*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.*

*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.The first special case of first order differential equations that we will look at is the linear first order differential equation.

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.

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.

Do not, at this point, worry about what this function is or where it came from.

Also note that we’re using \(k\) here because we’ve already used \(c\) and in a little bit we’ll have both of them in the same equation.

So, to avoid confusion we used different letters to represent the fact that they will, in all probability, have different values.

\[ = p\left( t \right)\] As with the process above all we need to do is integrate both sides to get.

\[\begin\ln \mu \left( t \right) k &= \int\ \ln \mu \left( t \right) & = \int k\end\] You will notice that the constant of integration from the left side, \(k\), had been moved to the right side and had the minus sign absorbed into it again as we did earlier.

Instead of memorizing the formula you should memorize and understand the process that I'm going to use to derive the formula.

Most problems are actually easier to work by using the process instead of using the formula.

## Comments Linear Equations Problem Solving

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