Verifying Solutions for Differential Equations
Help Questions
AP Calculus AB › Verifying Solutions for Differential Equations
Does $y=\dfrac{1}{x^3}$ satisfy $\dfrac{dy}{dx}=-\dfrac{3}{x}y$ for $x\ne 0$?
Yes, because $-\dfrac{3}{x}y=-\dfrac{3}{x^3}$ equals $y'$.
No, because $y'=-\dfrac{9}{x^4}$ but $-\dfrac{3}{x}y=-\dfrac{3}{x^4}$.
No, because $y'=-\dfrac{1}{x^3}$ but $-\dfrac{3}{x}y=-\dfrac{3}{x^4}$.
Yes, because $y'=-\dfrac{3}{x^4}$ and $-\dfrac{3}{x}y=-\dfrac{3}{x^4}$.
No, because $y'=\dfrac{3}{x^4}$ but $-\dfrac{3}{x}y=-\dfrac{3}{x^4}$.
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 = $-(3/x)(x^{-3}$) = $-3x^{-4}$ = -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.
Does $y=\dfrac{1}{x}$ satisfy $\dfrac{dy}{dx}=-y^2$ for all $x\ne 0$?
Yes, because $y'=-\dfrac{1}{x^2}$ and $-y^2=-\dfrac{1}{x}$.
No, because $y'=-\dfrac{2}{x^3}$ but $-y^2=-\dfrac{1}{x^2}$.
No, because $y'=\dfrac{1}{x^2}$ but $-y^2=-\dfrac{1}{x^2}$.
Yes, because $y'=-\dfrac{1}{x^2}$ and $-y^2=-\dfrac{1}{x^2}$.
No, because $y'=-\dfrac{1}{x}$ but $-y^2=-\dfrac{1}{x^2}$.
Explanation
To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For $y = \dfrac{1}{x} = x^{-1}$, 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 $y=\dfrac{1}{x}$ satisfy the differential equation $\dfrac{dy}{dx}=\dfrac{y}{x}$ for $x\ne 0$?
Yes, because $y'=\dfrac{1}{x^2}$ equals $\dfrac{y}{x}$.
No, because $y'=\dfrac{1}{x^2}$ and $\dfrac{y}{x}=\dfrac{1}{x}$.
Yes, because $y'=-\dfrac{1}{x^2}$ and $\dfrac{y}{x}=\dfrac{1}{x^2}$.
No, because $\dfrac{y}{x}=-\dfrac{1}{x^2}$ but $y'=\dfrac{1}{x^2}$.
No, because $y'=-\dfrac{1}{x^2}$ but $\dfrac{y}{x}=\dfrac{1}{x^2}$.
Explanation
To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For $y = \dfrac{1}{x}$, we find $y' = -\dfrac{1}{x^2}$ using the power rule. The differential equation $\dfrac{dy}{dx} = \dfrac{y}{x}$ requires that $y'$ equals $\dfrac{y}{x}$. We compute $\dfrac{y}{x} = \dfrac{\dfrac{1}{x}}{x} = \dfrac{1}{x^2}$. Since $y' = -\dfrac{1}{x^2}$ but $\dfrac{y}{x} = \dfrac{1}{x^2}$, 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.
Does $y=3e^{2x}$ satisfy the differential equation $\dfrac{dy}{dx}=2y$ for all $x$?
No, because $y'=6e^{x}$ but $2y=6e^{2x}$.
Yes, because $y'=6e^{2x}$ and $2y=6e^{2x}$.
No, because $y'=2e^{2x}$ but $2y=6e^{2x}$.
Yes, because $y'=3e^{2x}$ equals $2y$.
No, because $y'=6e^{2x}$ but $2y=3e^{2x}$.
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 = $2(3e^{2x}$) = $6e^{2x}$. 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 $y=\dfrac{1}{x-1}$ satisfy $\dfrac{dy}{dx}=-y^2$ for all $x\ne 1$?
No, because $y'=-\dfrac{1}{x-1}$ but $-y^2=-\dfrac{1}{(x-1)^2}$.
No, because $y'=\dfrac{1}{(x-1)^2}$ but $-y^2=-\dfrac{1}{(x-1)^2}$.
Yes, because $y'=-\dfrac{1}{(x-1)^2}$ and $-y^2=-\dfrac{1}{(x-1)^2}$.
Yes, because $-y^2=-\dfrac{1}{x-1}$ equals $y'$.
No, because $y'=-\dfrac{2}{(x-1)^3}$ but $-y^2=-\dfrac{1}{(x-1)^2}$.
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.
Does $y=\dfrac{1}{x^2+1}$ satisfy $\dfrac{dy}{dx}=-2xy^2$ for all real $x$?
Yes, because $y'=-\dfrac{2x}{x^2+1}$ equals $-2xy^2$.
No, because $y'=-\dfrac{2}{x^2+1}$ but $-2xy^2=-\dfrac{2x}{(x^2+1)^2}$.
Yes, because $y'=-\dfrac{2x}{(x^2+1)^2}$ and $-2xy^2=-\dfrac{2x}{(x^2+1)^2}$.
No, because $-2xy^2=-\dfrac{2x}{x^2+1}$, not $y'$.
No, because $y'=\dfrac{2x}{(x^2+1)^2}$ but $-2xy^2=-\dfrac{2x}{(x^2+1)^2}$.
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/(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.
Does $y=2x+5$ satisfy the differential equation $\dfrac{dy}{dx}=2$ for all $x$?
No, because $y'=0$ but the right side is $2$.
Yes, because $y'=2$ and the right side is $2$.
No, because $y'=2x$ but the right side is $2$.
No, because $y'=5$ but the right side is $2$.
Yes, because $y'=2x+5$ equals $2$.
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 = 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.
Does $y=x^2+1$ satisfy $\dfrac{dy}{dx}=2x$ at every $x$?
Yes, because $y'=x^2+1$ equals $2x$.
Yes, because $y'=2x$ and the right side is $2x$.
No, because $y'=2x+1$ but the right side is $2x$.
No, because $y'=2$ but the right side is $2x$.
No, because $y'=x^2$ but the right side is $2x$.
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 = 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.
Does $y=\dfrac{1}{3}x^3$ satisfy the differential equation $\dfrac{dy}{dx}=x^2$ for all $x$?
No, because $y'=3x^2$ but the right side is $x^2$.
No, because $y'=\dfrac{1}{3}x^3$ but the right side is $x^2$.
No, because $y'=x$ but the right side is $x^2$.
Yes, because $y'=\dfrac{1}{3}$ equals $x^2$.
Yes, because $y'=x^2$ and the right side is $x^2$.
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 = (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).
Does $y=e^{3x}$ satisfy the differential equation $\dfrac{dy}{dx}=3y$ for all $x$?
No, because $y'=e^{3x}$ but $3y=3e^{3x}$.
Yes, because $y'=3e^{3x}$ and $3y=3e^{3x}$.
Yes, because $y'=e^{3x}$ equals $3y$.
No, because $y'=3e^{x}$ but $3y=3e^{3x}$.
No, because $y'=9e^{3x}$ but $3y=3e^{3x}$.
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.