Verifying Solutions for Differential Equations
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AP Calculus BC › Verifying Solutions for Differential Equations
Does $y=x^2+1$ satisfy $\frac{dy}{dx}=2\sqrt{y-1}$ for all $x\ge 0$?
No, because $y'=2x$ but $2\sqrt{y-1}=\sqrt{2x}$.
Yes, because $y'=2x$ and $2\sqrt{x^2}=2x$ for $x\ge0$.
Yes, because $y'=2x+1$ and $2\sqrt{y-1}=2x+1$.
No, because $y'=x$ but $2\sqrt{y-1}=2x$.
No, because $y'=2x$ but $2\sqrt{y-1}=2|x|$.
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.
Does $y=\tan x$ satisfy the differential equation $\frac{dy}{dx}=1+y^2$ on its domain?
No, because $y'=\sec^2 x$ but $1+y^2=1+\sec^2 x$.
No, because $y'=\tan^2 x$ but $1+y^2=\sec^2 x$.
No, because $y'=\sec x$ but $1+y^2=\sec^2 x$.
Yes, because $y'=\sec^2 x$ and $1+y^2=\csc^2 x$.
Yes, because $y'=\sec^2 x$ and $1+\tan^2 x=\sec^2 x$.
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. The derivative of y = tan x is y' = sec² x. Substitute into the right side: 1 + y² = 1 + tan² x = sec² x by the Pythagorean identity. Since y' equals 1 + y² on the domain where tan x is defined, it satisfies the equation. A tempting distractor might claim y' = sec² x but 1 + y² = 1 + sec² x, which fails by adding sec² x instead of using the identity for tan² x. Always differentiate the proposed solution and plug into both sides of the equation to confirm equality as a general strategy for verification.
Is $y=x^3$ a solution to $\frac{dy}{dx}=\frac{3y}{x}$ for all $x \ne 0$?
No, because $y'=x^2$ but $\frac{3y}{x}=3x^2$.
Yes, because $y'=3x^2$ and $\frac{3y}{x}=\frac{3x^3}{x}=3x^2$.
No, because $y'=3x^2$ but $\frac{3y}{x}=3x^4$.
No, because $y'=3x^3$ but $\frac{3y}{x}=3x^2$.
Yes, because $y'=3x^2$ and $\frac{3y}{x}=\frac{3x^3}{x^2}=3x$.
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. Differentiate $y = x^3$ to obtain $y' = 3x^2$. Substitute into the right side: $3y/x = 3(x^3)/x = 3x^2$ for $x \ne 0$. Since $y'$ matches $3y/x$, it is a solution. A tempting distractor might suggest $3y/x = 3x^4$, which fails by incorrectly computing the power when dividing $x^3$ by x. Always differentiate the proposed solution and plug into both sides of the equation to confirm equality as a general strategy for verification.
Does $y=\ln x$ satisfy the differential equation $x\frac{dy}{dx}=1$ for all $x>0$?
Yes, because $y'=\frac{1}{x}$ so $xy'=1$.
No, because $y'=\frac{1}{x^2}$ so $xy'=\frac{1}{x}$.
No, because $y'=\ln x$ so $xy'=x\ln x$.
Yes, because $y'=1$ so $xy'=x$.
No, because $y'=\frac{1}{x}$ so $xy'=x$.
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. 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 $y=e^{2x}+3$ satisfy the differential equation $\frac{dy}{dx}=2y-6$ for all real $x$?
Yes, because $y'=2e^{2x}$ and $2y-6=2e^{2x}$.
Yes, because $y'=2e^{2x}+3$ and $2y-6=2e^{2x}+3$.
No, because $y'=2e^{x}$ but $2y-6=2e^{2x}$.
No, because $y'=e^{2x}$ but $2y-6=2e^{2x}$.
No, because $y'=2e^{2x}$ but $2y-6=2e^{2x}+6$.
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. To verify, first compute the derivative of y = $e^{2x}$ + 3, which is y' = $2e^{2x}$. Next, substitute into the right side of the equation: 2y - 6 = $2(e^{2x}$ + 3) - 6 = $2e^{2x}$ + 6 - 6 = $2e^{2x}$. Since y' equals 2y - 6 for all real x, y is indeed a solution. A tempting distractor might suggest y' = $2e^{2x}$ but 2y - 6 = $2e^{2x}$ + 6, which fails because it incorrectly adds rather than cancels the constant terms. 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 $x \ne 0$?
Yes, because $y'=-\frac{1}{x^2}$ and $-y^2=-\frac{1}{x^2}$.
No, because $y'=-\frac{2}{x^3}$ but $-y^2=-\frac{1}{x^2}$.
No, because $y'=-\frac{1}{x}$ but $-y^2=-\frac{1}{x^2}$.
No, because $y'=\frac{1}{x^2}$ but $-y^2=-\frac{1}{x^2}$.
Yes, because $y'=-\frac{1}{x^2}$ and $-y^2=-\frac{1}{x}$.
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. 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.
Does $y=\dfrac{1}{x+2}$ satisfy $\dfrac{dy}{dx}=-y^2$ on $x\ne -2$?
No, because $-y^2=-\dfrac{1}{x+2}$.
No, because $y^2=\dfrac{1}{x+2}$.
Yes, because $y'=-\dfrac{1}{(x+2)^2}$ and $-y^2=-\dfrac{1}{(x+2)^2}$.
Yes, because $y'=-\dfrac{1}{x+2}$.
No, because $y'=\dfrac{1}{(x+2)^2}$.
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.
For $\dfrac{dy}{dx}=y\cos x$, is $y=e^{\sin x}$ a solution for all $x$?
No, because $y'=\cos x$.
No, because $y'=e^{\sin x}\sin x$.
Yes, because $y'=e^{\sin x}\cos x=y\cos x$.
No, because $y\cos x=e^{\cos x}$.
Yes, because $y'=e^{x}\cos x$.
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.
Does $y=\dfrac{1}{x}$ satisfy $\dfrac{dy}{dx}=-y^2$ for $x\ne 0$?
No, because $-y^2=-\dfrac{1}{x}$.
No, because $y^2=\dfrac{1}{x}$ so $-y^2=-\dfrac{1}{x}$.
Yes, because $y'=-\dfrac{1}{x^2}$ and $-y^2=-\dfrac{1}{x^2}$.
No, because $y'=\dfrac{1}{x^2}$ but $-y^2=-\dfrac{1}{x^2}$.
Yes, because $y'=-\dfrac{1}{x}$ and $-y^2=-\dfrac{1}{x}$.
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.
For motion, velocity satisfies $\dfrac{dv}{dt}=4t$. Is $v(t)=2t^2+3$ a solution?
No, because $v'(t)=4t+3$.
Yes, because $\dfrac{dv}{dt}=8t$ and $v'(t)=8t$.
Yes, because $v'(t)=4t$ equals $\dfrac{dv}{dt}$.
No, because $4t=2t^2+3$ is not true for all $t$.
No, because $v'(t)=2t$.
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 $4t$ to $v(t) = 2t^2 + 3$ 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.