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I am having some confusion when it comes to solving for the inverse laplace transform. ( We are allowed the tables with the common values by the way).

Il give an example.

Take,

$$y''+4y'+8y=-6e^{-2t}cos(3t)$$ with $y(0)=-2$ and $y'(0)=-5$

Now, by invoking the linearity of the Laplace transform and using the tables, and letting $Y(s)=L[y]$

I get to the form,

$$(s^2+4s+8)Y(s)+2s+13=-6\frac{s+2}{(s+2)^2+3^2}$$

Now here is where I am having more trouble because I don't know the least messy way to get the RHS in such a form that I can use the tables to find the inverse transforms, using linearity of course.

I know I could get it in the form $$Y(s)=\frac{-6\frac{s+2}{(s+2)^2+3^2}-2s-13}{s^2+4s+8}$$ but that doesn't seem to be of any help.

I know also that we can usually use partial fractions, but how could I apply it in such a case when we have a rational in the numerator?

Does anyone have any hints and tricks to this type of thing?

Thanks!

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1 Answer 1

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Your work is correct. Gather everything under common denominator:

$$\frac{-2 s^3-21 s^2-84 s-181}{\left(s^2+4 s+8\right) \left(s^2+4 s+13\right)}=\frac{A s+B}{s^2+4 s+8}+\frac{C s+D}{s^2+4 s+13}$$

This will be super messy, I will only calculate $A$ and $B$. Multiply everything by ${s^2+4 s+8}$

$$\frac{-2 s^3-21 s^2-84 s-181}{s^2+4 s+13}=(A s+B)+\frac{(s^2+4 s+8)(C s+D)}{s^2+4 s+13}$$

As usual, we want to set $s^2+4 s+8=0$. This makes the second term in $ RHS =0$. Inserting $s^2=-4s-8$, LHS becomes:

$$\frac{1}{5} \left(8 s^2+16 s-13\right)$$

Insert $s^2=-4s-8$ once more:

$$\frac{1}{5} (-16 s-77)=A\,s+B$$

You do the same for $s^2+4 s+13=0$ and you'll get:

$$\frac{6 s+12}{5 \left(s^2+4 s+13\right)}+\frac{-16 s-77}{5 \left(s^2+4 s+8\right)}=\frac{6 s+12}{5 \left((s+2)^2+3^2\right)}+\frac{-16 s-77}{5 \left((s+2)^2+2^2\right)}$$

we get:

$$\frac 6 5 \frac{s+2}{(s+2)^2+3^3}\rightarrow\frac 6 5 e^{-2t}\cos (3t)$$

and:

$$\frac{-16 s-77}{5 \left((s+2)^2+2^2\right)}=-\frac{16}{5}\frac{s+2}{(s+2)^2+4}-\frac{45}{10}\frac{2}{(s+2)^2+4}\\ \rightarrow-\frac{16}{5}e^{-2t}\cos (2t)-\frac{45}{10}e^{-2t}\sin (2t)$$

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  • $\begingroup$ Thanks But I am still having trouble with the final form, i can't seem to figure out how to do the inverse of this even with charts $\endgroup$
    – Quality
    Commented May 27, 2015 at 19:56
  • $\begingroup$ @Quality Updated $\endgroup$ Commented May 27, 2015 at 20:06

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