Can I Simplify This Diff EQ by Using Laplace Transform and Partial Fractions?

[kdinser]

This is actually part of a larger problem of solving a diff EQ by use of a laplce transform, but it's the partial fraction part that I'm stuck on.

$$\frac{1}{s^2(s^2+w^2)} = \frac{A}{s}+\frac{B}{s^2}+\frac{C*s+D}{s^2+w^2}$$

Good so far?

If I then multiply through by the common denominator I get

EQ.2$$1 = A*s(s^2+w^2) + B(s^2+w^2) + s^2(C*s+D)$$

So then I set s=0

$$1=B*w^2 \longrightarrow B=1/w^2$$

Then using the quadratic equation, I find that the roots of s^2+w^2 are +/- jw

letting s=jw I end up with

$$1 = -jC*w^3 - Dw^2$$

this is where I get stuck, I'm not sure how to find C and D at this point.

Note: I just noticed a mistake in EQ.2,

 

[FunkyDwarf]

you can equate the various bits, ie the coefficients of the quadratic linear and constant terms. you cut out a bit of that with s=0 which is a good trick but u need the other way to get all the constants (if only in terms of w)

ie $${s^2}(C*s + D) = 0$$

i think

 

[kdinser]

Thanks for the reply, but I'm not really sure what you are saying.

 

[jpr0]

You can get two equations for $C$ and $D$ by letting $s=\pm i\omega$. You can then solve for $C$ and $D$.

By the way if you want to know a quick way to do this you can say the following:

[tex] \frac{1}{xy} = \frac{1}{x-y}\left\{\frac{1}{y}-\frac{1}{x}\right\}<br /> [/itex]<br /> <br /> So if I choose $x=s^2+\omega^2$, and $y=s^2$ then your initial fraction can be cast directly into the form on the RHS of your equation.[/tex]