AP Calculus AB Quiz: Verifying Solutions For Differential Equations
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Verifying Solutions For Differential EquationsQuestion 1 of 20

Does y=1x3y=\dfrac{1}{x^3} satisfy dydx=−3xy\dfrac{dy}{dx}=-\dfrac{3}{x}y for x≠0x\ne 0?

Yes, because y′=−3x4y'=-\dfrac{3}{x^4} and −3xy=−3x4-\dfrac{3}{x}y=-\dfrac{3}{x^4}.
No, because y′=3x4y'=\dfrac{3}{x^4} but −3xy=−3x4-\dfrac{3}{x}y=-\dfrac{3}{x^4}.
No, because y′=−1x3y'=-\dfrac{1}{x^3} but −3xy=−3x4-\dfrac{3}{x}y=-\dfrac{3}{x^4}.
Yes, because −3xy=−3x3-\dfrac{3}{x}y=-\dfrac{3}{x^3} equals y′y'.
No, because y′=−9x4y'=-\dfrac{9}{x^4} but −3xy=−3x4-\dfrac{3}{x}y=-\dfrac{3}{x^4}.
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AP Calculus AB Quiz

AP Calculus AB Quiz: Verifying Solutions For Differential Equations

Practice Verifying Solutions For Differential Equations in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Verifying Solutions For Differential Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Does y=1x3y=\dfrac{1}{x^3} satisfy dydx=−3xy\dfrac{dy}{dx}=-\dfrac{3}{x}y for x≠0x\ne 0?

  1. Yes, because y′=−3x4y'=-\dfrac{3}{x^4} and −3xy=−3x4-\dfrac{3}{x}y=-\dfrac{3}{x^4}. (correct answer)
  2. No, because y′=3x4y'=\dfrac{3}{x^4} but −3xy=−3x4-\dfrac{3}{x}y=-\dfrac{3}{x^4}.
  3. No, because y′=−1x3y'=-\dfrac{1}{x^3} but −3xy=−3x4-\dfrac{3}{x}y=-\dfrac{3}{x^4}.
  4. Yes, because −3xy=−3x3-\dfrac{3}{x}y=-\dfrac{3}{x^3} equals y′y'.
  5. No, because y′=−9x4y'=-\dfrac{9}{x^4} but −3xy=−3x4-\dfrac{3}{x}y=-\dfrac{3}{x^4}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = 1/x³ = x^{-3}, we find y' = -3x^{-4} = -3/x⁴ using the power rule. The differential equation dy/dx = -(3/x)y requires that y' equals -(3/x)y. We compute -(3/x)y = -(3/x)(x−3x^{-3}) = -3x^{-4} = -3/x⁴. Since y' = -3/x⁴ and -(3/x)y = -3/x⁴, both sides are equal. Choice B has the wrong sign, stating y' = 3/x⁴ instead of -3/x⁴. When applying the power rule to negative powers, the result includes a negative coefficient.

Question 2

Does y=1xy=\dfrac{1}{x} satisfy dydx=−y2\dfrac{dy}{dx}=-y^2 for all x≠0x\ne 0?

  1. Yes, because y′=−1x2y'=-\dfrac{1}{x^2} and −y2=−1x2-y^2=-\dfrac{1}{x^2}. (correct answer)
  2. No, because y′=1x2y'=\dfrac{1}{x^2} but −y2=−1x2-y^2=-\dfrac{1}{x^2}.
  3. No, because y′=−1xy'=-\dfrac{1}{x} but −y2=−1x2-y^2=-\dfrac{1}{x^2}.
  4. Yes, because y′=−1x2y'=-\dfrac{1}{x^2} and −y2=−1x-y^2=-\dfrac{1}{x}.
  5. No, because y′=−2x3y'=-\dfrac{2}{x^3} but −y2=−1x2-y^2=-\dfrac{1}{x^2}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y=1x=x−1y = \dfrac{1}{x} = x^{-1}, we find y′=−x−2=−1x2y' = -x^{-2} = -\dfrac{1}{x^2} using the power rule. The differential equation dydx=−y2\dfrac{dy}{dx} = -y^2 requires that y′y' equals −y2-y^2. We compute −y2=−(1x)2=−1x2-y^2 = -\left(\dfrac{1}{x}\right)^2 = -\dfrac{1}{x^2}. Since y′=−1x2y' = -\dfrac{1}{x^2} and −y2=−1x2-y^2 = -\dfrac{1}{x^2}, both sides are equal. Choice B has the wrong sign for y', stating y′=1x2y' = \dfrac{1}{x^2} instead of −1x2-\dfrac{1}{x^2}. To verify solutions involving negative powers, carefully apply the power rule and check signs throughout.

Question 3

Does y=1xy=\dfrac{1}{x} satisfy the differential equation dydx=yx\dfrac{dy}{dx}=\dfrac{y}{x} for x≠0x\ne 0?

  1. Yes, because y′=−1x2y'=-\dfrac{1}{x^2} and yx=1x2\dfrac{y}{x}=\dfrac{1}{x^2}.
  2. No, because y′=−1x2y'=-\dfrac{1}{x^2} but yx=1x2\dfrac{y}{x}=\dfrac{1}{x^2}. (correct answer)
  3. No, because y′=1x2y'=\dfrac{1}{x^2} and yx=1x\dfrac{y}{x}=\dfrac{1}{x}.
  4. Yes, because y′=1x2y'=\dfrac{1}{x^2} equals yx\dfrac{y}{x}.
  5. No, because yx=−1x2\dfrac{y}{x}=-\dfrac{1}{x^2} but y′=1x2y'=\dfrac{1}{x^2}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y=1xy = \dfrac{1}{x}, we find y′=−1x2y' = -\dfrac{1}{x^2} using the power rule. The differential equation dydx=yx\dfrac{dy}{dx} = \dfrac{y}{x} requires that y′y' equals yx\dfrac{y}{x}. We compute yx=1xx=1x2\dfrac{y}{x} = \dfrac{\dfrac{1}{x}}{x} = \dfrac{1}{x^2}. Since y′=−1x2y' = -\dfrac{1}{x^2} but yx=1x2\dfrac{y}{x} = \dfrac{1}{x^2}, the two sides have opposite signs and are not equal. The function does not satisfy the differential equation. Choice A incorrectly claims they are equal, missing the sign difference. Always check signs carefully when verifying solutions, as sign errors are common sources of mistakes.

Question 4

Does y=3e2xy=3e^{2x} satisfy the differential equation dydx=2y\dfrac{dy}{dx}=2y for all xx?

  1. No, because y′=6e2xy'=6e^{2x} but 2y=3e2x2y=3e^{2x}.
  2. Yes, because y′=6e2xy'=6e^{2x} and 2y=6e2x2y=6e^{2x}. (correct answer)
  3. Yes, because y′=3e2xy'=3e^{2x} equals 2y2y.
  4. No, because y′=2e2xy'=2e^{2x} but 2y=6e2x2y=6e^{2x}.
  5. No, because y′=6exy'=6e^{x} but 2y=6e2x2y=6e^{2x}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = 3e^{2x}, we find y' = 3 · 2e^{2x} = 6e^{2x}. The differential equation dy/dx = 2y requires that y' equals 2y. We compute 2y = 2(3e2x3e^{2x}) = 6e^{2x}. Since y' = 6e^{2x} and 2y = 6e^{2x}, both sides are equal. Choice A incorrectly calculates 2y as 3e^{2x} instead of 6e^{2x}. To verify any solution, calculate the derivative, substitute both y and y' into the equation, and confirm both sides match.

Question 5

Does y=1x−1y=\dfrac{1}{x-1} satisfy dydx=−y2\dfrac{dy}{dx}=-y^2 for all x≠1x\ne 1?

  1. Yes, because y′=−1(x−1)2y'=-\dfrac{1}{(x-1)^2} and −y2=−1(x−1)2-y^2=-\dfrac{1}{(x-1)^2}. (correct answer)
  2. No, because y′=1(x−1)2y'=\dfrac{1}{(x-1)^2} but −y2=−1(x−1)2-y^2=-\dfrac{1}{(x-1)^2}.
  3. No, because y′=−1x−1y'=-\dfrac{1}{x-1} but −y2=−1(x−1)2-y^2=-\dfrac{1}{(x-1)^2}.
  4. Yes, because −y2=−1x−1-y^2=-\dfrac{1}{x-1} equals y′y'.
  5. No, because y′=−2(x−1)3y'=-\dfrac{2}{(x-1)^3} but −y2=−1(x−1)2-y^2=-\dfrac{1}{(x-1)^2}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = 1/(x-1) = (x-1)^{-1}, we find y' = -(x-1)^{-2} = -1/(x-1)² using the chain rule. The differential equation dy/dx = -y² requires that y' equals -y². We compute -y² = -(1/(x-1))² = -1/(x-1)². Since y' = -1/(x-1)² and -y² = -1/(x-1)², both sides are equal. Choice B has the wrong sign, stating y' = 1/(x-1)² instead of -1/(x-1)². When differentiating (x-a)^{-1}, the chain rule gives -(x-a)^{-2}, which includes the negative sign.

Question 6

Does y=1x2+1y=\dfrac{1}{x^2+1} satisfy dydx=−2xy2\dfrac{dy}{dx}=-2xy^2 for all real xx?

  1. Yes, because y′=−2x(x2+1)2y'=-\dfrac{2x}{(x^2+1)^2} and −2xy2=−2x(x2+1)2-2xy^2=-\dfrac{2x}{(x^2+1)^2}. (correct answer)
  2. No, because y′=2x(x2+1)2y'=\dfrac{2x}{(x^2+1)^2} but −2xy2=−2x(x2+1)2-2xy^2=-\dfrac{2x}{(x^2+1)^2}.
  3. No, because y′=−2x2+1y'=-\dfrac{2}{x^2+1} but −2xy2=−2x(x2+1)2-2xy^2=-\dfrac{2x}{(x^2+1)^2}.
  4. Yes, because y′=−2xx2+1y'=-\dfrac{2x}{x^2+1} equals −2xy2-2xy^2.
  5. No, because −2xy2=−2xx2+1-2xy^2=-\dfrac{2x}{x^2+1}, not y′y'.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = 1/(x²+1), we find y' = -2x/(x²+1)² using the quotient rule or chain rule. The differential equation dy/dx = -2xy² requires that y' equals -2xy². We compute -2xy² = -2x(1/(x²+1))² = -2x/(x²+1)². Since y' = -2x/(x²+1)² and -2xy² = -2x/(x²+1)², both sides are equal. Choice B has the wrong sign, stating y' = 2x/(x²+1)² instead of -2x/(x²+1)². When differentiating rational functions, carefully apply the quotient rule and track signs throughout.

Question 7

Does y=2x+5y=2x+5 satisfy the differential equation dydx=2\dfrac{dy}{dx}=2 for all xx?

  1. Yes, because y′=2y'=2 and the right side is 22. (correct answer)
  2. No, because y′=2xy'=2x but the right side is 22.
  3. No, because y′=5y'=5 but the right side is 22.
  4. Yes, because y′=2x+5y'=2x+5 equals 22.
  5. No, because y′=0y'=0 but the right side is 22.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = 2x + 5, we find y' = 2 using the power rule (derivative of 2x is 2, derivative of constant 5 is 0). The differential equation dy/dx = 2 requires that y' equals 2. Substituting our derivative: y' = 2, which exactly matches the right side 2. Since both sides are identical, the function satisfies the equation. Choice B incorrectly states y' = 2x, confusing the original function with its derivative. For linear functions y = mx + b, the derivative is always the slope m.

Question 8

Does y=x2+1y=x^2+1 satisfy dydx=2x\dfrac{dy}{dx}=2x at every xx?

  1. Yes, because y′=2xy'=2x and the right side is 2x2x. (correct answer)
  2. No, because y′=x2y'=x^2 but the right side is 2x2x.
  3. No, because y′=2x+1y'=2x+1 but the right side is 2x2x.
  4. Yes, because y′=x2+1y'=x^2+1 equals 2x2x.
  5. No, because y′=2y'=2 but the right side is 2x2x.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = x² + 1, we find y' = 2x using the power rule. The differential equation dy/dx = 2x requires that y' equals 2x. Substituting our derivative: y' = 2x, which exactly matches the right side 2x. Since both sides are identical, the function satisfies the equation. Choice B incorrectly states y' = x², confusing the original function with its derivative. Always compute derivatives carefully and verify both sides of the equation are equal.

Question 9

Does y=13x3y=\dfrac{1}{3}x^3 satisfy the differential equation dydx=x2\dfrac{dy}{dx}=x^2 for all xx?

  1. Yes, because y′=x2y'=x^2 and the right side is x2x^2. (correct answer)
  2. No, because y′=3x2y'=3x^2 but the right side is x2x^2.
  3. No, because y′=13x3y'=\dfrac{1}{3}x^3 but the right side is x2x^2.
  4. Yes, because y′=13y'=\dfrac{1}{3} equals x2x^2.
  5. No, because y′=xy'=x but the right side is x2x^2.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = (1/3)x³, we find y' = (1/3)(3x²) = x² using the power rule. The differential equation dy/dx = x² requires that y' equals x². Substituting our derivative: y' = x², which exactly matches the right side x². Since both sides are identical, the function satisfies the equation. Choice B incorrectly states y' = 3x², missing the coefficient 1/3 in the original function. When differentiating with constant factors, multiply the power rule result by the constant: d/dx[cf(x)] = c·f'(x).

Question 10

Does y=e3xy=e^{3x} satisfy the differential equation dydx=3y\dfrac{dy}{dx}=3y for all xx?

  1. No, because y′=e3xy'=e^{3x} but 3y=3e3x3y=3e^{3x}.
  2. Yes, because y′=3e3xy'=3e^{3x} and 3y=3e3x3y=3e^{3x}. (correct answer)
  3. No, because y′=3exy'=3e^{x} but 3y=3e3x3y=3e^{3x}.
  4. Yes, because y′=e3xy'=e^{3x} equals 3y3y.
  5. No, because y′=9e3xy'=9e^{3x} but 3y=3e3x3y=3e^{3x}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = e^{3x}, we find y' = 3e^{3x} using the chain rule. The differential equation dy/dx = 3y requires that y' equals 3y. We compute 3y = 3e^{3x}. Since y' = 3e^{3x} and 3y = 3e^{3x}, both sides are equal. Choice A incorrectly states y' = e^{3x}, missing the factor of 3 from the chain rule. When differentiating e^{kx}, multiply by the coefficient k of the exponent using the chain rule.

Question 11

Does y=ln⁡(x2)y=\ln(x^2) satisfy the differential equation dydx=2x\dfrac{dy}{dx}=\dfrac{2}{x} for x≠0x\ne 0?

  1. No, because y′=1xy'=\dfrac{1}{x} but the right side is 2x\dfrac{2}{x}.
  2. Yes, because y′=2xy'=\dfrac{2}{x} and the right side is 2x\dfrac{2}{x}. (correct answer)
  3. No, because y′=2x2y'=\dfrac{2}{x^2} but the right side is 2x\dfrac{2}{x}.
  4. Yes, because y′=ln⁡(x2)y'=\ln(x^2) equals 2x\dfrac{2}{x}.
  5. No, because y′=2xy' = 2x but the right side is 2x\dfrac{2}{x}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = ln(x²), we can rewrite this as y = 2ln(x) and find y' = 2/x using the chain rule or logarithm properties. Alternatively, using the chain rule directly: y' = (1/x²)(2x) = 2/x. The differential equation dy/dx = 2/x requires that y' equals 2/x. Since y' = 2/x matches the right side exactly, the function satisfies the equation. Choice A incorrectly states y' = 1/x, missing the factor of 2 from the chain rule. When differentiating ln(x²), use either the chain rule or the property ln(x²) = 2ln(x).

Question 12

Does y=x2y=x^2 satisfy the differential equation dydx=2yx\dfrac{dy}{dx}=\dfrac{2y}{x} for x≠0x\ne 0?

  1. No, because y′=2xy'=2x but 2yx=2x2x=x\dfrac{2y}{x}=\dfrac{2x^2}{x}=x.
  2. Yes, because y′=2xy'=2x and 2yx=2x2x=2x\dfrac{2y}{x}=\dfrac{2x^2}{x}=2x. (correct answer)
  3. No, because y′=x2y'=x^2 but 2yx=2x\dfrac{2y}{x}=2x.
  4. Yes, because y′=2y'=2 and 2yx=2x\dfrac{2y}{x}=2x.
  5. No, because 2yx=2x2\dfrac{2y}{x}=\dfrac{2}{x^2}, not 2x2x.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = x², we find y' = 2x using the power rule. The differential equation dy/dx = (2y)/x requires that y' equals (2y)/x. We compute (2y)/x = (2x²)/x = 2x for x ≠ 0. Since y' = 2x and (2y)/x = 2x, both sides are equal. Choice A incorrectly calculates (2y)/x as x instead of 2x, making an algebraic error. When simplifying rational expressions, carefully cancel common factors and verify the algebra is correct.

Question 13

Does y=12e4xy=\dfrac{1}{2}e^{4x} satisfy the differential equation dydx=4y\dfrac{dy}{dx}=4y for all xx?

  1. Yes, because y′=2e4xy'=2e^{4x} and 4y=2e4x4y=2e^{4x}. (correct answer)
  2. No, because y′=12e4xy'=\dfrac{1}{2}e^{4x} but 4y=2e4x4y=2e^{4x}.
  3. No, because y′=2exy'=2e^{x} but 4y=2e4x4y=2e^{4x}.
  4. Yes, because y′=4e4xy'=4e^{4x} equals 4y4y.
  5. No, because 4y=12e4x4y=\dfrac{1}{2}e^{4x}, not y′y'.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = (1/2)e^{4x}, we find y' = (1/2)(4e4x4e^{4x}) = 2e^{4x} using the chain rule. The differential equation dy/dx = 4y requires that y' equals 4y. We compute 4y = 4((1/2)e^{4x}) = 2e^{4x}. Since y' = 2e^{4x} and 4y = 2e^{4x}, both sides are equal. Choice B incorrectly states y' = (1/2)e^{4x}, missing the factor of 4 from the chain rule. When differentiating exponential functions with chain rule, multiply by the derivative of the exponent.

Question 14

Does y=1xy=\dfrac{1}{\sqrt{x}} satisfy dydx=−12xy\dfrac{dy}{dx}=-\dfrac{1}{2x}y for x>0x>0?

  1. Yes, because y′=−12x−3/2y'=-\dfrac{1}{2}x^{-3/2} and −12xy=−12x−3/2-\dfrac{1}{2x}y=-\dfrac{1}{2}x^{-3/2}. (correct answer)
  2. No, because y′=12x−3/2y'=\dfrac{1}{2}x^{-3/2} but −12xy=−12x−3/2-\dfrac{1}{2x}y=-\dfrac{1}{2}x^{-3/2}.
  3. No, because y′=−x−1/2y'=-x^{-1/2} but −12xy=−12x−3/2-\dfrac{1}{2x}y=-\dfrac{1}{2}x^{-3/2}.
  4. Yes, because −12xy=−12x−1/2-\dfrac{1}{2x}y=-\dfrac{1}{2}x^{-1/2} equals y′y'.
  5. No, because y′=−32x−5/2y'=-\dfrac{3}{2}x^{-5/2} but −12xy=−12x−3/2-\dfrac{1}{2x}y=-\dfrac{1}{2}x^{-3/2}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = 1/√x = x^{-1/2}, we find y' = (-1/2)x^{-3/2} = -1/(2x3/22x^{3/2}) using the power rule. The differential equation dy/dx = -(1/2x)y requires that y' equals -(1/2x)y. We compute -(1/2x)y = -(1/2x)(x−1/2x^{-1/2}) = (-1/2)x^{-3/2}. Since y' = (-1/2)x^{-3/2} and -(1/2x)y = (-1/2)x^{-3/2}, both sides are equal. Choice B has the wrong sign, stating y' = (1/2)x^{-3/2} instead of (-1/2)x^{-3/2}. When differentiating negative fractional powers, carefully track the negative sign from the power rule.

Question 15

Does y=ex+e−xy=e^{x}+e^{-x} satisfy the differential equation dydx=ex−e−x\dfrac{dy}{dx}=e^{x}-e^{-x} for all xx?

  1. No, because y′=ex+e−xy'=e^{x}+e^{-x} but the right side is ex−e−xe^{x}-e^{-x}.
  2. Yes, because y′=ex−e−xy'=e^{x}-e^{-x} and the right side matches. (correct answer)
  3. No, because y′=−ex+e−xy'=-e^{x}+e^{-x} but the right side is ex−e−xe^{x}-e^{-x}.
  4. Yes, because y′=exy'=e^{x} equals ex−e−xe^{x}-e^{-x}.
  5. No, because y′=−e−xy'= -e^{-x} but the right side is ex−e−xe^{x}-e^{-x}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = e^x + e^{-x}, we find y' = e^x + (-1)e^{-x} = e^x - e^{-x} using the chain rule. The differential equation dy/dx = e^x - e^{-x} requires that y' equals e^x - e^{-x}. Substituting our derivative: y' = e^x - e^{-x}, which exactly matches the right side e^x - e^{-x}. Since both sides are identical, the function satisfies the equation. Choice A incorrectly states y' = e^x + e^{-x}, missing the negative sign from differentiating e^{-x}. When differentiating e^{-x}, the chain rule gives -e^{-x}.

Question 16

Does y=cos⁡xy=\cos x satisfy dydx=−sin⁡x\dfrac{dy}{dx}=-\sin x for all xx?

  1. Yes, because y′=−sin⁡xy'=-\sin x and the right side is −sin⁡x-\sin x. (correct answer)
  2. No, because y′=sin⁡xy'=\sin x but the right side is −sin⁡x-\sin x.
  3. No, because y′=−cos⁡xy'=-\cos x but the right side is −sin⁡x-\sin x.
  4. Yes, because y′=cos⁡xy'=\cos x equals −sin⁡x-\sin x.
  5. No, because y′=−sin⁡2xy'=-\sin^2 x but the right side is −sin⁡x-\sin x.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = cos x, we find y' = -sin x using standard trigonometric derivatives. The differential equation dy/dx = -sin x requires that y' equals -sin x. Substituting our derivative: y' = -sin x, which exactly matches the right side -sin x. Since both sides are identical, the function satisfies the equation. Choice C incorrectly states y' = -cos x, which would be the derivative of sin x, not cos x. Remember that d/dx[cos x] = -sin x and d/dx[sin x] = cos x.

Question 17

Does y=12x−2y=\dfrac{1}{2}x^{-2} satisfy the differential equation dydx=−2xy\dfrac{dy}{dx}=-\dfrac{2}{x}y for x≠0x\ne 0?

  1. No, because y′=−x−3y'=-x^{-3} but −2xy=−2x−3-\dfrac{2}{x}y=-2x^{-3}.
  2. Yes, because y′=−x−3y'=-x^{-3} and −2xy=−x−3-\dfrac{2}{x}y=-x^{-3}. (correct answer)
  3. No, because y′=x−3y'=x^{-3} but −2xy=−x−3-\dfrac{2}{x}y=-x^{-3}.
  4. Yes, because y′=12x−2y'=\dfrac{1}{2}x^{-2} equals −2xy-\dfrac{2}{x}y.
  5. No, because −2xy=−1x2-\dfrac{2}{x}y=-\dfrac{1}{x^2}, not y′y'.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y=12x−2=12x2y = \dfrac{1}{2} x^{-2} = \dfrac{1}{2 x^2}, we find y′=12(−2)x−3=−x−3=−1x3y' = \dfrac{1}{2} (-2) x^{-3} = -x^{-3} = -\dfrac{1}{x^3} using the power rule. The differential equation dydx=−2xy\dfrac{dy}{dx} = -\dfrac{2}{x} y requires that y′y' equals −2xy-\dfrac{2}{x} y. We compute −2xy=−2x(12x−2)=−x−3=−1x3-\dfrac{2}{x} y = -\dfrac{2}{x} \left( \dfrac{1}{2} x^{-2} \right) = -x^{-3} = -\dfrac{1}{x^3}. Since y′=−1x3y' = -\dfrac{1}{x^3} and −2xy=−1x3-\dfrac{2}{x} y = -\dfrac{1}{x^3}, both sides are equal. Choice C has the wrong sign, stating y′=x−3y' = x^{-3} instead of −x−3-x^{-3}. When applying the power rule to negative powers, carefully track the negative coefficient.

Question 18

Does y=ln⁡xy=\ln x satisfy the differential equation dydx=1x\dfrac{dy}{dx}=\dfrac{1}{x} for x>0x>0?

  1. No, because y′=1ln⁡xy'=\dfrac{1}{\ln x} but the right side is 1x\dfrac{1}{x}.
  2. Yes, because y′=1xy'=\dfrac{1}{x} and the right side is 1x\dfrac{1}{x}. (correct answer)
  3. No, because y′=1x2y'=\dfrac{1}{x^2} but the right side is 1x\dfrac{1}{x}.
  4. Yes, because y′=ln⁡xy'=\ln x equals 1x\dfrac{1}{x}.
  5. No, because y′=xy'=x but the right side is 1x\dfrac{1}{x}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = ln x, we find y' = 1/x using the standard logarithmic derivative. The differential equation dy/dx = 1/x requires that y' equals 1/x. Substituting our derivative: y' = 1/x, which exactly matches the right side 1/x. Since both sides are identical, the function satisfies the equation. Choice A incorrectly states y' = 1/(ln x), confusing the derivative of ln x with some other expression. Remember that d/dx[ln x] = 1/x for x > 0.

Question 19

Does y=ln⁡(1+x)y=\ln(1+x) satisfy dydx=11+x\dfrac{dy}{dx}=\dfrac{1}{1+x} for x>−1x>-1?

  1. No, because y′=1xy'=\dfrac{1}{x} but the right side is 11+x\dfrac{1}{1+x}.
  2. Yes, because y′=11+xy'=\dfrac{1}{1+x} and the right side matches. (correct answer)
  3. No, because y′=1ln⁡(1+x)y'=\dfrac{1}{\ln(1+x)} but the right side is 11+x\dfrac{1}{1+x}.
  4. Yes, because y′=ln⁡(1+x)y'=\ln(1+x) equals 11+x\dfrac{1}{1+x}.
  5. No, because y′=1(1+x)2y'=\dfrac{1}{(1+x)^2} but the right side is 11+x\dfrac{1}{1+x}.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = ln(1+x), we find y' = 1/(1+x) using the chain rule for logarithms. The differential equation dy/dx = 1/(1+x) requires that y' equals 1/(1+x). Substituting our derivative: y' = 1/(1+x), which exactly matches the right side 1/(1+x). Since both sides are identical, the function satisfies the equation. Choice A incorrectly states y' = 1/x, missing the chain rule application for the inner function (1+x). When differentiating ln(u), the result is u'/u where u' is the derivative of the inner function.

Question 20

Does y=sin⁡xy=\sin x satisfy the differential equation dydx=cos⁡x\dfrac{dy}{dx}=\cos x for all xx?

  1. No, because y′=−sin⁡xy'=-\sin x but the right side is cos⁡x\cos x.
  2. Yes, because y′=cos⁡xy'=\cos x and the right side is cos⁡x\cos x. (correct answer)
  3. No, because y′=sin⁡xy'=\sin x but the right side is cos⁡x\cos x.
  4. Yes, because y′=−cos⁡xy'=-\cos x equals cos⁡x\cos x.
  5. No, because y′=cos⁡2xy'=\cos^2 x but the right side is cos⁡x\cos x.
Explanation: To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = sin x, we find y' = cos x using standard trigonometric derivatives. The differential equation dy/dx = cos x requires that y' equals cos x. Substituting our derivative: y' = cos x, which exactly matches the right side cos x. Since both sides are identical, the function satisfies the equation. Choice A incorrectly states y' = -sin x, which would be the derivative of cos x, not sin x. Always apply derivative formulas correctly and verify both sides of the equation match.