AP Calculus BC 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables - Exam Style Questions - MCQs - New Syllabus
Question
Which of the following satisfies the differential equation \(\displaystyle \frac{dy}{dx}=\cos\!\left(\frac{\pi}{2}x^{2}\right)\) with the initial condition \(y(0)=2\)?
A) \(y=2+\sin\!\left(\frac{\pi}{2}x^{2}\right)\)
B) \(y=2+\dfrac{1}{\pi x}\sin\!\left(\frac{\pi}{2}x^{2}\right)\)
C) \(y=\displaystyle \int_{2}^{x}\cos\!\left(\frac{\pi}{2}t^{2}\right)\,dt\)
D) \(y=2+\displaystyle \int_{0}^{x}\cos\!\left(\frac{\pi}{2}t^{2}\right)\,dt\)
B) \(y=2+\dfrac{1}{\pi x}\sin\!\left(\frac{\pi}{2}x^{2}\right)\)
C) \(y=\displaystyle \int_{2}^{x}\cos\!\left(\frac{\pi}{2}t^{2}\right)\,dt\)
D) \(y=2+\displaystyle \int_{0}^{x}\cos\!\left(\frac{\pi}{2}t^{2}\right)\,dt\)
▶️ Answer/Explanation
Detailed solution
For \(y'(x)=f(x)\) and \(y(a)=y_{0}\), a particular solution is \(y(x)=y_{0}+\displaystyle\int_{a}^{x} f(t)\,dt\).
Here \(f(x)=\cos\!\left(\frac{\pi}{2}x^{2}\right)\), \(a=0\), \(y_{0}=2\). Thus \[ y(x)=2+\int_{0}^{x}\cos\!\left(\tfrac{\pi}{2}t^{2}\right)\,dt, \] which satisfies both the differential equation and \(y(0)=2\).
✅ Correct: D