Laplace transforms and partial fractions

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SUMMARY

The discussion focuses on finding the inverse Laplace transform of the expression (-2e^-s)/(s(s+4)(s+1)). The user initially expresses confusion about whether to use partial fractions, which is confirmed as the correct approach. The rational part is factored into A/s + B/(s+4) + C/(s+1), with values A=1/4, B=1/12, and C=-1/3 derived through substitution. The use of Laplace transform tables is recommended for solving the individual components after determining A, B, and C.

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  • Familiarity with partial fraction decomposition techniques.
  • Knowledge of solving ordinary differential equations (ODEs).
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morry
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Hey guys, I am supposed to find the Laplace transform of a set of ODEs.

Ive broken it down a bit and I am left with finding the Laplace transform of:

(-2e^-s)/(s(s+4)(s+1))

Is this something I have to use partial fractions for? Or is there another way? I am a bit confused.
 
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Do you mean the inverse Laplace transform of (-2e^-s)/(s(s+4)(s+1))?
 
Ahh, yeah that would be correct. oops
 
If so, factor out the exponential and just deal with the rational part:

-\frac{2e^{-s}}{s(s+4)(s+1)}=-2e^{-s}\frac{1}{s(s+4)(s+1)}

So \frac{1}{s(s+4)(s+1)}=\frac{A}{s}+\frac{B}{s+4}+\frac{C}{s+1}
 
Ok, so after finding values for A,B,C, do I just use the tables in order to work out the individual bits?
ie ((-2e^-s)*A)/s etc?
 
Cross-multiply to get:

1=A(s+4)(s+1)+Bs(s+1)+Cs(s+4)

then plug, say, s=0,-4,-1 into the above to solve for A,B, and C
 
yep, use tables if you got 'em.
 
benorin said:
Cross-multiply to get:

1=A(s+4)(s+1)+Bs(s+1)+Cs(s+4)

then plug, say, s=0,-4,-1 into the above to solve for A,B, and C


s=0 gives 1=4A or A=\frac{1}{4}

s=-4 gives 1=12B or B=\frac{1}{12}

and s=-1 gives 1=-3C or C=-\frac{1}{3}
 
Cheers Benorin, partial fractions are no worries, just that exponential that was confusing me, seems pretty simple now.
 

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