Struggling with AP Multiple Choice: ky Differential Equation

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SUMMARY

The discussion focuses on solving the differential equation \(\frac{dy}{dt} = ky\), where \(k\) is a nonzero constant. The correct approach to find the solution involves separating variables and integrating, leading to the general solution \(y = Ae^{kt}\). Participants suggest two methods: plugging in the answer choices to check for validity or differentiating each option to identify the correct one. The solution confirms that the only viable answer is \(b) 2e^{kt}\).

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  • Understanding of differential equations, specifically first-order linear equations.
  • Familiarity with the method of separation of variables.
  • Knowledge of natural logarithms and exponential functions.
  • Basic calculus skills, including differentiation and integration.
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  • Study the method of separation of variables in differential equations.
  • Learn how to apply integration techniques to solve first-order linear differential equations.
  • Explore the properties of exponential functions and their applications in differential equations.
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Students preparing for AP Calculus exams, educators teaching differential equations, and anyone looking to strengthen their understanding of first-order differential equations.

tandoorichicken
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Heres an AP multiple choice problem that's giving me some trouble:

If [tex]\frac{\,dy}{\,dt} = ky[/tex] and k is a nonzero constant, then y could be

a) [itex]2e^{kty}[/itex]
b) [itex]2e^{kt}[/itex]
c) [itex]e^{kt} +3[/itex]
d) [itex]kty + 5[/itex]
e) [itex]\frac{1}{2}ky^2 + \frac{1}{2}[/itex]

Dont quite know where to begin?
 
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You have two choices.

#1 (probably the hardest way, but useful if you forgot how to solve it): Take all of the answers and plug them in. See which one works.

#2: Separate the variables and integrate.

dy/y = k*dt
ln(y) = kt + C
y = e^(kt + c) = Ae^(kt)

cookiemonster
 
Actually, for this problem, I suspect that most people would just differentiate each of the answers.
 

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