Verifying Solutions for Differential Equations - AP Calculus AB

Does $y=\sqrt{x}$ satisfy the differential equation $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}$ for $x>0$ ?
Answer: Yes, because $y'=\dfrac{1}{2\sqrt{x}}$ and the right side matches. To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For $y = \sqrt{x} = x^{1/2}$ , we find $y' = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$ using the power rule. The differential equation $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ requires that $y'$ equals $\frac{1}{2\sqrt{x}}$ . Substituting our derivative: $y' = \frac{1}{2\sqrt{x}}$ , which exactly matches the right side $\frac{1}{2\sqrt{x}}$ . Since both sides are identical, the function satisfies the equation. Choice A incorrectly states y' = 1/√x, missing the factor of 1/2 from the power rule. Remember that $\frac{d}{dx}[x^{1/2}] = \frac{1}{2} x^{-1/2}$ .
Does $y=\dfrac{1}{x+2}$ satisfy $\dfrac{dy}{dx}=-y^2$ for all $x\ne -2$ ?
Answer: Yes, because $y'=-\dfrac{1}{(x+2)^2}$ and $-y^2=-\dfrac{1}{(x+2)^2}$ . 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+2) = $(x+2)^{-1}$ , we find y' = - $(x+2)^{-2}$ = -1/(x+2)² using the chain rule. The differential equation dy/dx = -y² requires that y' equals -y². We compute -y² = -(1/(x+2))² = -1/(x+2)². Since y' = -1/(x+2)² and -y² = -1/(x+2)², both sides are equal. Choice A has the wrong sign, stating y' = 1/(x+2)² instead of -1/(x+2)². When differentiating negative powers using the chain rule, carefully track the negative sign.
Does $y=\dfrac{1}{x^2}$ satisfy $\dfrac{dy}{dx}=-\dfrac{2y}{x}$ for $x\ne 0$ ?
Answer: Yes, because $y'=-\dfrac{2}{x^3}$ and $-\dfrac{2y}{x}=-\dfrac{2}{x^3}$ . 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 = $x^{-2}$ , we find y' = - $2x^{-3}$ = -2/x³ using the power rule. The differential equation dy/dx = -(2y)/x requires that y' equals -(2y)/x. We compute -(2y)/x = -(2)( $x^{-2}$ )/x = - $2x^{-3}$ = -2/x³. Since y' = -2/x³ and -(2y)/x = -2/x³, both sides are equal. Choice B has the wrong sign, stating y' = 2/x³ instead of -2/x³. When applying the power rule to negative powers, the negative exponent produces a negative coefficient in the derivative.
Does $y=e^{x}+1$ satisfy the differential equation $\dfrac{dy}{dx}=y-1$ for all $x$ ?
Answer: Yes, because $y'=e^{x}$ and $y-1=e^{x}$ . To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = $e^x$ + 1, we find y' = $e^x$ using the fact that the derivative of $e^x$ is $e^x$ and the derivative of constant 1 is 0. The differential equation dy/dx = y - 1 requires that y' equals y - 1. We compute y - 1 = ( $e^x$ + 1) - 1 = $e^x$ . Since y' = $e^x$ and y - 1 = $e^x$ , both sides are equal. Choice A incorrectly states y' = $e^x$ + 1, adding the derivative of the constant term. Remember that the derivative of a constant is always zero.
Does $y=e^{3x}$ satisfy the differential equation $\dfrac{dy}{dx}=3y$ for all $x$ ?
Answer: Yes, because $y'=3e^{3x}$ and $3y=3e^{3x}$ . 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.
Does $y=\dfrac{1}{2}x^2$ satisfy the differential equation $\dfrac{dy}{dx}=\sqrt{2y}$ for $x\ge 0$ ?
Answer: Yes, because $y'=x$ and for $x\ge 0$ , $\sqrt{2y}=\sqrt{x^2}=x$ . 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)x², we find y' = x using the power rule. The differential equation dy/dx = √(2y) requires that y' equals √(2y). We compute √(2y) = √(2·(1/2)x²) = √(x²) = |x|. For x ≥ 0, |x| = x, so y' = x equals √(2y) = x in this domain. Choice A incorrectly states that √(x²) = |x| creates a problem, but the domain restriction x ≥ 0 resolves this. When dealing with square roots of squares, consider domain restrictions and absolute value carefully.
Does $y=e^{2x}$ satisfy the differential equation $\dfrac{dy}{dx}=y^2$ for all $x$ ?
Answer: No, because $y'=2e^{2x}$ but $y^2=e^{4x}$ . To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = $e^{2x}$ , we find y' = $2e^{2x}$ using the chain rule. The differential equation dy/dx = y² requires that y' equals y². We compute y² = ( $e^{2x}$ )² = $e^{4x}$ . Since y' = $2e^{2x}$ but y² = $e^{4x}$ , the two sides are not equal ( $2e^{2x}$ ≠ $e^{4x}$ ). Choice A incorrectly claims they are equal, stating y² = $2e^{2x}$ instead of $e^{4x}$ . When squaring exponential functions, use the rule ( $e^a$ )² = $e^{2a}$ , and verify both sides match exactly.
Does $y=\dfrac{1}{2}x^{-2}$ satisfy the differential equation $\dfrac{dy}{dx}=-\dfrac{2}{x}y$ for $x\ne 0$ ?
Answer: Yes, because $y'=-x^{-3}$ and $-\dfrac{2}{x}y=-x^{-3}$ . To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For $y = \dfrac{1}{2} x^{-2} = \dfrac{1}{2 x^2}$ , we find $y' = \dfrac{1}{2} (-2) x^{-3} = -x^{-3} = -\dfrac{1}{x^3}$ using the power rule. The differential equation $\dfrac{dy}{dx} = -\dfrac{2}{x} y$ requires that $y'$ equals $-\dfrac{2}{x} y$ . We compute $-\dfrac{2}{x} y = -\dfrac{2}{x} \left( \dfrac{1}{2} x^{-2} \right) = -x^{-3} = -\dfrac{1}{x^3}$ . Since $y' = -\dfrac{1}{x^3}$ and $-\dfrac{2}{x} y = -\dfrac{1}{x^3}$ , both sides are equal. Choice C has the wrong sign, stating $y' = x^{-3}$ instead of $-x^{-3}$ . When applying the power rule to negative powers, carefully track the negative coefficient.
Does $y=3e^{2x}$ satisfy the differential equation $\dfrac{dy}{dx}=2y$ for all $x$ ?
Answer: Yes, because $y'=6e^{2x}$ and $2y=6e^{2x}$ . 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( $3e^{2x}$ ) = $6e^{2x}$ . 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.
Does $y=\dfrac{1}{x}$ satisfy $\dfrac{dy}{dx}=-y^2$ for all $x\ne 0$ ?
Answer: Yes, because $y'=-\dfrac{1}{x^2}$ and $-y^2=-\dfrac{1}{x^2}$ . To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For $y = \dfrac{1}{x} = x^{-1}$ , we find $y' = -x^{-2} = -\dfrac{1}{x^2}$ using the power rule. The differential equation $\dfrac{dy}{dx} = -y^2$ requires that $y'$ equals $-y^2$ . We compute $-y^2 = -\left(\dfrac{1}{x}\right)^2 = -\dfrac{1}{x^2}$ . Since $y' = -\dfrac{1}{x^2}$ and $-y^2 = -\dfrac{1}{x^2}$ , both sides are equal. Choice B has the wrong sign for y', stating $y' = \dfrac{1}{x^2}$ instead of $-\dfrac{1}{x^2}$ . To verify solutions involving negative powers, carefully apply the power rule and check signs throughout.
Does $y=\dfrac{x^2}{2}$ satisfy the differential equation $\dfrac{dy}{dx}=x$ for all $x$ ?
Answer: Yes, because $y'=x$ and the right side is $x$ . To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = x²/2, we find y' = (1/2)(2x) = x using the power rule and constant factor rule. The differential equation dy/dx = x requires that y' equals x. Substituting our derivative: y' = x, which exactly matches the right side x. Since both sides are identical, the function satisfies the equation. Choice B incorrectly states y' = 2x, forgetting to account for the coefficient 1/2 in the original function. Always apply the constant factor rule when differentiating: d/dx[cf(x)] = c·f'(x).
Does $y=\dfrac{1}{x}$ satisfy the differential equation $x\dfrac{dy}{dx}+y=0$ for $x\ne 0$ ?
Answer: Yes, because $y'=-\dfrac{1}{x^2}$ so $x y'+y=-\dfrac{1}{x}+\dfrac{1}{x}=0$ . To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For $y = \dfrac{1}{x}$ , we find $y' = -\dfrac{1}{x^2}$ using the power rule. The differential equation $x \dfrac{dy}{dx} + y = 0$ can be rewritten as $x \cdot y' + y = 0$ . We substitute: $x(-\dfrac{1}{x^2}) + (\dfrac{1}{x}) = -\dfrac{1}{x} + \dfrac{1}{x} = 0$ . Since the left side equals 0, the equation is satisfied. Choice B incorrectly states $y' = \dfrac{1}{x^2}$ , missing the negative sign from differentiating $x^{-1}$ . When verifying solutions to equations in this form, substitute both $y$ and $y'$ and simplify to check if the result equals zero.
Does $y=\tan x$ satisfy $\dfrac{dy}{dx}=\sec^2 x$ for all $x$ where defined?
Answer: Yes, because $y'=\sec^2 x$ and the right side is $\sec^2 x$ . To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = tan x, we find y' = sec² x using the standard trigonometric derivative. The differential equation dy/dx = sec² x requires that y' equals sec² x. Substituting our derivative: y' = sec² x, which exactly matches the right side sec² x. Since both sides are identical, the function satisfies the equation. Choice A incorrectly states y' = sec x, missing the square on sec and not applying the correct derivative formula. Remember that d/dx[tan x] = sec² x.
Does $y=x^3$ satisfy the differential equation $\dfrac{dy}{dx}=3x^2$ for all $x$ ?
Answer: Yes, because $y'=3x^2$ and the right side is $3x^2$ . To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = x³, we find y' = 3x² using the power rule. The differential equation dy/dx = 3x² requires that y' equals 3x². Substituting our derivative: y' = 3x², which exactly matches the right side 3x². Since both sides are identical, the function satisfies the equation. Choice A incorrectly states y' = x², missing the coefficient 3 from the power rule. Always apply the power rule completely: d/dx[ $x^n$ ] = $nx^{n-1}$ , and verify both sides match exactly.
Does $y=\sin x+\cos x$ satisfy $\dfrac{dy}{dx}=\cos x-\sin x$ for all $x$ ?
Answer: Yes, because $y'=\cos x-\sin x$ and the right side matches. 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 + cos x, we find y' = cos x - sin x using standard trigonometric derivatives. The differential equation dy/dx = cos x - sin x requires that y' equals cos x - sin x. Substituting our derivative: y' = cos x - sin x, which exactly matches the right side cos x - sin x. Since both sides are identical, the function satisfies the equation. Choice B incorrectly states y' = cos x + sin x, adding instead of subtracting the derivatives. Remember that d/dx[sin x] = cos x and d/dx[cos x] = -sin x.
Does $y=\sec x$ satisfy the differential equation $\dfrac{dy}{dx}=y\tan x$ where defined?
Answer: Yes, because $y'=\sec x\tan x$ and $y\tan x=\sec x\tan x$ . To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = sec x, we find y' = sec x tan x using the standard trigonometric derivative. The differential equation dy/dx = y tan x requires that y' equals y tan x. We compute y tan x = (sec x)(tan x) = sec x tan x. Since y' = sec x tan x and y tan x = sec x tan x, both sides are equal. Choice A incorrectly states y' = sec x, missing the tan x factor from the derivative formula. Remember that d/dx[sec x] = sec x tan x, not just sec x.
Does $y=x+\dfrac{1}{x}$ satisfy $\dfrac{dy}{dx}=1-\dfrac{1}{x^2}$ for $x\ne 0$ ?
Answer: Yes, because $y'=1-\dfrac{1}{x^2}$ and the right side matches. To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For y = x + 1/x, we find y' = 1 + (-1) $x^{-2}$ = 1 - 1/x² using the power rule on each term. The differential equation dy/dx = 1 - 1/x² requires that y' equals 1 - 1/x². Substituting our derivative: y' = 1 - 1/x², which exactly matches the right side 1 - 1/x². Since both sides are identical, the function satisfies the equation. Choice B incorrectly states y' = 1 + 1/x², adding instead of subtracting the second term. Remember that d/dx[ $x^{-1}$ ] = - $x^{-2}$ , which gives the negative sign.
Does $y=\sin x$ satisfy the differential equation $\dfrac{dy}{dx}=\cos x$ for all $x$ ?
Answer: Yes, because $y'=\cos x$ and the right side is $\cos x$ . 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.
Does $y=\cos x$ satisfy $\dfrac{dy}{dx}=-\sin x$ for all $x$ ?
Answer: Yes, because $y'=-\sin x$ and the right side is $-\sin x$ . 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.
Does $y=\dfrac{1}{2}x^2+\dfrac{1}{2}$ satisfy $\dfrac{dy}{dx}=x$ for all $x$ ?
Answer: Yes, because $y'=x$ and the right side is $x$ . 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)x² + (1/2), we find y' = (1/2)(2x) + 0 = x using the power rule. The differential equation dy/dx = x requires that y' equals x. Substituting our derivative: y' = x, which exactly matches the right side x. Since both sides are identical, the function satisfies the equation. Choice A incorrectly states y' = x + 3, adding a constant that should not appear since the derivative of any constant is zero. Remember that constants disappear when differentiating.

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