At this point we know:[br][br][list][*]what a differential equation is[/*][*]how to visualize a single first order differential equation with slope fields -- and how to show slope fields in GeoGebra with [code]slopefield( )[/code][/*][*]We know about one class of differential equations and how to solve them: first order separable differential equations. [/*][/list][br]For example, this first order differential equation[br][br][math]\frac{dP}{dt}=5P[/math][br][br]is separable. Note: [i]P[/i] is the dependent variable (you might be accustomed to [i]y[/i] or [i]f[/i] instead), and [i]t[/i] is the independent variable (you might be accustomed to [i]x[/i] instead). This is in fact separable. Divide both sides by [i]P[/i]:[br][br][math]\frac{dP}{dt}\cdot\frac{1}{P}=5P\cdot\frac{1}{P}[/math][br][br][math]\frac{1}{P}\frac{dP}{dt}=5[/math][br][br]In this form, [math]N(P)=1/P[/math] and [math]M(t)=5[/math]. To solve we'll follow the process for solving separable differential equations, starting with multiplying both sides by [i]dt[/i].[br][br][math]\frac{1}{P}\frac{dP}{dt}\cdot dt=5\cdot dt[/math][br][br][math]\frac{1}{P}\cdot dP=5\cdot dt[/math][br][br]Now integrate both sides. Note that we can use a single constant of integration to the right side since adding one to each side is just as arbitrary as adding one to just the right side. [br][br][math]\int\frac{1}{P}\cdot dP=\int5\cdot dt[/math][br][br][math]\ln\left(\left|P\right|\right)=5t+c[/math][br][br]The final step is to solve for [i]P[/i] (the dependent variable)[br][br][math]e^{\ln\left(\left|P\right|\right)}=\left|P\right|=e^{5t+c}[/math][br][br]In other words all of these infinitely many functions [br][br][math]P\left(t\right)=\pm e^{5t+c}[/math][br][br]are a solution to the differential equation. The slope field of the differential equation, and the infinite family of solutions are plotted in GeoGebra applet below.

An [b]initial condition [/b]of a differential equation is a way of selecting a specific solution function from the infinite family of solutions of a differential equation. For example we can add an initial condition to the differential equation like so[br][br][math]\frac{dP}{dt}=5P;P\left(0\right)=3[/math][br][br]To be clear: the initial condition is [math]P(0)=3[/math], and it means that the investigator (you) need to find a particular solution function amongst the infinite family represented by the general solution that returns a [i]P[/i] value of 3 when t is equal to 0. [br][br]Visually, this means you are on the hunt for a solution function that passes through the point (0,3). Try to find it by sliding [i]C[/i] in the applet below. You may have to zoom in on (0,3) to see how close you can get it! It's ok if you can't get [i]C[/i] exactly correct.

If a differential equation is stated with an initial condition, and the family of solutions has already been found, then it is an algebra 1 or 2 level task to find the particular solution. [br][br]For instance, let's revisit [br][br][math]\frac{dP}{dt}=5P;P\left(0\right)=3[/math][br][br]We already know that [br][br][br][math]P\left(t\right)=\pm e^{5t+c}[/math][br][br]is an infinite family of solutions of the differential equation. In the applet above, it appeared that when [i]C[/i] was about 1.1, that resulted in [math]P_1(t)[/math] passing through (0,3). Let's use algebra to find the exact value of [i]C[/i].[br][br]The initial condition [math]P(0)=3[/math] means that when [i]t[/i] is 0, that [i]P[/i] should be equal to 3. Algebraically this means[br][br][math]P\left(0\right)=\pm e^{\left(5\cdot0+c\right)}=3[/math][br][br]Just like we did geometrically above, we must find a value of [i]c[/i] that makes this equation true. In other words, we must solve[br][br][math]\pm e^{\left(5\cdot0+c\right)}=3[/math][br][br]for c. This is an algebraic task. It has nothing to do with calculus. Nonetheless, it isn't easy! First, notice that we MUST be in the positive of the +/- since 3 is positive, e only ever returns positive numbers. [br][br][math]+e^{\left(5\cdot0+c\right)}=3[/math][br][br]This means that in the graph above, we know that we are looking in the realm of [math]P_1(t)[/math]. Now, use the natural logarithm to solve an equation with the unknown in an exponent. [br][br][math]\ln\left(e^{\left(5\cdot0+c\right)}\right)=\ln\left(3\right)[/math][br][br]The natural log of [i]e[/i]^(something) is just "something". So this simplifies to[br][br][math]5\cdot0+c=\ln\left(3\right)[/math][br][br]And of course, 5*0 is 0, so really we're done![br][br][math]c=\ln\left(3\right)\approx1.0986[/math] [br][br][br]The initial condition has specified the following specific solution:[br][br][math]P\left(t\right)=e^{5t+\ln\left(3\right)}[/math][br][br]Check it out in applet below. Check that when [i]t[/i] is 0, that [i]P[/i] is in fact 3 by typing [i]P[/i](0) in the input bar. The result should be 3.

The final topic for today is realize that differential equations can be thought of as dynamic models of changing quantities. For instance, in a biological setting, we might know that the growth of the population of some bacteria per hour in an experiment is growing at a rate of 5 times the current population of bacteria. [br][br]Let's re-write this sentence a few times[br][br]"The growth of the population of some bacteria per hour in an experiment is growing at a rate of 5 times the current population of bacteria."[br][br]"The rate of change of a population is equal to 5 times the population."[br][br]"The derivative of [i]P[/i] is equal to 5 times [i]P[/i]."[br][br][math]\frac{dP}{dt}=5P[/math][br][br]Does this look familiar? I sure hope so! This is the differential equation we've been studying in this activity! There's nothing to do here at the moment, but notice that statements of how the growth of a quantity depend on the quantity (or an independent variable that also controls the quantity) are inherently differential equation models. [br][br]We'll be returning to this in it's own right in future classes.