Apentix/AP Calculus AB/Unit 7: Differential Equations/Question

AP Calculus ABmediummcq1 pt

What is the particular solution of $\dfrac{dy}{dx} = \dfrac{x}{y}$ that satisfies $y(0) = 1$ ?

A.

y = \sqrt{x^2 + 1}

B.

y = \sqrt{x^2 - 1}

C.

y = x^2 + 1

D.

y = \dfrac{x^2}{2} + 1

Explanation

Core Concept

Separating gives $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$ , so $y^2 = x^2 + 2C$ ; with $y(0)=1$ , $2C = 1$ .

Correct Answer

A $y = \sqrt{x^2 + 1}$

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