Damnit I am terrible at Partial Fractions

[Saladsamurai]

Homework Statement

Solve y"+4y'=sin 3t subject to y(0)=y'(0)=0 using Laplace Transform

The Attempt at a Solution

So I got:

$$s^2Y(s)-sy(0)-y'(0)+4[sY(s)-y(0)]=\frac{3}{s^2+9}$$

$$\Rightarrow Y(s)=\frac{3}{(s^2+9)(s^2+4)}$$

Now it looks like two irreducible quadratics, which I know should not be too bad, but I have never dealt with more than one.

Now am I correct to say that

$$\frac{3}{(s^2+9)(s^2+4)}=\frac{Ax+B}{s^2+9}+\frac{Cx+D}{s^2+4}$$

This is where I think I have the problem... the notation.

Thanks!

 

[dynamicsolo]

Given this transform:

$$s^2Y(s)-sy(0)-y'(0)+4[sY(s)-y(0)]=\frac{3}{s^2+9}$$ ,

shouldn't this be

$$\Rightarrow Y(s)=\frac{3}{(s^2+9)(s^2+4*s*)}$$


Also, when you go to solve the partial fractions, you want to have 's' in the numerators:

$$\frac{3}{(s^2+9)(s^2+4s)}=\frac{As+B}{s^2+9}+...$$[/QUOTE]

 

[Saladsamurai]

Oh crap...

Yes, so I get:

$$\frac{As+B}{s^2+9}+\frac{C}{s}+\frac{D}{s+4}$$

Thanks! !

 

[Saladsamurai]

Oh crap...

Yes, so I get:

$$\frac{As+B}{s^2+9}+\frac{C}{s}+\frac{D}{s+4}$$

Thanks! !

Anyone know of a quick way to do this? I let s=0 and -4 to solve for C and D... but what about A and B? Do I have to distribute this whole mess out? Or is there a more expedient way?

 

[Saladsamurai]

Guess not.

So now I have

$$Y(s)=\frac{-4/75s-3/25}{s^2+9}+\frac{1}{12s}+\frac{3}{100(s+4)}$$

How do I simplify the 1st term?

I can see that it looks like cosine. But...How do I get rid of all the crap?

Hmm I guess I could...Oh! Break it up! I think that will work!