Home / AP Calculus AB 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables – MCQs

AP Calculus AB 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables - MCQs - Exam Style Questions

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No-Calc Question

Let \(y=f(x)\) be the particular solution to the differential equation \(\displaystyle \frac{dy}{dx}=\frac{x+1}{y}\) with the initial condition \(f(0)=-2\). Which of the following is an expression for \(f(x)\)?

(A) \(-2-\sqrt{x^{2}+2x}\)
(B) \(-2+\sqrt{x^{2}+2x}\)
(C) \(-\sqrt{x^{2}+2x+4}\)
(D) \(\sqrt{x^{2}+2x+4}\)
▶️ Answer/ExplanationSeparate variables: \(y\,dy=(x+1)\,dx\).
Integrate both sides: \(\displaystyle \frac{y^{2}}{2}=\frac{x^{2}}{2}+x+C\).
Apply \(y(0)=-2\): \(\displaystyle \frac{(-2)^{2}}{2}=0+0+C \Rightarrow 2=C\).
Substitute \(C=2\) and multiply by \(2\): \(y^{2}=x^{2}+2x+4\).
Thus \(y=\pm\sqrt{x^{2}+2x+4}\). Using \(y(0)=-2\) selects the negative branch: \(y=-\sqrt{x^{2}+2x+4}\).
✅ Answer: (C) \(-\sqrt{x^{2}+2x+4}\)
Question
Population \( y \) grows according to the equation \(\frac{dy}{dt} = ky\), where \( k \) is a constant and \( t \) is measured in years. If the population doubles every 10 years, then the value of \( k \) is
(A) 0.069
(B) 0.200
(C) 0.301
(D) 3.322
(E) 5.000
▶️ Answer/Explanation
Solution
The solution to \(\frac{dy}{dt} = ky\) is \( y(t) = y(0) e^{kt} \).
Given the population doubles every 10 years, \( y(10) = 2y(0) \).
Substitute: \( 2y(0) = y(0) e^{10k} \).
Divide by \( y(0) \) (assuming \( y(0) \neq 0 \)): \( 2 = e^{10k} \).
Take natural log: \( \ln(2) = 10k \).
Solve: \( k = \frac{\ln(2)}{10} \approx \frac{0.693}{10} = 0.069 \).
✅ Answer: A
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