Hello,

I have the DE:

with IC's:

hs

If I convert this to a system of 1st order ODES is it just:

How about if the DE is:

and the same IC's.

Thanks

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- May 30th 2010, 04:32 PMRoyalFlushConvert 2nd order ode to first order ODES
Hello,

I have the DE:

with IC's:

hs

If I convert this to a system of 1st order ODES is it just:

How about if the DE is:

and the same IC's.

Thanks - May 30th 2010, 07:55 PMAllanCuz
The first question is a second order homogenious equation. We find the roots by computing the characteristic equation.

Clearly this has the imaginary roots and

So,

Compute and sub in initial conditions.

The second question can be computed by method of comparing co-efficients (we can also use variation of paramters but co-efficients is easiast here).

For the general solution is the solution from question 1 plus the particular solution.

For the particular solution,

Let

So,

Subbing into the equation we get,

And - May 30th 2010, 08:15 PMProve It
- May 31st 2010, 05:36 PMAllanCuz
- June 1st 2010, 03:02 AMHallsofIvy
That's very clever, guys, but it doesn't actually answer the question asked, does it?

Royal Flush, yes, if you let and , then we have immediately that and then so that so that

And your initial conditions are and .

This can also be written as the "matrix" equation:

subject to the initial condition .

In the same way, the equation y"+ y= sin(2t) becomes the pair of equations and with the same intial conditions.

That also can be written as a "matrix" equation:

with initial condition .

Of course, the eigenvalues of the matrix are i and -i which lead to the complex exponentials (sine and cosine) solutions Prove It and AllenCuz show. - June 1st 2010, 05:43 AMAllanCuz