I didn’t get started on the unit test from Differential Equations this week, but I did manage to get through the remaining videos and exercises in it. It took me three days to get through the 2^nd and final exercise in the section Particular Solutions to Differential Equations. The calculus concepts from this exercise weren’t difficult to understand but I found the algebra fairly difficult. Once I got through that exercise, I began the final section of the unit, Exponential Models, on Friday and was happily surprised that it wasn’t too tough to get through. That being said, I don’t fully understand what was taught in this section but I think with a bit more practice it won’t be too hard to figure it out. Overall, I’m happy with how the week went and that I’ll be able to get straight to the unit test at the beginning of next week. 🥳

 Since most of my week was spent working on exercises, this post will mostly be examples of the questions that I worked through. Here are four questions that I worked on in the second exercises of Particular Solutions to Differential Equations:

Question 1:

• Givens:
  • f’(x) = 2 * f(x)
  • f(1) = 5
  • f(3) = m*eⁿ
• Question:
  • m = ?
  • n = ?
• Solution:
  • f'(x) = dy/dx
    • dy/dx = 2 * y
    • 1/y * dy = 2 * x⁰ * dx 
    • ∫ y^-1 * dy = ∫ 2x⁰ * dx
    • ln|y| = 2x + C
      • Input: f(1) = 5
    • ln|5| = 2(1) + C
    • ln|5| = 2 + C
      • C = ln|5| – 2
        • Input: C = ln|5| – 2 into ln|y| = 2x + C at f(3)
    • ln|y| = 2(3) + (ln|5| – 2)
    • ln|y| = 2(3) + (ln|5| – 2)
    • ln|y| = 6 + ln|5| – 2
    • e^ln|y| = e^{6 + ln|5| – 2}
    • |y| = e⁴ * e^ln|5|
    • |y| = e⁴ * 5
    • y = 5e⁴
      • (Note: Since 5e⁴ is a positive number, you can take the absolute-value brackets off of y.)
• Conclusion:
  • y = 5e⁴
    • m = 5
    • n = 4

Question 2:

Question 3:

Question 4:

I began the second section of the unit, Exponential Models, on Friday. Right now, I don’t have a strong understanding of the concepts that were covered in the section. As far as I understand it, this section covers how to use calculus to solve for variables in exponential functions. Once I figured out the pattern of how to solve the questions from the 2 exercise I worked through, the math wasn’t difficult, but I don’t have an intuitive understanding for how/why the calculus actually works. The silver lining is that in the past ~2.5 years this often happens; first I learn the mechanics of how to solve a set of questions then the intuition for what’s happening eventually comes. The following are 3 example questions I worked on. The first question came from the first exercise and the following 2 questions were from the second exercise:

Question 1:

• Givens:
  • dy/dt = 2y
  • Point at (0, 8)
• Question:
  • Solve for y.
• Solution:
  • dy/dt = 2y
    • y^-1 * dy = 2 * t⁰ * dt
    • ∫ y^-1 * dy = ∫2t⁰ * dt
    • ln|y| = 2t + C
      • Input (0, 8)
    • ln|8| = 2(0) + C
    • ln|8| = C
      • Input C = ln|8| back into ln|y| = 2t + C
    • ln|y| = 2t + (ln|8|)
    • e^ln|y| = e^{2t + (ln|8|)}
    • y = e^2t * e^ln|8|
    • y = e^2t * 8
    • y = 8 * e^2t
• Conclusion:
  • y = 8 * e^2t

Question 2:

• Givens:
  • Mass decreases at a rate proportional to the mass at a given time
    • i.e. dM/dt = K * M
      • (Note: This part it confusing, but it describes an exponential equation in that the mass of the isotope is a function of time and the rate at which the mass decreases is constant but relative to the mass at any given time.)
  • Initial Mass = 40 grams
    • i.e. (0, 40)
  • Mass after 16 days = 10 grams 
    • i.e. (16, 10)
• Question:
  • What is the mass after 20 days?
• Solution:
  • dM/dt = K * M
  • M^-1 * dM = K * t⁰ * dt
  • ∫ M^-1 * dM = ∫ K * t⁰ * dt
  • ln|M| = K * t + C
  • e^ln|M| = e^{K * t + C}
  • M(t) = e^Kt * e^C
  • M(t) = C * e^Kt
    • Input (0, 40) and solve for C.
  • M(0) = 40 = C * e^K(0)
    • 40 = C * (1)
      • C = 40
    • Solve for K using (16, 10) in M(t) = C * e^Kt
  • M(16) = 10 = 40 * e^K(16)
    • 10/40 = e^k(16)
    • ln(1/4) = ln(e^K(16))
    • ln(1/4) = K * 16
      • K = ln(1/4)/16
    • Input t = 20 into M(t) = 40 * e^{(ln(1/4))/16) * t}
  • M(20) = 40 * e^{(ln(1/4))/16) * 20}
    • = ~7.07107
  • Conclusion:
    • After 20 days the mass of the isotope is approximately 7 grams.

Question 3:

Typing out the solutions to the questions above, I’m realizing just how confusing these concepts are to me. I don’t think it’s going to take me too long to figure it all out, but right now it seems very abstract and hard to grasp…

I’m hoping it won’t take me too many attempts to get through the unit test this coming week. Assuming I get through it, that means I will have gotten through Differential Equations (880/1,100 M.P.) in ~3.5 weeks! As I said in my last post, there’s essentially no chance that I’ll get through the final unit in Calculus 1, Applications of Integrals (0/1,900 M.P.), before the end of the month but that’s ok. I can definitely see the light at the end of the tunnel in regards to getting through all of calculus, especially considering I’m already 41% of the way through the course Calculus 2! If I can somehow manage to get through Calc 2 before September, I will have gone from learning arithmetic to understanding calculus in 3 years. That’s a lot longer than the <1 year I thought it was going to take (I was so naïve lol) but still a solid accomplishment!