We've come to the end of this introductory course on Differential Equations. There's so much more out there to learn, but this is a good survey of the basics, and so it's a good place to stop. We haven't dipped our toe into partial differential equations unfortunately, but the principles we've discussed at length here apply to that as well. It's just new methods and new data structures to keep track of.[br][br]With so much content both from this course, and what you might study outside this course, it can be tricky to organize it all, so I thought I'd share a mantra of mine for keeping all the material and theory of differential equations straight:[br][br][quote]Represent visually; go with the flow; look up algebraic/numerical methods.[/quote]What does this mean though? Let's take it piece by piece.

This first part of the mantra is a reminder that every differential equation has a geometric representation. [br][br]Let's take for instance[br][br][math]xy'=y+2x^3;y\left(1\right)=2[/math][br][br]This differential equation and initial condition has a visual representation. Put it in slope field form, and then use GeoGebra to visualize.

The second part of the mantra means that even if you can't find the algebraic solution, you can use the slopefield to get a sense of what the specific solution is likely to be. In this case, we can tell that the specific solution appears to be something like a cubic function just by "going with the flow" of the slope field. [br][br]It's not a bad idea to sketch an estimated solution right away. Also, don't forget, numerical methods all rely on this principle so doing a sketch helps set your mind up for being in the right place to implement numerical methods.[br][br]Here's a sketch of the solution.

The last part of the mantra is a reminder to yourself to permit yourself to look up algebraic or numerical methods. For that, refer to this book, or others, for examples of how to use an algebraic method to solve a differential equation. [br][br]In this particular example, this particular differential equation screams "linearity". The key to recognizing that a differential equation is linear is to see echos of the standard linear form in its current expression. The standard linear form is[br][br][math]y'+p(x)y=g(x)[/math][br][br]You can go back and review the method here: https://www.geogebra.org/m/cxgtwkqa#material/dpguwczw[br][br]For this particular differential equation, we can put it in standard linear form as[br][br][math]y'-\frac{1}{x}y=2x^2;y(1)=2[/math][br][br]Written this way, we see [math]p(x)=-1/x[/math] and [math]g(x)=2x^2[/math]. Now we apply the algebraic method (in particular the formula) for linear differential equations.[br][br]1. [math]\int p\left(x\right)dx=\int-\frac{1}{x}dx=-\ln\left(x\right)=\ln\left(x^{-1}\right)[/math][br]2. [math]e^{\int p\left(x\right)dx}=e^{\ln\left(x^{-1}\right)}=x^{-1}[/math][br]3.,4.,5. [math]y\left(x\right)=\frac{\int e^{\int p\left(x\right)dx}g\left(x\right)dx+c}{e^{\int p\left(x\right)dx}}=\frac{\int x^{-1}\cdot2x^2dx+c}{x^{-1}}=\frac{\int2x^{ }dx+c}{x^{-1}}[/math][br][math]=\frac{x^2+c}{x^{-1}}=x^3+c\cdot x[/math][br]6. The general solution is [math]y(x)=x^3+c\cdot x[/math]. The initial condition is that when [math]x[/math] is 1 [math]y[/math] should be 2. Thus [math]y(1)=1^3+c\cdot1=2[/math] which implies that [math]c=1[/math], and the specific solution is therefore [math]y(x)=x^3+x[/math].[br][br]Don't forget to visualize it in GeoGebra!

The best thing to do is to spend time practicing. If you're interested, here's a sheet of several differential equations that you can use to practice.