AP Calculus AB: Separation of Variables (Drill 1)

Question 1. Which of the following differential equations is not separable?

• A) \( \dfrac{dy}{dx} = xy \)
• B) \( \dfrac{dy}{dx} = \dfrac{\cos x}{y^2} \)
• C) \( \dfrac{dy}{dx} = x + y \) ✓
• D) \( \dfrac{dy}{dx} = x^2 e^{-y} \)

Explanation: Choice C is correct. A separable equation must be writable as \( g(y)\,dy = f(x)\,dx \), meaning the right-hand side must factor into a product of a function of x alone and a function of y alone. The expression \( x + y \) is a sum and cannot be factored this way, so the equation is not separable. Choice A is incorrect because \( \dfrac{dy}{dx} = xy \) is separable: \( \dfrac{dy}{y} = x\,dx \). Choice B is incorrect because \( \dfrac{dy}{dx} = \dfrac{\cos x}{y^2} \) is separable: \( y^2\,dy = \cos x\,dx \). Choice D is incorrect because \( \dfrac{dy}{dx} = x^2 e^{-y} \) is separable: \( e^y\,dy = x^2\,dx \).

Question 2. If \( \dfrac{dy}{dx} = 2xy \) and \( y > 0 \), which of the following is the general solution?

• A) \( y = e^{x^2} + C \)
• B) \( y = Ce^{x^2} \) ✓
• C) \( y = Ce^{2x} \)
• D) \( y = x^2 + C \)

Explanation: Choice B is correct. Separating variables: \( \dfrac{dy}{y} = 2x\,dx \). Integrating both sides: \( \ln|y| = x^2 + C_1 \), so \( y = Ce^{x^2} \) where \( C > 0 \) since \( y > 0 \). Choice A is incorrect because it places the constant of integration outside the exponential, the correct form absorbs the constant into the coefficient, not as an additive term. Choice C is incorrect because it results from integrating \( 2x \) as though it were the constant \( 2 \), giving \( e^{2x} \) instead of \( e^{x^2} \). Choice D is incorrect because it treats the equation as \( \dfrac{dy}{dx} = 2x \), ignoring the factor of \( y \) entirely and producing a polynomial rather than an exponential solution.

Question 3. Consider the differential equation \( \dfrac{dy}{dx} = \dfrac{3x^2}{y} \) with initial condition \( y(0) = 4 \). What is the particular solution?

• A) \( y = \sqrt{2x^3 + 16} \) ✓
• B) \( y = \sqrt{2x^3 + 4} \)
• C) \( y = 2x^3 + 16 \)
• D) \( y = \sqrt{x^3 + 16} \)

Explanation: Choice A is correct. Separating variables: \( y\,dy = 3x^2\,dx \). Integrating: \( \dfrac{y^2}{2} = x^3 + C \), so \( y^2 = 2x^3 + C_1 \). Applying \( y(0) = 4 \): \( 16 = 0 + C_1 \), so \( C_1 = 16 \). Since \( y(0) = 4 > 0 \), take the positive root: \( y = \sqrt{2x^3 + 16} \). Choice B is incorrect because the student substituted \( y(0) = 4 \) directly as the constant rather than squaring it first, giving \( C_1 = 4 \) instead of 16. Choice C is incorrect because the student correctly found \( y^2 = 2x^3 + 16 \) but forgot to take the square root, reporting \( y^2 \) as the final answer instead of \( y \). Choice D is incorrect because the student dropped the coefficient of 2 when solving \( \dfrac{y^2}{2} = x^3 + C \), failing to multiply both sides by 2, and then applied the initial condition to the uncorrected equation, giving \( y = \sqrt{x^3 + 16} \).

Question 4. A student solves the separable equation \( \dfrac{dy}{dx} = \dfrac{x}{y} \) with initial condition \( y(0) = -3 \) and obtains \( y = -\sqrt{x^2 + 9} \). Which of the following best explains why the negative square root is chosen rather than the positive square root?

• A) The negative square root produces a smaller value, which minimizes the solution.
• B) The initial condition \( y(0) = -3 \) requires the solution to be negative at \( x = 0 \), ruling out the positive branch. ✓
• C) Taking the square root of both sides of \( y^2 = x^2 + 9 \) always produces a \( \pm \), and the negative sign must be kept because \( x^2 \) is always positive.
• D) Separating variables introduces a negative sign that must be carried into the final answer.

Explanation: Choice B is correct. After separating and integrating, the general solution is \( y = \pm\sqrt{x^2 + 9} \). Since \( y(0) = -3 < 0 \) and a continuous solution cannot switch branches, the negative root \( y = -\sqrt{x^2 + 9} \) must be selected to satisfy the initial condition. Choice A is incorrect because minimizing the solution value is not a mathematical criterion for branch selection, the choice is determined entirely by the initial condition. Choice C is incorrect because the sign of \( x^2 \) inside the radical has no bearing on which branch to choose; the \( \pm \) is resolved by applying the initial condition, not by examining the form of the expression under the square root. Choice D is incorrect because no extra negative sign is introduced by the algebraic process of separating variables; the \( \pm \) arises from taking the square root of \( y^2 \).

Question 5. The differential equation \( \dfrac{dy}{dx} = \dfrac{2x+1}{3y^2} \) has the initial condition \( y(1) = 2 \). What is the value of \( y(0) \)?

• A) \( \sqrt[3]{6} \) ✓
• B) \( \sqrt[3]{10} \)
• C) \( \sqrt[3]{5} \)
• D) \( \sqrt[3]{7} \)

Explanation: Choice A is correct. Separating variables: \( 3y^2\,dy = (2x+1)\,dx \). Integrating: \( y^3 = x^2 + x + C \). Applying \( y(1) = 2 \): \( 8 = 1 + 1 + C \), so \( C = 6 \). The particular solution is \( y^3 = x^2 + x + 6 \). At \( x = 0 \): \( y^3 = 6 \), so \( y = \sqrt[3]{6} \). Choice B is incorrect because the student swapped the initial and evaluation points, applying \( y(0) = 2 \) gives \( C = 8 \), then \( y(1)^3 = 1 + 1 + 8 = 10 \), yielding \( \sqrt[3]{10} \). Choice C is incorrect because the student made an arithmetic error when solving for \( C \), computing \( 8 - 3 = 5 \) instead of \( 8 - 1 - 1 = 6 \), perhaps treating \( x^2 + x \) at \( x = 1 \) as a single value of 3 rather than evaluating each term separately. Choice D is incorrect because the student dropped the \( +x \) term when integrating, treating \( \int(2x+1)\,dx \) as \( x^2 \) only. With \( y^3 = x^2 + C \) and \( y(1) = 2 \): \( C = 7 \), giving \( y(0) = \sqrt[3]{7} \).