AP Calculus BC Quiz: Verifying Solutions For Differential Equations

20 questions · exam conditions

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Verifying Solutions For Differential EquationsQuestion 1 of 20

Does $y=x^2+1$ satisfy $\frac{dy}{dx}=2\sqrt{y-1}$ for all $x\ge 0$ ?

No, because

y'=x

but

2\sqrt{y-1}=2x

Yes, because

y'=2x

and

2\sqrt{x^2}=2x

for

x\ge0

No, because

y'=2x

but

2\sqrt{y-1}=2|x|

Yes, because

y'=2x+1

and

2\sqrt{y-1}=2x+1

No, because

y'=2x

but

2\sqrt{y-1}=\sqrt{2x}

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AP Calculus BC Quiz

AP Calculus BC Quiz: Verifying Solutions For Differential Equations

Practice Verifying Solutions For Differential Equations in AP Calculus BC 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 BC.

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 $y=x^2+1$ satisfy $\frac{dy}{dx}=2\sqrt{y-1}$ for all $x\ge 0$ ?

No, because $y'=x$ but $2\sqrt{y-1}=2x$ .
Yes, because $y'=2x$ and $2\sqrt{x^2}=2x$ for $x\ge0$ . (correct answer)
No, because $y'=2x$ but $2\sqrt{y-1}=2|x|$ .
Yes, because $y'=2x+1$ and $2\sqrt{y-1}=2x+1$ .
No, because $y'=2x$ but $2\sqrt{y-1}=\sqrt{2x}$ .

Explanation: Verifying whether a proposed function satisfies a differential equation is a key skill in differential equations, ensuring the solution meets the equation's conditions across the specified domain. Start by finding y' for y = x² + 1, which is y' = 2x. Substitute into the right side: 2√(y - 1) = 2√(x²) = 2|x|, and for x ≥ 0, this simplifies to 2x. Since y' equals 2√(y - 1) for x ≥ 0, the function satisfies the equation. A tempting distractor might state 2√(y - 1) = 2|x| without considering the domain restriction, which could mislead for x < 0, but here the domain is x ≥ 0. Always differentiate the proposed solution and plug into both sides of the equation to confirm equality as a general strategy for verification.

Question 2

Does $y=\dfrac{1}{x+2}$ satisfy $\dfrac{dy}{dx}=-y^2$ on $x\ne -2$ ?

No, because $y'=\dfrac{1}{(x+2)^2}$ .
Yes, because $y'=-\dfrac{1}{(x+2)^2}$ and $-y^2=-\dfrac{1}{(x+2)^2}$ . (correct answer)
No, because $-y^2=-\dfrac{1}{x+2}$ .
Yes, because $y'=-\dfrac{1}{x+2}$ .
No, because $y^2=\dfrac{1}{x+2}$ .

Explanation: Verifying solutions requires computing derivatives using the chain rule and checking algebraic relationships. For

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

where

x \ne -2

, we have

y' = -\dfrac{1}{(x+2)^2}

using the chain rule on

(x+2)^{-1}

. The right side

-y^2

becomes

-\left[ \dfrac{1}{x+2} \right]^2 = -\dfrac{1}{(x+2)^2}

. Since both

y'

and

-y^2

equal

-\dfrac{1}{(x+2)^2}

, the solution is verified. Choice A has the wrong sign for the derivative, while choice C makes an error in computing

-y^2

. The key verification approach is to apply the chain rule correctly to expressions of the form

(ax+b)^n

and carefully handle negative signs in algebraic expressions.

Question 3

For $\dfrac{dy}{dx}=y\cos x$ , is $y=e^{\sin x}$ a solution for all $x$ ?

Yes, because $y'=e^{\sin x}\cos x=y\cos x$ . (correct answer)
No, because $y'=e^{\sin x}\sin x$ .
No, because $y'=\cos x$ .
Yes, because $y'=e^{x}\cos x$ .
No, because $y\cos x=e^{\cos x}$ .

Explanation: Solution verification for composite exponential functions requires applying the chain rule correctly. For y = e^(sin x), we compute y' = e^(sin x) · cos x by the chain rule. The right side y cos x becomes e^(sin x) · cos x. Since both y' and y cos x equal e^(sin x) cos x, the solution is verified for all x. Choice B incorrectly computes the derivative, while choice E makes an error in the exponential expression. The key verification approach is to apply the chain rule carefully when differentiating composite functions, especially exponentials with trigonometric arguments.

Question 4

Does $y=\dfrac{1}{x}$ satisfy $\dfrac{dy}{dx}=-y^2$ for $x\ne 0$ ?

Yes, because $y'=-\dfrac{1}{x^2}$ and $-y^2=-\dfrac{1}{x^2}$ . (correct answer)
No, because $y'=\dfrac{1}{x^2}$ but $-y^2=-\dfrac{1}{x^2}$ .
No, because $-y^2=-\dfrac{1}{x}$ .
Yes, because $y'=-\dfrac{1}{x}$ and $-y^2=-\dfrac{1}{x}$ .
No, because $y^2=\dfrac{1}{x}$ so $-y^2=-\dfrac{1}{x}$ .

Explanation: Verifying solutions involves computing derivatives using appropriate rules and checking algebraic identities. For y = 1/x where x ≠ 0, we compute y' = -1/x² using the power rule. The right side becomes -y² = -(1/x)² = -1/x². Since y' = -1/x² and -y² = -1/x², both sides are equal, confirming the solution. Choice B has the wrong sign for the derivative, while choice C incorrectly computes -y². The systematic verification approach is to differentiate using standard rules, substitute into the equation, and verify that both sides are algebraically identical.

Question 5

For motion, velocity satisfies $\dfrac{dv}{dt}=4t$ . Is $v(t)=2t^2+3$ a solution?

No, because $v'(t)=4t+3$ .
Yes, because $v'(t)=4t$ equals $\dfrac{dv}{dt}$ . (correct answer)
No, because $v'(t)=2t$ .
Yes, because $\dfrac{dv}{dt}=8t$ and $v'(t)=8t$ .
No, because $4t=2t^2+3$ is not true for all $t$ .

Explanation: Verifying whether a proposed function satisfies a differential equation is a key skill in differential equations. To check if

v(t) = 2t^2 + 3

solves

\dfrac{dv}{dt} = 4t

, compute

v'(t) = 4t

. The right-hand side is simply

4t

. Since

v'(t)

equals

4t

, it satisfies the equation. Choice E is tempting but wrong because it compares

4t

v(t) = 2t^2 + 3

instead of to

v'(t)

. Always substitute the proposed solution into both sides of the differential equation and simplify to check for equality across the domain.

Question 6

For $\dfrac{dy}{dx}=\dfrac{y}{x}$ , is $y=3x$ a solution on $x\ne 0$ ?

Yes, because $y'=3$ and $\dfrac{y}{x}=\dfrac{3x}{x}=3$ . (correct answer)
No, because $y'=3x$ but $\dfrac{y}{x}=3$ .
No, because $\dfrac{y}{x}=\dfrac{3}{x}$ is not constant.
Yes, because $y'=1$ and $\dfrac{y}{x}=1$ .
No, because $y'=0$ and $\dfrac{y}{x}=3$ .

Explanation: To verify a solution, substitute both the function and its derivative into the differential equation. For y = 3x, we compute y' = 3. The differential equation dy/dx = y/x becomes 3 = (3x)/x = 3, which is true for all x ≠ 0. Choice B incorrectly states y' = 3x instead of y' = 3. Choice E incorrectly claims y' = 0, which would only be true for a constant function. The systematic approach is to differentiate the proposed solution, substitute both y and y' into the equation, and verify the identity holds on the specified domain.

Question 7

For $\dfrac{dy}{dx}=\dfrac{y}{x}+x$ , is $y=x^2$ a solution on $x\ne 0$ ?

Yes, because $y'=2x$ and $\dfrac{y}{x}+x=\dfrac{x^2}{x}+x=2x$ . (correct answer)
No, because $\dfrac{y}{x}+x=\dfrac{x^2}{x^2}+x=1+x$ .
No, because $y'=x$ .
Yes, because $y'=2$ .
No, because $\dfrac{y}{x}+x=\dfrac{x}{x}+x=1+x$ .

Explanation: Solution verification involves computing derivatives and careful algebraic manipulation of combined terms. For y = x² where x ≠ 0, we have y' = 2x. The right side y/x + x becomes x²/x + x = x + x = 2x. Since both y' and y/x + x equal 2x, the solution is verified. Choice B incorrectly computes y/x as x²/x² = 1 instead of x²/x = x. Choice E makes a similar error. The systematic approach is to substitute the proposed solution carefully into each term of sum or difference expressions and simplify algebraically before comparing sides.

Question 8

A curve is defined by $\dfrac{dy}{dx}=2xy$ . Is $y=e^{x^2}$ a solution for all real $x$ ?

No, because $y'=2x$ but $2xy=2xe^{x^2}$ .
Yes, because $y'=2xe^{x^2}$ and $2xy=2xe^{x^2}$ . (correct answer)
No, because $y'=e^{x^2}$ but $2xy=2xe^{x^2}$ .
Yes, because $y'=e^{2x}$ and $2xy=e^{2x}$ .
No, because $2xy=2x^2e^{x^2}$ , which does not equal $y'$ .

Explanation: Verifying whether a proposed function satisfies a differential equation is a key skill in differential equations. To check if y = e^{x^2} solves dy/dx = 2xy, compute y' using the chain rule, resulting in 2x e^{x^2}. Next, calculate the right-hand side 2xy, which is 2x times e^{x^2}, or 2x e^{x^2}. Since y' equals 2xy for all real x, it is a solution. A distractor like choice A fails by claiming y' = 2x, forgetting the chain rule that multiplies by e^{x^2}. Always substitute the proposed solution into both sides of the differential equation and simplify to check for equality across the domain.

Question 9

Does $y=\tan x$ satisfy $\dfrac{dy}{dx}=1+y^2$ where defined?

Yes, because $y'=\sec^2 x$ and $1+\tan^2 x=\sec^2 x$ . (correct answer)
No, because $y'=\sec x\tan x$ .
No, because $1+y^2=1+\sec^2 x$ .
Yes, because $y'=\cos^2 x$ .
No, because $y'=1-\tan^2 x$ .

Explanation: Verifying solutions for trigonometric differential equations requires applying derivatives correctly and using trigonometric identities. For y = tan x, we compute y' = sec²x. The right side 1 + y² becomes 1 + tan²x. Using the trigonometric identity 1 + tan²x = sec²x, both sides are equal, confirming the solution where tan x is defined. Choice B gives an incorrect derivative (sec x tan x is the derivative of sec x, not tan x). Choice C makes an error in the identity. Always use standard trigonometric derivatives and identities when verifying solutions involving trigonometric functions.

Question 10

For $\dfrac{dy}{dx}=4x^3$ , is $y=x^4+7$ a solution for all $x$ ?

No, because $y'=x^3$ .
Yes, because $y'=4x^3$ equals the right-hand side. (correct answer)
No, because $y'=4x^4$ .
Yes, because $y'=4x^3+7$ .
No, because the constant $7$ changes $\dfrac{dy}{dx}$ .

Explanation: Solution verification for polynomial functions involves straightforward differentiation and substitution. For y = x⁴ + 7, we compute y' = 4x³. The differential equation dy/dx = 4x³ becomes 4x³ = 4x³, which is clearly an identity for all x. The constant term 7 in the solution does not affect the derivative since the derivative of a constant is zero. Choice A incorrectly states the derivative as x³. Choice E incorrectly suggests the constant affects the derivative. The key verification principle is that adding constants to solutions doesn't change derivatives, so constants don't affect verification of first-order differential equations.

Question 11

A cooling object follows $\dfrac{dT}{dt}=-2(T-10)$ . Is $T(t)=10+6e^{-2t}$ a solution?

Yes, because $T'(t)=-12e^{-2t}$ and $-2(T-10)=-12e^{-2t}$ . (correct answer)
No, because $T'(t)=12e^{-2t}$ , which has the wrong sign.
No, because $T-10=6e^{2t}$ so $-2(T-10)=-12e^{2t}$ .
Yes, because $T'(t)=-2(10+6e^{-2t})$ .
No, because $-2(T-10)=-2(10+6e^{-2t})$ .

Explanation: Verifying whether a proposed function satisfies a differential equation is a key skill in differential equations. To check if T(t) = 10 + 6e^{-2t} solves dT/dt = -2(T - 10), first compute T'(t) by differentiating, yielding -12e^{-2t} via the chain rule. Then, substitute into the right-hand side: T - 10 = 6e^{-2t}, so -2(T - 10) = -12e^{-2t}. Since both sides match, the function satisfies the equation. Choice C is a common distractor because it mistakenly uses e^{2t} instead of e^{-2t}, leading to an incorrect sign and exponent. Always substitute the proposed solution into both sides of the differential equation and simplify to check for equality across the domain.

Question 12

Does $y=\dfrac{1}{x^2}$ satisfy $\dfrac{dy}{dx}=-\dfrac{2}{x}y$ on $x\ne 0$ ?

Yes, because $y'=-\dfrac{2}{x^3}$ and $-\dfrac{2}{x}y=-\dfrac{2}{x}\cdot\dfrac{1}{x^2}=-\dfrac{2}{x^3}$ . (correct answer)
No, because $y'=\dfrac{2}{x^3}$ .
No, because $-\dfrac{2}{x}y=-\dfrac{2}{x^2}$ .
Yes, because $y'=-\dfrac{2}{x^2}$ and RHS is $-\dfrac{2}{x^2}$ .
No, because $y'=-\dfrac{1}{x^3}$ .

Explanation: Verifying solutions requires computing derivatives using the power rule and careful substitution. For

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

where

x \ne 0

, we have

y' = -\dfrac{2}{x^3}

using the power rule on

x^{-2}

. The right side

-\dfrac{2}{x} y

becomes

-\dfrac{2}{x} \cdot \dfrac{1}{x^2} = -\dfrac{2}{x^3}

. Since both

y'

and

-\dfrac{2}{x} y

equal

-\dfrac{2}{x^3}

, the solution is verified. Choice B has the wrong sign for the derivative, while choice C incorrectly computes the right-hand side. The systematic approach is to apply the power rule correctly to negative exponents and carefully substitute the proposed solution into rational expressions.

Question 13

For $\dfrac{dy}{dx}=\dfrac{1}{1+x^2}$ , is $y=\arctan x +5$ a solution?

Yes, because $y'=\dfrac{1}{1+x^2}$ . (correct answer)
No, because $y'=\dfrac{1}{1+(\arctan x)^2}$ .
No, because $y'=\dfrac{1}{(1+x^2)^2}$ .
Yes, because $y'=\dfrac{1}{1+x^2}+5$ .
No, because the constant $5$ makes $y'$ undefined.

Explanation: Solution verification for inverse trigonometric functions involves using standard derivatives and recognizing constant terms. For y = arctan x + 5, we compute y' = 1/(1+x²) + 0 = 1/(1+x²), since the derivative of a constant is zero. The differential equation dy/dx = 1/(1+x²) becomes 1/(1+x²) = 1/(1+x²), which is clearly an identity for all real x. Choice B incorrectly suggests the derivative involves arctan x, while choice D incorrectly includes the constant 5 in the derivative. Constants in solutions don't affect verification since their derivatives are zero.

Question 14

Does $y=2\sin x$ satisfy $\dfrac{dy}{dx}=\sqrt{4-y^2}$ on $0\le x\le \pi/2$ ?

Yes, because $y'=2\cos x$ and $\sqrt{4-4\sin^2 x}=2\cos x$ on $[0,\pi/2]$ . (correct answer)
No, because $\sqrt{4-y^2}=\sqrt{4-2\sin x}$ .
No, because $y'=\cos x$ .
Yes, because $\sqrt{4-y^2}=2\sin x$ .
No, because $\sqrt{4-4\sin^2 x}=2|\sin x|$ .

Explanation: Verifying trigonometric solutions requires careful differentiation and attention to domain restrictions. For y = 2sin x on [0, π/2], we have y' = 2cos x. The right side √(4-y²) becomes √(4-4sin²x) = √(4(1-sin²x)) = √(4cos²x) = 2|cos x|. On the interval [0, π/2], cos x ≥ 0, so |cos x| = cos x, making √(4-y²) = 2cos x. Since y' = 2cos x and √(4-y²) = 2cos x, the solution is verified. Choice B incorrectly substitutes the expression, while choice E incorrectly handles the absolute value. Always consider domain restrictions when evaluating expressions involving absolute values.

Question 15

Does $y=\sin x$ satisfy $\dfrac{dy}{dx}=\sqrt{1-y^2}$ on the interval $0\le x\le \pi$ ?

Yes, because $y'=\cos x$ and $\sqrt{1-y^2}=\sqrt{1-\sin^2 x}=\cos x$ .
Yes, because $y'=-\cos x$ and $\sqrt{1-y^2}=\cos x$ .
No, because $\sqrt{1-\sin^2 x}=|\cos x|$ which equals $\cos x$ only for $0\le x\le \tfrac{\pi}{2}$ . (correct answer)
No, because $y'=\sin x$ but $\sqrt{1-y^2}=\cos x$ .
No, because $\sqrt{1-y^2}=\sqrt{1-\sin x}$ after substitution.

Explanation: We must verify if y = sin x satisfies dy/dx = √(1 - y²) on the interval [0, π]. First, we find y' = cos x. Then we compute √(1 - y²) = √(1 - sin²x) = √(cos²x) = |cos x|. The key insight is that √(cos²x) = |cos x|, not simply cos x. On the interval [0, π], cos x is positive only for 0 ≤ x < π/2 and negative for π/2 < x ≤ π. Therefore, y' = cos x equals √(1 - y²) = |cos x| only when cos x ≥ 0, which is not true for the entire interval. When verifying solutions involving square roots, always consider the absolute value that arises from √(a²) = |a|.

Question 16

Given $\frac{dy}{dx}+2y=4$ and $y=2+Ce^{-2x}$ , does the family satisfy the differential equation?

No, because $y'=-2Ce^{-2x}$ so $y'+2y=4+2Ce^{-2x}$ .
Yes, because $y'=2Ce^{-2x}$ so $y'+2y=4$ .
Yes, because $y'=-2Ce^{-2x}$ so $y'+2y=4$ . (correct answer)
No, because $y'=-2Ce^{-x}$ so $y'+2y=4$ .
No, because $2y=4+Ce^{-2x}$ so $y'+2y\ne4$ .

Explanation: This problem requires verifying if y = 2 + Ce^(-2x) satisfies dy/dx + 2y = 4. First, we find y' = d/dx(2 + Ce^(-2x)) = 0 + C · e^(-2x) · (-2) = -2Ce^(-2x). Now we compute the left side: y' + 2y = -2Ce^(-2x) + 2(2 + Ce^(-2x)) = -2Ce^(-2x) + 4 + 2Ce^(-2x) = 4. This equals the right side, so the family satisfies the equation. Students might forget the chain rule and miss the -2 factor, or make algebraic errors when expanding 2y. When verifying linear differential equations, carefully track all terms through substitution and simplification.

Question 17

For $\frac{dy}{dx}=y^2$ with $y\ne0$ , is $y=-\frac{1}{x}$ a solution on its domain?

Yes, because $y'=\frac{1}{x^2}$ and $y^2=\frac{1}{x^2}$ . (correct answer)
No, because $y'=-\frac{1}{x^2}$ and $y^2=\frac{1}{x^2}$ .
Yes, because $y'=-\frac{1}{x^2}$ and $y^2=-\frac{1}{x^2}$ .
No, because $y'=\frac{1}{x}$ and $y^2=\frac{1}{x^2}$ .
No, because $y^2=\frac{1}{x}$ .

Explanation: We need to verify if y = -1/x satisfies dy/dx = y² on its domain. First, we differentiate: y' = d/dx(-1/x) = d/dx(-x^(-1)) = -(-1)x^(-2) = 1/x². For the right side, we compute y² = (-1/x)² = 1/x². Since both sides equal 1/x², the proposed function is a solution. A common mistake is getting the wrong sign when differentiating -1/x, thinking y' = -1/x². Remember that d/dx(

x^n

) = nx^(n-1), and when n = -1, the derivative includes a sign change.

Question 18

For $\dfrac{dy}{dx}=2y$ , is the function $y=e^{2x}$ a solution on all real $x$ ?

Yes, because $y'=2e^{2x}$ and $2y=2e^{2x}$ . (correct answer)
Yes, because $y'=e^{2x}$ and $2y=2e^{2x}$ .
No, because $y'=2e^{x}$ and $2y=2e^{2x}$ .
No, because $2y=e^{2x}$ but $y'=2e^{2x}$ .
No, because $y'=2x e^{2x}$ and $2y=2e^{2x}$ .

Explanation: This problem requires verifying whether y = e^(2x) satisfies dy/dx = 2y. To verify, we differentiate y = e^(2x) using the chain rule: y' = 2e^(2x). We also calculate 2y = 2(e^(2x)) = 2e^(2x). Since y' = 2e^(2x) equals 2y = 2e^(2x), the equation is satisfied for all real x. A common mistake (choice B) is forgetting the chain rule and writing y' = e^(2x), missing the factor of 2 from the inner derivative. For exponential solutions to differential equations, always remember to apply the chain rule when the exponent contains a coefficient.

Question 19

Does $y=x\cos x$ satisfy $\dfrac{dy}{dx}=\cos x-x\sin x$ for all real $x$ ?

Yes, because $y'=\cos x-x\sin x$ by the product rule. (correct answer)
Yes, because $y'=\cos x+x\sin x$ matches the right side.
No, because $y'=\cos x-x\sin x$ only when $x=0$ .
No, because $y'=x(-\sin x)$ , so $y'=-x\sin x$ .
No, because $y'=\sin x-x\cos x$ by the product rule.

Explanation: We need to verify if y = x cos x satisfies dy/dx = cos x - x sin x. Using the product rule on y = x cos x: y' = (1)(cos x) + x(-sin x) = cos x - x sin x. This exactly matches the right side of the differential equation, so y = x cos x is indeed a solution. A common error (choice D) would be to forget the product rule and only differentiate one part, getting y' = -x sin x. When verifying solutions involving products, always apply the product rule: (uv)' = u'v + uv', taking care to include both terms.

Question 20

For $\dfrac{dy}{dx}=y(1-y)$ , is $y=\dfrac{1}{1+e^{-x}}$ a solution on all real $x$ ?

Yes, because $y'=\dfrac{e^{-x}}{(1+e^{-x})^2}$ and $y(1-y)=\dfrac{e^{-x}}{(1+e^{-x})^2}$ . (correct answer)
Yes, because $y'=\dfrac{-e^{-x}}{1+e^{-x}}$ and $y(1-y)=\dfrac{e^{-x}}{(1+e^{-x})^2}$ .
No, because $y'=\dfrac{e^{-x}}{1+e^{-x}}$ but $y(1-y)=\dfrac{e^{-x}}{(1+e^{-x})^2}$ .
No, because $y(1-y)=\dfrac{1}{1+e^{-x}}\left(1-\dfrac{1}{1+e^{-x}}\right)=\dfrac{1}{(1+e^{-x})^2}$ .
No, because $y'=\dfrac{e^{x}}{(1+e^{-x})^2}$ but $y(1-y)=\dfrac{e^{-x}}{(1+e^{-x})^2}$ .

Explanation: To verify if y = 1/(1 + e^(-x)) satisfies dy/dx = y(1 - y), we need both sides. Using the chain rule and quotient rule: y' = e^(-x)/(1 + e^(-x))². For the right side, we have y = 1/(1 + e^(-x)) and 1 - y = 1 - 1/(1 + e^(-x)) = e^(-x)/(1 + e^(-x)). Therefore, y(1 - y) = [1/(1 + e^(-x))] · [e^(-x)/(1 + e^(-x))] = e^(-x)/(1 + e^(-x))². Since both sides equal e^(-x)/(1 + e^(-x))², this is a solution. The key insight is recognizing that 1 - 1/(1 + e^(-x)) simplifies to e^(-x)/(1 + e^(-x)) by finding a common denominator. For logistic-type solutions, careful algebraic manipulation is essential for verification.