Verifying Solutions for Differential Equations
Help Questions
AP Calculus AB › Verifying Solutions for Differential Equations
Does $y=\dfrac{1}{2}e^{4x}$ satisfy the differential equation $\dfrac{dy}{dx}=4y$ for all $x$?
No, because $4y=\dfrac{1}{2}e^{4x}$, not $y'$.
Yes, because $y'=4e^{4x}$ equals $4y$.
No, because $y'=\dfrac{1}{2}e^{4x}$ but $4y=2e^{4x}$.
Yes, because $y'=2e^{4x}$ and $4y=2e^{4x}$.
No, because $y'=2e^{x}$ but $4y=2e^{4x}$.
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' = $(1/2)(4e^{4x}$) = $2e^{4x}$ 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.
Does $y=\cos x$ satisfy $\dfrac{dy}{dx}=-\sin x$ for all $x$?
No, because $y'=-\cos x$ but the right side is $-\sin x$.
No, because $y'=-\sin^2 x$ but the right side is $-\sin x$.
No, because $y'=\sin x$ but the right side is $-\sin x$.
Yes, because $y'=\cos x$ equals $-\sin x$.
Yes, because $y'=-\sin x$ and the right side is $-\sin x$.
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.
Does $y=e^{x}+e^{-x}$ satisfy the differential equation $\dfrac{dy}{dx}=e^{x}-e^{-x}$ for all $x$?
No, because $y'=-e^{x}+e^{-x}$ but the right side is $e^{x}-e^{-x}$.
No, because $y'=e^{x}+e^{-x}$ but the right side is $e^{x}-e^{-x}$.
Yes, because $y'=e^{x}-e^{-x}$ and the right side matches.
No, because $y'= -e^{-x}$ but the right side is $e^{x}-e^{-x}$.
Yes, because $y'=e^{x}$ equals $e^{x}-e^{-x}$.
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 $-e^{-x}$.
Does $y=x^2$ satisfy the differential equation $\dfrac{dy}{dx}=\dfrac{2y}{x}$ for $x\ne 0$?
No, because $y'=2x$ but $\dfrac{2y}{x}=\dfrac{2x^2}{x}=x$.
No, because $\dfrac{2y}{x}=\dfrac{2}{x^2}$, not $2x$.
Yes, because $y'=2x$ and $\dfrac{2y}{x}=\dfrac{2x^2}{x}=2x$.
No, because $y'=x^2$ but $\dfrac{2y}{x}=2x$.
Yes, because $y'=2$ and $\dfrac{2y}{x}=2x$.
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, 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.
Does $y=\dfrac{1}{x^2+1}$ satisfy $\dfrac{dy}{dx}=-2xy^2$ for all real $x$?
No, because $y'=\dfrac{2x}{(x^2+1)^2}$ but $-2xy^2=-\dfrac{2x}{(x^2+1)^2}$.
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 $y'$.
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 $y=\sqrt{x}$ satisfy the differential equation $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}$ for $x>0$?
No, because $y'=\dfrac{1}{2x}$ but the right side is $\dfrac{1}{2\sqrt{x}}$.
No, because $y'=\dfrac{1}{\sqrt{x}}$ but the right side is $\dfrac{1}{2\sqrt{x}}$.
Yes, because $y'=\dfrac{1}{2\sqrt{x}}$ and the right side matches.
Yes, because $y'=\sqrt{x}$ equals $\dfrac{1}{2\sqrt{x}}$.
No, because $y'=2\sqrt{x}$ but the right side is $\dfrac{1}{2\sqrt{x}}$.
Explanation
To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For $y = \sqrt{x} = x^{1/2}$, 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 $y=x^4$ satisfy the differential equation $\dfrac{dy}{dx}=4x^3$ for all $x$?
No, because $y'=12x^2$ but the right side is $4x^3$.
Yes, because $y'=4x^3$ and the right side is $4x^3$.
Yes, because $y'=x^4$ equals $4x^3$.
No, because $y'=4x^4$ but the right side is $4x^3$.
No, because $y'=x^3$ but the right side is $4x^3$.
Explanation
To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For $y = x^4$, we find $y' = 4x^3$ using the power rule. The differential equation $\dfrac{dy}{dx} = 4x^3$ requires that $y'$ equals $4x^3$. Substituting our derivative: $y' = 4x^3$, which exactly matches the right side $4x^3$. Since both sides are identical, the function satisfies the equation. Choice A incorrectly states $y' = x^3$, missing the coefficient 4 from the power rule. When applying the power rule $\dfrac{d}{dx}[x^n] = n x^{n-1}$, don't forget to bring down the exponent as a coefficient.
Does $y=xe^{x}$ satisfy the differential equation $\dfrac{dy}{dx}=y+e^{x}$ for all $x$?
No, because $y'=x e^{x-1}$ but $y+e^{x}=xe^{x}+e^{x}$.
No, because $y'=xe^{x}$ but $y+e^{x}=xe^{x}+e^{x}$.
No, because $y'=e^{x}$ but $y+e^{x}=xe^{x}+e^{x}$.
Yes, because $y'=e^{x}(x-1)$ equals $y+e^{x}$.
Yes, because $y'=xe^{x}+e^{x}$ and $y+e^{x}=xe^{x}+e^{x}$.
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 = $xe^x$, we find y' = $e^x$ + $xe^x$ = $e^x$(1 + x) using the product rule. The differential equation dy/dx = y + $e^x$ requires that y' equals y + $e^x$. We compute y + $e^x$ = $xe^x$ + $e^x$ = $e^x$(x + 1). Since y' = $e^x$(1 + x) and y + $e^x$ = $e^x$(x + 1), both sides are equal. Choice C incorrectly states y' = $xe^x$, forgetting to apply the product rule completely. When differentiating products, use the product rule: d/dx[uv] = u'v + uv'.
Does $y=\dfrac{1}{x}$ satisfy the differential equation $\dfrac{dy}{dx}=\dfrac{y}{x}$ for $x\ne 0$?
Yes, because $y'=-\dfrac{1}{x^2}$ and $\dfrac{y}{x}=\dfrac{1}{x^2}$.
No, because $y'=\dfrac{1}{x^2}$ and $\dfrac{y}{x}=\dfrac{1}{x}$.
Yes, because $y'=\dfrac{1}{x^2}$ equals $\dfrac{y}{x}$.
No, because $y'=-\dfrac{1}{x^2}$ but $\dfrac{y}{x}=\dfrac{1}{x^2}$.
No, because $\dfrac{y}{x}=-\dfrac{1}{x^2}$ but $y'=\dfrac{1}{x^2}$.
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 $y=e^{x^2}$ satisfy the differential equation $\dfrac{dy}{dx}=2xy$ for all $x$?
No, because $y'=e^{x^2}$ but $2xy=2xe^{x^2}$.
Yes, because $y'=2x$ equals $2xy$.
Yes, because $y'=2xe^{x^2}$ and $2xy=2xe^{x^2}$.
No, because $y'=2x^2e^{x^2}$ but $2xy=2xe^{x^2}$.
No, because $2xy=2x^2e^{x^2}$, not $y'$.
Explanation
To verify if a function satisfies a differential equation, we substitute the function and its derivative into the equation to check equality. For $y = e^{x^2}$, we find $y' = e^{x^2} \cdot 2x = 2x e^{x^2}$ using the chain rule. The differential equation $\frac{dy}{dx} = 2xy$ requires that $y'$ equals $2xy$. We compute $2xy = 2x \cdot e^{x^2} = 2x e^{x^2}$. Since $y' = 2x e^{x^2}$ and $2xy = 2x e^{x^2}$, both sides are equal. Choice B incorrectly states $y' = e^{x^2}$, missing the factor 2x from the chain rule. When differentiating composite exponential functions, multiply by the derivative of the exponent.