Trig derivative questions.

Splint

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

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

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

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

$$\displaystyle Solution$$

$$\displaystyle y=a\ sin\ bx$$

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

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

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

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

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

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

Random Variable

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

Splint

dwsmith

MHF Hall of Honor
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 $$\displaystyle y=C_1cos(bt)+C_2sin(bt)$$?

Random Variable

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 $$\displaystyle C_{2} = a$$ ).

dwsmith

dwsmith

MHF Hall of Honor
That's the general solution. What's given is a particular solution (when $$\displaystyle C_{1}=0$$ and $$\displaystyle C_{2} = a$$ ).
Ok