Home / AP Calculus BC 7.6 Finding General Solutions Using Separation of Variables – Exam Style Questions – MCQs

AP Calculus BC 7.6 Finding General Solutions Using Separation of Variables - MCQs - Exam Style Questions

AP Calculus BC-MCQs and Answers - All Topics

Question

Which of the following is the solution to the differential equation \( \dfrac{dy}{dx}=x\big(y-2\big) \) with the initial condition \( y(2)=3 \)?
(A) \( y=2-\mathrm e^{\tfrac{x^{2}}{2}-2} \)
(B) \( y=2+\mathrm e^{\tfrac{x^{2}}{2}-2} \)
(C) \( y=2+\mathrm e^{\,2-\tfrac{x^{2}}{2}} \)
(D) \( y=3-\mathrm e^{2}+\mathrm e^{\tfrac{x^{2}}{2}} \)
▶️ Answer/Explanation
Detailed solution
Separate variables: \[ \frac{dy}{y-2}=x\,dx \quad\Rightarrow\quad \ln|y-2|=\frac{x^{2}}{2}+C. \] Hence \( y-2=K\,\mathrm e^{x^{2}/2} \). Use \( y(2)=3 \Rightarrow 1=K\,\mathrm e^{2}\Rightarrow K=\mathrm e^{-2} \). So \( y=2+\mathrm e^{\,x^{2}/2-2} \). ✅ Answer: (B)

No-Calc Question

If \(\displaystyle \frac{dy}{dx}=2-y\), and \(y=1\) when \(x=1\), then \(y=\)
(A) \(2-e^{x-1}\)
(B) \(2-e^{\,1-x}\)
(C) \(2-e^{-x}\)
(D) \(2+e^{-x}\)
▶️ Answer/Explanation
Detailed solution

Solve \(\dfrac{dy}{dx}=2-y\Rightarrow \dfrac{dy}{dx}+y=2\).
Homogeneous solution: \(y_h=C\,e^{-x}\). Particular solution: \(y_p=2\).
General solution: \(y=2+C\,e^{-x}\).

Apply \(y(1)=1\): \(1=2+C\,e^{-1}\Rightarrow C=-e\).
Therefore \(y=2- e\cdot e^{-x}=2-e^{\,1-x}\).

✅ Answer: (B)

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