*The transform becomes, \[\begin F\left( s \right) & = 6\frac \frac\ & = 6\frac \frac\frac\end\] Taking the inverse transform gives, \[f\left( t \right) = 6\cos \left( \right) \frac\sin \left( \right)\] In this case the first term will be a sine once we factor a 3 out of the denominator, while the second term appears to be a hyperbolic sine (#17).Again, be careful with the difference between these two.*

*The transform becomes, \[\begin F\left( s \right) & = 6\frac \frac\ & = 6\frac \frac\frac\end\] Taking the inverse transform gives, \[f\left( t \right) = 6\cos \left( \right) \frac\sin \left( \right)\] In this case the first term will be a sine once we factor a 3 out of the denominator, while the second term appears to be a hyperbolic sine (#17).Again, be careful with the difference between these two.*

However, the numerator doesn’t match up to either of these in the table. We will just split up the transform into two terms and then do inverse transforms.

A cosine wants just an \(s\) in the numerator with at most a multiplicative constant, while a sine wants only a constant and no \(s\) in the numerator. \[\begin F\left( s \right) & = \frac - \frac\\ f\left( t \right) & = 6\cos \left( \right) - \frac\sin \left( \right)\end\] Do not get too used to always getting the perfect squares in sines and cosines that we saw in the first set of examples.

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It is an exponential, but in this case, we’ll need to factor a 3 out of the denominator before taking the inverse transform. Whenever a numerator is off by a multiplicative constant, as in this case, all we need to do is put the constant that we need in the numerator.

The denominator of the third term appears to be #3 in the table with \(n = 4\). There is currently a 7 in the numerator and we need a \(4! We will just need to remember to take it back out by dividing by the same constant. \[\begin H\left( s \right) & = \frac - \frac \frac\ & = 19\frac - \frac\frac \frac\frac\end\] So, what did we do here? We factored the 3 out of the denominator of the second term since it can’t be there for the inverse transform and in the third term we factored everything out of the numerator except the 4!

As you will see this can be a more complicated and lengthy process than taking transforms.

In these cases we say that we are finding the Inverse Laplace Transform of \(F(s)\) and use the following notation.

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## Comments Laplace Transform Solved Problems Pdf

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