How do you turn an implicit solution into an explicit function in Differential Equations? In this 46-second clip, we walk through the exponentiation algebra used to transform ln|y| = -cos(t) + C1 into the explicit general solution y(t) = C * e^(-cos(t)).
Step-by-Step Breakdown:
Exponentiating Both Sides: To clear the natural log, we raise e to the power of both sides: e^(ln|y|) = e^(-cos(t) + C1).
Using Exponent Rules: Splitting the right side yields e^(C1) * e^(-cos(t)).
Absorbing Constants & Absolute Values: Since e^(C1) is just a positive constant, we can absorb both it and the plus/minus sign from dropping the absolute value |y| into a single arbitrary real constant, C.
The Explicit General Solution: We arrive at our final explicit form: y(t) = C * e^(-cos(t)).
Mastering how e^C1 absorbs signs and constants is a critical skill for calculus 2 and differential equations exams!
📚 Watch the Full 80-Minute Exam Review:
• Full Review Video:
#DifferentialEquations #Calculus2 #AlgebraTricks #Integration #Exponentiation #CollegeMath #STEM #ExamPrep
Pinned Comment: That step where e^(C1) turns into a single arbitrary constant C (and swallows the plus/minus from the absolute value) trips up so many students! 🧠⚡ It’s totally valid and makes working with the final solution much cleaner.
WATCH THE FULL 80-MINUTE EXAM REVIEW HERE:
Does your professor let you skip straight to writing y = C * e^(-cos(t)), or do they make you show all the steps with e^(C1) first? Let me know below! 👇
📖 Differential Equations, by Blanchard, Devaney, and Hall:
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That step where e^(C1) turns into a single arbitrary constant C (and swallows the plus/minus from the absolute value) trips up so many students! 🧠⚡ It's totally valid and makes working with the final solution much cleaner.
WATCH THE FULL 80-MINUTE EXAM REVIEW HERE: https://www.youtube.com/watch?v=fmZBPvn2X6s
Does your professor let you skip straight to writing y = C * e^(-cos(t)), or do they make you show all the steps with e^(C1) first? Let me know below! 👇