# 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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