# Linear Equations Problem Solving

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.

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.

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It's sometimes easy to lose sight of the goal as we go through this process for the first time.

In order to solve a linear first order differential equation we MUST start with the differential equation in the form shown below.

As we will see, provided $$p(t)$$ is continuous we can find it. $\begin\mu \left( t \right)\frac \mu '\left( t \right)y = \mu \left( t \right)g\left( t \right) \label \end$ At this point we need to recognize that the left side of $$\eqref$$ is nothing more than the following product rule.

$\mu \left( t \right)\frac \mu '\left( t \right)y =$ So we can replace the left side of $$\eqref$$ with this product rule.

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.

So, now that we’ve got a general solution to $$\eqref$$ we need to go back and determine just what this magical function $$\mu \left( t \right)$$ is.

This is actually an easier process than you might think. $\mu \left( t \right)p\left( t \right) = \mu '\left( t \right)$ Divide both sides by $$\mu \left( t \right)$$, $\frac = p\left( t \right)$ Now, hopefully you will recognize the left side of this from your Calculus I class as nothing more than the following derivative.

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.