Verifying Solutions for Differential Equations - AP Calculus BC

Does $y=\dfrac{1}{x^2}$ satisfy $\dfrac{dy}{dx}=-\dfrac{2}{x}y$ on $x\ne 0$ ?
Answer: Yes, because $y'=-\dfrac{2}{x^3}$ and $-\dfrac{2}{x}y=-\dfrac{2}{x}\cdot\dfrac{1}{x^2}=-\dfrac{2}{x^3}$ . 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.
For $\dfrac{dy}{dx}=\dfrac{2y}{x}$ , is $y=x^2$ a solution on $x \neq 0$ ?
Answer: Yes, because $y'=2x$ and $\dfrac{2y}{x}=\dfrac{2x^2}{x}=2x$ . Solution verification requires careful differentiation and algebraic substitution with rational expressions. For $y = x^2$ on $x \neq 0$ , we have $y' = 2x$ . The right side $\dfrac{2y}{x}$ becomes $2(x^2)/x = 2x$ . Since $y' = 2x$ and $\dfrac{2y}{x} = 2x$ , both sides are equal, confirming the solution. Choice B incorrectly substitutes $y = x$ instead of $y = x^2$ . Choice E incorrectly computes $\dfrac{2y}{x}$ as $2x^2$ instead of $2x$ . The systematic verification approach is to differentiate the proposed solution, substitute it into both sides of the equation, and simplify to check for equality.
For the differential equation $\dfrac{dy}{dx}=3x^2-4$ , is $y=x^3-4x+7$ a solution?
Answer: Yes, because $y'=3x^2-4$ matches the differential equation. This problem asks us to verify whether y = x³ - 4x + 7 is a solution to the differential equation dy/dx = 3x² - 4. To verify a solution, we differentiate the proposed function and check if it matches the given differential equation. Taking the derivative of y = x³ - 4x + 7, we get y' = 3x² - 4, which exactly matches the right side of the differential equation. A common error would be to include the constant 7 in the derivative (as in choice D), but constants disappear when differentiating. To verify any solution to a differential equation, always differentiate the proposed solution and check for an exact match with the given equation.
For $\dfrac{dy}{dx}=4x^3$ , is $y=x^4+7$ a solution for all $x$ ?
Answer: Yes, because $y'=4x^3$ equals the right-hand side. 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.
A cooling object follows $\dfrac{dT}{dt}=-2(T-10)$ . Is $T(t)=10+6e^{-2t}$ a solution?
Answer: Yes, because $T'(t)=-12e^{-2t}$ and $-2(T-10)=-12e^{-2t}$ . 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.
For $\dfrac{dy}{dx}=\dfrac{y}{x}+x$ , is $y=x^2$ a solution on $x\ne 0$ ?
Answer: Yes, because $y'=2x$ and $\dfrac{y}{x}+x=\dfrac{x^2}{x}+x=2x$ . 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.
Does $y=\dfrac{1}{2}x^2$ satisfy $\dfrac{dy}{dx}=x$ for all real $x$ ?
Answer: Yes, because $y'=x$ . Solution verification for polynomial functions involves straightforward differentiation and comparison. For y = (1/2)x², we compute y' = (1/2) · 2x = x. The differential equation dy/dx = x becomes x = x, which is clearly an identity for all real x. Choice A incorrectly states the derivative as x/2, while choice C incorrectly doubles it to 2x. Choice D incorrectly claims the derivative is x². The verification strategy is to apply the power rule correctly, remembering that the derivative of $ax^n$ is $nax^{n-1}$ , and check that both sides of the equation are identical.
For $\dfrac{dy}{dx}=\dfrac{2}{x}y$ , is $y=x^2$ a solution on $x \ne 0$ ?
Answer: Yes, because $y'=2x$ and $\dfrac{2}{x}y=2x$ . Solution verification involves computing derivatives and careful substitution into rational expressions. For $y = x^2$ where $x \neq 0$ , we have $y' = 2x$ . The right side $\dfrac{2}{x} y$ becomes $\dfrac{2}{x} \cdot x^2 = 2x$ . Since both $y'$ and $\dfrac{2}{x} y$ equal $2x$ , the solution is verified. Choice A incorrectly suggests they differ, possibly making an algebraic error in the simplification. The systematic verification approach is to differentiate the proposed solution, substitute it into rational expressions, simplify algebraically, and verify that both sides are identical.
Does $y=\sqrt{1-x^2}$ satisfy $\dfrac{dy}{dx}=-\dfrac{x}{y}$ on $-1<x<1$ ?
Answer: Yes, because $y'=-\dfrac{x}{\sqrt{1-x^2}}$ and $-\dfrac{x}{y}=-\dfrac{x}{\sqrt{1-x^2}}$ . Solution verification involves differentiating radical functions and checking relationships with rational expressions. For y = √(1-x²) where -1 < x < 1, we compute y' = (1/2) $(1-x²)^{-1/2}$ · (-2x) = -x/√(1-x²) using the chain rule. The right side -x/y becomes -x/√(1-x²). Since both y' and -x/y equal -x/√(1-x²), the solution is verified. Choice B has the wrong sign for the derivative, while choice C makes an error in the substitution. The systematic approach is to apply the chain rule correctly to expressions involving $(1-x²)^{1/2}$ and verify reciprocal relationships through substitution.
For $\dfrac{dy}{dx}=y\tan x$ , is $y=\sec x$ a solution where defined?
Answer: Yes, because $y'=\sec x\tan x$ and $y\tan x=\sec x\tan x$ . Solution verification requires proper differentiation and trigonometric identity application. For y = sec x, we compute y' = sec x tan x. The equation dy/dx = y tan x becomes sec x tan x = (sec x)(tan x) = sec x tan x, which is an identity where sec x is defined. Choice B incorrectly states the derivative as sec² x, which is the derivative of tan x, not sec x. Choice E incorrectly expands y tan x, leading to confusion. The key verification approach is to use correct differentiation rules for trigonometric functions and recognize when both sides simplify to the same expression.
For $\dfrac{dy}{dx}=2xy$ , is $y=e^{x^2}$ a solution for all $x$ ?
Answer: Yes, because $y'=2xe^{x^2}$ and $2xy=2x e^{x^2}$ . Solution verification requires applying the chain rule correctly and matching both sides of the equation. For y = $e^{x²}$ , we compute y' = $e^{x²}$ · 2x = $2xe^{x²}$ by the chain rule. The right side 2xy becomes 2x · $e^{x²}$ = $2xe^{x²}$ . Since both sides equal $2xe^{x²}$ , the solution is verified for all x. Choice C incorrectly computes the derivative as $e^{x²}$ , omitting the chain rule factor of 2x. Choice E makes an error in the substitution of y. Always apply the chain rule when differentiating composite functions and ensure careful substitution of the proposed solution.
Does $y=\ln x$ satisfy the differential equation $x\frac{dy}{dx}=1$ for all $x>0$ ?
Answer: Yes, because $y'=\frac{1}{x}$ so $xy'=1$ . 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. For y = ln x, the derivative is y' = 1/x. Multiply by x: x y' = x * (1/x) = 1. Since this matches the right side for x > 0, it satisfies the equation. A tempting distractor might claim x y' = x, which fails because it incorrectly assumes y' = 1 instead of 1/x. Always differentiate the proposed solution and plug into both sides of the equation to confirm equality as a general strategy for verification.
Does $y=\cos(2x)$ satisfy $\dfrac{dy}{dx}=-2\sin(2x)$ for all $x$ ?
Answer: Yes, because $y'=-2\sin(2x)$ by the chain rule. Verifying trigonometric solutions requires applying the chain rule to composite trigonometric functions. For y = cos(2x), we compute y' = -sin(2x) · 2 = -2sin(2x) by the chain rule. The differential equation dy/dx = -2sin(2x) becomes -2sin(2x) = -2sin(2x), which is clearly an identity for all x. Choice A incorrectly omits the factor of 2 from the chain rule, while choice C has the wrong sign. The key verification approach is to apply the chain rule correctly when differentiating composite trigonometric functions, remembering that d/dx[f(g(x))] = f'(g(x)) · g'(x).
For $\dfrac{dy}{dx}=x^2y$ , is $y=e^{x^3/3}$ a solution for all $x$ ?
Answer: Yes, because $y'=x^2 e^{x^3/3}=x^2y$ . Solution verification involves applying the chain rule to exponential functions and checking the resulting equation. For y = $e^{x³/3}$ , we compute y' = $e^{x³/3}$ · (d/dx)(x³/3) = $e^{x³/3}$ · x² = x² $e^{x³/3}$ by the chain rule. The right side x²y becomes x² · $e^{x³/3}$ = x² $e^{x³/3}$ . Since both sides equal x² $e^{x³/3}$ , the solution is verified for all x. Choice B incorrectly applies the chain rule, missing a factor. The systematic approach is to carefully apply the chain rule when differentiating composite exponential functions and verify that substitution yields identical expressions on both sides.
Does $y=\dfrac{1}{2}e^{2x}$ satisfy $\dfrac{dy}{dx}=2y$ for all $x$ ?
Answer: Yes, because $y'=e^{2x}$ and $2y=e^{2x}$ . Solution verification involves computing derivatives correctly and checking coefficient relationships. For y = (1/2) $e^{2x}$ , we compute y' = (1/2) · $e^{2x}$ · 2 = $e^{2x}$ by the chain rule. The right side 2y becomes 2 · (1/2) $e^{2x}$ = $e^{2x}$ . Since both y' and 2y equal $e^{2x}$ , the solution is verified for all x. Choice B incorrectly states the derivative as $2e^{2x}$ , missing that the coefficient 1/2 affects the final result. The key verification approach is to apply the chain rule correctly and track how constants in the solution affect the derivative computation.
Does $y=\ln x$ satisfy $x\dfrac{dy}{dx}=1$ for all $x>0$ ?
Answer: Yes, because $y'=\dfrac{1}{x}$ so $x\dfrac{dy}{dx}=x\cdot\dfrac{1}{x}=1$ . We must verify if y = ln x satisfies x(dy/dx) = 1 for x > 0. First, we find the derivative: y' = 1/x. Then we compute x(dy/dx) = x · (1/x) = 1. Since this equals 1 as required, y = ln x is indeed a solution for all x > 0. A common mistake (choice C) would be to incorrectly differentiate ln x as 1/(ln x), confusing the derivative with a reciprocal. Remember that the derivative of ln x is always 1/x, and when verifying solutions, substitute the derivative before simplifying the entire expression.
Is $y=\sin x$ a solution to the differential equation $\frac{dy}{dx}=\sqrt{1-y^2}$ on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ ?
Answer: Yes, because $y'=\cos x$ and $\sqrt{1-\sin^2 x}=\cos x$ on the interval. 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. To verify, compute the derivative of $y = \sin x$ , yielding $y' = \cos x$ . Then, substitute into the right side: $\sqrt{1 - y^2} = \sqrt{1 - \sin^2 x} = \sqrt{\cos^2 x} = |\cos x|$ , and on $[-\frac{\pi}{2}, \frac{\pi}{2}]$ , $\cos x \geq 0$ , so it equals $\cos x$ . Since $y'$ matches $\sqrt{1 - y^2}$ on the interval, y is a solution. A tempting distractor might claim $\sqrt{1 - \sin^2 x} = |\sin x|$ , which fails because it confuses the identity with sin x instead of cos x. Always differentiate the proposed solution and plug into both sides of the equation to confirm equality as a general strategy for verification.
If $\dfrac{dy}{dx}=\dfrac{2x}{1+y^2}$ , does $y=\tan(x^2)$ satisfy the differential equation where defined?
Answer: No; $y'=2x\sec^2(x^2)$ and $\dfrac{2x}{1+y^2}=\dfrac{2x}{\sec^2(x^2)}=2x\cos^2(x^2)$ . This problem asks us to verify if y = tan(x²) satisfies dy/dx = 2x/(1 + y²) where defined. To verify, we compute the derivative and check if it equals the right side after substitution. For y = tan(x²), using the chain rule: y' = sec²(x²) · 2x = 2x sec²(x²). For the right side, we need 1 + y² = 1 + tan²(x²) = sec²(x²) (using the trigonometric identity). Therefore, 2x/(1 + y²) = 2x/sec²(x²) = 2x cos²(x²). Since y' = 2x sec²(x²) ≠ 2x cos²(x²), the equation is not satisfied. A common error is to incorrectly apply the identity 1 + tan²θ = sec²θ or to confuse sec²θ with cos²θ. When verifying solutions involving trigonometric functions, carefully apply identities and remember that sec²θ = 1/cos²θ.
Does $y=x^2+1$ satisfy $\frac{dy}{dx}=2\sqrt{y-1}$ for all $x\ge 0$ ?
Answer: Yes, because $y'=2x$ and $2\sqrt{x^2}=2x$ for $x\ge0$ . 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.
Is $y=\frac{1}{x}$ a solution to $\frac{dy}{dx}=-y^2$ on the domain $x \ne 0$ ?
Answer: Yes, because $y'=-\frac{1}{x^2}$ and $-y^2=-\frac{1}{x^2}$ . 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. Compute the derivative of y = 1/x, which is $y' = -\frac{1}{x^2}$ . Substitute into the right side: $-y^2 = -(1/x)^2 = -\frac{1}{x^2}$ . Since $y'$ equals $-y^2$ for $x \ne 0$ , it is a solution. A tempting distractor might suggest $y' = \frac{1}{x^2}$ but $-y^2 = -\frac{1}{x^2}$ , which fails because the sign of the derivative is positive instead of negative. Always differentiate the proposed solution and plug into both sides of the equation to confirm equality as a general strategy for verification.

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