This week was one of the most satisfying weeks I’ve had in a long time. I got through 36% of the unit Differential Equations and understood everything I learned this week without much difficulty. In total I made it through 8 videos, 2 articles, and 4 exercises. The hardest part of what I worked on this week was the Algebra that came up in some of the questions, but it was good practice for me and was somewhat fun in a puzzle-solving type of way. Considering how long and difficult the last unit, Integrals, was, I’m a bit surprised at how easy I’m finding this unit so far. I only have a few more vids/exercises left in the unit so hopefully I didn’t just jinx it!

I started the week off by watching the first video in the section Separation of Variables. I learned that this is the first technique that should be used when tackling differential equations.

As you can see in from my notes, this technique only works when you’re able to separate the variables and get y and dy to one side of the equation and x and dx to the other side. As I worked through this section, Sal mentioned a few times that in order to use this technique you have to be able to get the equation into some form of dy/dx = g(y)h(x).

Here are a couple of examples of equations where the ‘separation-of-variables’ technique would work:

It took me until Thursday to get through the 4 videos and 2 articles in this section. Working through the 3 exercises in this section really helped me to wrap my head around how separable differential equations work and, as I said up top, gave me some good practice with some tricky algebra. Here are two questions I worked on from the exercises in this section:

• dy/dx = –10x⁴y
  • (1/y) * dy = dx * –10x⁴
  • y^-1 * dy = dx * –10x⁴
  • ∫ y^-1 * dy = ∫ –10x⁴ * dx
  • ln|y| = –10x⁵/5 + C
  • ln|y| = –2x⁵ + C
    • (Note: The following step took me a minute to wrap my head around. I still don’t have an intuitive feel for how/when to use log rules to isolate the argument which is what I needed to do to isolate y. As you can see from the right side of my notes, I used the example of log₂(8) to work out how set it as the power on base 2 in order to isolate the argument 8. Once I figured it out, I then did the same thing to ln|y| and set it at the power to e^ln|y| to isolate y.)
  • e^ln|y| = e^{–2x^5 + C}
    • (I also had to review exponent rules to remember that when a base has exponents which are being added to each other, it’s the same thing as the base being raised to both exponents individually and then multiplying both arguments.)
  • y = e^{–2x^5} * e^C_1
    • (In the step above you can see that I labelled C as C_1 which denotes C₁. The following concept took me awhile to understand but I think now understand how it works. C is an arbitrary number that, depending on the (x, y) input, is going to shift the function up or down the y-axis. Raising e to the power of C₁ isn’t going to affect the shape of the function, it will simply shift the function further up or further down the y-axis. Because of this, you can relabel e^C_1 as C₂ since e^C_1 will end up being some arbitrary number.) 
  • y = e^{–2x^5} * C₂

I didn’t start the following section, Particular Solutions to Separable Equations, until Saturday but still managed to get through 4 videos and 1 exercise in the section. It took me a bit of time to figure out what this section was teaching but, once I figured it out, it made everything regarding C seem much more clear to me. In this section, I worked on questions where Sal gave me the derivative of a function, f’(x), a solution to the function, f(x) = y (as an example from below, f(1) = 1), and then asked me to find the y-coordinate to function at another x input. (Reading that back, that sounds really confusing but I’m pretty sure I’m saying that all correctly.) Here are two examples:

• Givens:
  • f'(x) = 8x³ – 12x
  • f(1) = 1
  • f(–2) = ?
• Step 1 – Find antiderivative of f’(x), a.k.a. the equation for f(x):
  • ∫ f’(x) dx = ∫ 8x³ – 12x * dx
    • = 2x⁴ – 6x² + C = f(x)
• Step 2 – Knowing that f(1) = 1, use this information to solve for C from f(x) = 2x⁴ – 6x² + C:
  • f(1) = 1 = 2x⁴ – 6x² + C
    • 1 = 2(1)⁴ – 6(1)² + C
    • 1 = 2 – 6 + C
    • 1 = –4 + C
    • 5 = C
• Step 3 – Knowing that C = 5 in f(x), solve for f(–2):
  • f(x) = 2x⁴ – 6x² + 5
    • f(–2) = 2(–2)⁴ – 6(–2)² + 5
      • = 2(16) – 6(4) + 5
      • = 32 – 24 + 5
      • = 8 + 5
      • = 13
• Conclusion:
  • f(–2) = 13

I’m now 58% of the way through Differential Equations (640/1,100 M.P.). I only have 3 videos and 3 exercises left to get through in the entire unit so I’m hoping I might be able to get through it all, and even the unit test, by the end of this week. Considering that I’ve only been working on this unit for a week and a half and thinking about how long it took me to get through Integrals, it’s crazy to think that I might be able to finish this unit off in ~2.5 weeks. If I do, I still think it’ll be tough for me to reach my goal of getting through Calculus 1 by May 5^th as I would still need to fly through the following unit, which is pretty big, plus get through the Course Challenge. Nonetheless, I’m a lot closer to realizing that goal than I thought I’d be which I’m definitely happy about! 😊