Home / AP Calculus BC : 7.2 Verifying Solutions for  Differential Equations- Exam Style questions with Answer- MCQ

AP Calculus BC : 7.2 Verifying Solutions for  Differential Equations- Exam Style questions with Answer- MCQ

Question

If \(\frac{\mathrm{d} y}{\mathrm{d} x}=-x\cos (x^{2})\) and y = 2 when x = 0, then a solution to the differential equation is:
(A) \(y=-\frac{1}{2}\sin (x^{2})\)
(B) \(y=-\frac{1}{2}(\sin (x))^{2}+2\)
(C) \(y=-\frac{1}{2}\cos (x^{2})+2\)
(D) \(y=-\frac{1}{2}\sin (x^{2})+2\)

Answer/Explanation

Ans:(D)

To solve the differential equation, separate the variables out on two sides and integrate to solve for y.
\(\frac{\mathrm{d} y}{\mathrm{d} x}=-x\cos (x^{2})\)
\(dy=-x\cos (x^{2})dx\)
\(\int dy=\int -x\cos (x^{2})dx\)
To solve the integral on the right use substitution with \(u=x^{2}\) and du=2xdx
\(y=-\int \frac{du}{2}\cos (u)\)
\(y=-\frac{1}{2}\int \cos (u)du\)
\(y=-\frac{1}{2}\sin (u)+C\)
\(y=-\frac{1}{2}\sin (x^{2})+C\)
C can be solved for using our initial information that at x = 0, y = 2.
\(2=-\frac{1}{2}\sin (0)+C\Rightarrow C=2\)
\(y=-\frac{1}{2}\sin (x^{2})+2\)

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