Linear Equations Problem Solving

Now, we are going to assume that there is some magical function somewhere out there in the world, \(\mu \left( t \right)\), called an integrating factor.

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\[\begin\mu \left( t \right)y\left( t \right) & = \int - c\\ y\left( t \right) & = \frac\end\] Now, from a notational standpoint we know that the constant of integration, \(c\), is an unknown constant and so to make our life easier we will absorb the minus sign in front of it into the constant and use a plus instead.

This will NOT affect the final answer for the solution. \[\beginy\left( t \right) = \frac \label \end\] Again, changing the sign on the constant will not affect our answer.

We are going to assume that whatever \(\mu \left( t \right)\) is, it will satisfy the following.

\[\begin\mu \left( t \right)p\left( t \right) = \mu '\left( t \right) \label \end\] Again do not worry about how we can find a \(\mu \left( t \right)\) that will satisfy \(\eqref\).

Recall that a quick and dirty definition of a continuous function is that a function will be continuous provided you can draw the graph from left to right without ever picking up your pencil/pen.

In other words, a function is continuous if there are no holes or breaks in it.

We will figure out what \(\mu \left( t \right)\) is once we have the formula for the general solution in hand.

So, now that we have assumed the existence of \(\mu \left( t \right)\) multiply everything in \(\eqref\) by \(\mu \left( t \right)\). \[\begin\mu \left( t \right)\frac \mu \left( t \right)p\left( t \right)y = \mu \left( t \right)g\left( t \right) \label \end\] Now, this is where the magic of \(\mu \left( t \right)\) comes into play.

If the differential equation is not in this form then the process we’re going to use will not work.

\[\begin\frac p\left( t \right)y = g\left( t \right) \label \end\] Where both \(p(t)\) and \(g(t)\) are continuous functions.


Comments Linear Equations Problem Solving

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