin bt \ \frac{d}{dt} (bt) = - ab^{2} \sin bt$
• May 18th 2010, 09:11 PM
dwsmith
Quote:

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 $y=C_1cos(bt)+C_2sin(bt)$?
• May 18th 2010, 09:18 PM
Random Variable
Quote:

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 $C_{2} = a$ ).
• May 18th 2010, 09:29 PM
dwsmith
Quote:

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