No solution for an arbitrary constant.

**MaverickUK82** · March 25th 2011, 04:01 PM

Say I have found a general solution to an equation of motion, which is, simple harmonic motion.

The general solution I have found is of the form:

$x(t) = L+B\cdot cos(\omega t)+C\cdot sin(\omega t)$

When inserting some (supplied) inital conditions, I am able to solve to find $B$ but cannot find a solution for $C$ , does it follow that $C=0$ ?

My initial conditions contain only the position when $t=0$ , $x$ and no derivatives of it - this being the case

$C\cdot sin(\omega t)=C\cdot sin(\omega \cdot 0)=0$

I hope it does, because I'm stumped if not - If the question above is too vague, I can supply the full question. However, I want to solve it myself and this is why I have used generic formulae above.

**Ackbeet** · March 25th 2011, 04:11 PM

What exactly are the conditions? Are they initial conditions?

**MaverickUK82** · March 25th 2011, 10:54 PM

Yes. They are initial conditions.

The formula is for a spring, I am given the height it is released from at $t=0$ only.

**MaverickUK82** · March 26th 2011, 04:21 AM

I have worked this one out now. Thanks for the help.

**Ackbeet** · April 1st 2011, 04:41 AM

You're welcome for whatever help I could provide.

Thread: No solution for an arbitrary constant.

LinkBack

Thread Tools

Display

No solution for an arbitrary constant.

Similar Math Help Forum Discussions

Why ode solution tends to a constant?

Eliminating the arbitrary constant

Question involving arbitrary constant

Proving Indeterminate Forms can equal an arbitrary constant

Eliminating the arbitrary constant

Search Tags