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# 2015 AP Calculus AB/BC 4cd

Solutions for the differential equation.

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• Sal, I'm not sure if you did the problem wrong or if this is an oversight by the college board (or, more likely, I'm not understanding this.)

if y = 2x-2, the initial condition of f(2)=3 is incorrect.
• Part c, the f(x) is not necessary a linear function. Part d function must be a linear function, as it is in the form y=mx+b. So they could be different functions, and so the conditions in part c cannot apply to part d.
• To get the linear equation, I tried plugging in the initial conditions f(2) = 3 into the linear equation using m=2 to find the y intercept. 3 = 2(2) + b, so b = -1. Why doesn't that approach work? Thanks, ben
• Well, the general solution to this differential equation is
y = f(x) = 2x - 2 + Ce^(-x), where C ∈ ℝ.
(Don't worry if you don't know how to get this solution.)
With the initial condition f(2) = 3 the constant C is not 0. (It is actually e².)
Because C ≠ 0, we can't just ignore that exponential part.
However, if the initial condition was f(2) = 2, your method would work, because in this case C = 0.
This is a good question, but I don't know how to explain this without talking about the general solution.
(1 vote)
• Why can't I understand most of this.....oh wait, I'm still an 8th grader