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Damnit! I am terrible at Partial Fractions!

2,981
2
1. Homework Statement

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




3. The Attempt at a Solution

So I got:

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

[tex]\Rightarrow Y(s)=\frac{3}{(s^2+9)(s^2+4)}[/tex]

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

[tex]\frac{3}{(s^2+9)(s^2+4)}=\frac{Ax+B}{s^2+9}+\frac{Cx+D}{s^2+4}[/tex]

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

Thanks!
 

Answers and Replies

dynamicsolo
Homework Helper
1,648
4
Given this transform:

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

shouldn't this be

[tex]\Rightarrow Y(s)=\frac{3}{(s^2+9)(s^2+4*s*)}[/tex]


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

[tex]\frac{3}{(s^2+9)(s^2+4s)}=\frac{As+B}{s^2+9}+...[/tex][/QUOTE]
 
2,981
2
Oh crap...

Yes, so I get:

[tex]\frac{As+B}{s^2+9}+\frac{C}{s}+\frac{D}{s+4}[/tex]

Thanks!!!! !
 
2,981
2
Oh crap...

Yes, so I get:

[tex]\frac{As+B}{s^2+9}+\frac{C}{s}+\frac{D}{s+4}[/tex]

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?
 
2,981
2
Guess not.

So now I have

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

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!
 

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