Solving Initial Value Problems Examples

Solving Initial Value Problems Examples-82
\[\left( \right)Y\left( s \right) s - 12 = \frac\] Solve for \(Y(s)\).

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Let’s take a look at another fairly simple problem.

As with the first example, let’s first take the Laplace transform of all the terms in the differential equation.

These are going to be invaluable skills for the next couple of sections so don’t forget what we learned there.

Before proceeding into differential equations we will need one more formula.

However, if we combine the two terms up we will only be doing partial fractions once.

Not only that, but the denominator for the combined term will be identical to the denominator of the first term.\[\mathcal\left\ - 10\mathcal\left\ 9\mathcal\left\ = \mathcal\left\] Using the appropriate formulas from our table of Laplace transforms gives us the following.\[Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - 10\left( \right) 9Y\left( s \right) = \frac\] Plug in the initial conditions and collect all the terms that have a \(Y(s)\) in them.While Laplace transforms are particularly useful for nonhomogeneous differential equations which have Heaviside functions in the forcing function we’ll start off with a couple of fairly simple problems to illustrate how the process works.The first step in using Laplace transforms to solve an IVP is to take the transform of every term in the differential equation.So, in order to find the solution all that we need to do is to take the inverse transform.Before doing that let’s notice that in its present form we will have to do partial fractions twice.It’s now time to get back to differential equations.We’ve spent the last three sections learning how to take Laplace transforms and how to take inverse Laplace transforms.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.

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