Solving Initial Value Problems Examples

The point of this problem however, was to show how we would use Laplace transforms to solve an IVP.

There are a couple of things to note here about using Laplace transforms to solve an IVP.

Combining the two terms gives, \[Y\left( s \right) = \frac\] The partial fraction decomposition for this transform is, \[Y\left( s \right) = \frac \frac \frac \frac\] Setting numerators equal gives, \[5 12 - = As\left( \right)\left( \right) B\left( \right)\left( \right) C\left( \right) D\left( \right)\] Picking appropriate values of \(s\) and solving for the constants gives, \[\begin & s = 0 & 5 & = 9B & \Rightarrow \hspace B & = \frac\\ & s = 1 & 16 & = - 8D & \Rightarrow \hspace D & = - 2\\ & s = 9 & 248 & = 648C & \Rightarrow \hspace C & = \frac\\ & s = 2 & 45 & = - 14A \frac & \Rightarrow \hspace A & = \frac\end\] Plugging in the constants gives, \[Y\left( s \right) = \frac \frac \frac - \frac\] Finally taking the inverse transform gives us the solution to the IVP.

\[y\left( t \right) = \frac \fract \frac - 2\] That was a fair amount of work for a problem that probably could have been solved much quicker using the techniques from the previous chapter.

In many of the later problems Laplace transforms will make the problems significantly easier to work than if we had done the straight forward approach of the last chapter.

Also, as we will see, there are some differential equations that simply can’t be done using the techniques from the last chapter and so, in those cases, Laplace transforms will be our only solution.

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First, using Laplace transforms reduces a differential equation down to an algebra problem.

In the case of the last example the algebra was probably more complicated than the straight forward approach from the last chapter. The algebra, while still very messy, will often be easier than a straight forward approach.

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