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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 12

Lesson 1: AP Calculus AB 2017 free response

# 2017 AP Calculus AB/BC 4c

Potato problem from 2017 AP exam (Question 4, part c). Finding a particular solution for a separable differential equation.

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• where is the C for the integral with (G-27)^(-2/3)?
• The constants of integration are entirely arbitrary, so we don't need one on each side.

Let's, for argument's sake, have one on both sides ie
3(G - 27)^(1/3) + C₁ = -t + C₂
We could take C₁ from both sides:
3(G - 27)^(1/3) = -t + C₂ - C₁
But since they are arbitrary we can let C = C₂ - C₁ and so we have
3(G - 27)^(1/3) = -t + C
• What if the negative sign on (G-27) was not transferred to t? Wouldn't the constant be negative 12 instead? How did Sal know to move the negative over with the t?
• It doesn't matter. He could have moved it before, during, or after integration; G(t) would still be the same in all cases.
• At , how did Sal turn `dG/((G-27)^2/3)` into `(G-27)^2/3 dG` ?
• Sal turned dG/((G-27)^2/3) into (G-27)^-2/3 dG , not (G-27)^2/3. A negative exponent turns it into a denominator of a fraction
• Why doesn't Sal just substitute G(0)=91 in the dG/dt equation?
• The question is to evaluate at t=3, so it's G(3)