Verifying Solutions for Differential Equations
Help Questions
AP Calculus BC › Verifying Solutions for Differential Equations
Does $y=x^2+1$ satisfy $\frac{dy}{dx}=2\sqrt{y-1}$ for all $x\ge 0$?
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 $2\sqrt{y-1}=\sqrt{2x}$.
Yes, because $y'=2x$ and $2\sqrt{x^2}=2x$ for $x\ge0$.
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 $y=\dfrac{1}{x+2}$ satisfy $\dfrac{dy}{dx}=-y^2$ on $x\ne -2$?
No, because $-y^2=-\dfrac{1}{x+2}$.
No, because $y'=\dfrac{1}{(x+2)^2}$.
Yes, because $y'=-\dfrac{1}{x+2}$.
No, because $y^2=\dfrac{1}{x+2}$.
Yes, because $y'=-\dfrac{1}{(x+2)^2}$ and $-y^2=-\dfrac{1}{(x+2)^2}$.
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 $x$?
Yes, because $y'=e^{x}\cos x$.
No, because $y'=\cos x$.
No, because $y'=e^{\sin x}\sin x$.
No, because $y\cos x=e^{\cos x}$.
Yes, because $y'=e^{\sin x}\cos x=y\cos x$.
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 $y'=\dfrac{1}{x^2}$ but $-y^2=-\dfrac{1}{x^2}$.
Yes, because $y'=-\dfrac{1}{x^2}$ and $-y^2=-\dfrac{1}{x^2}$.
No, because $y^2=\dfrac{1}{x}$ so $-y^2=-\dfrac{1}{x}$.
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 $v'(t)=2t$.
Yes, because $\dfrac{dv}{dt}=8t$ and $v'(t)=8t$.
No, because $v'(t)=4t+3$.
Yes, because $v'(t)=4t$ equals $\dfrac{dv}{dt}$.
No, because $4t=2t^2+3$ is not true for all $t$.
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.
For $\dfrac{dy}{dx}=\dfrac{y}{x}$, is $y=3x$ a solution on $x\ne 0$?
No, because $\dfrac{y}{x}=\dfrac{3}{x}$ is not constant.
Yes, because $y'=3$ and $\dfrac{y}{x}=\dfrac{3x}{x}=3$.
No, because $y'=0$ and $\dfrac{y}{x}=3$.
Yes, because $y'=1$ and $\dfrac{y}{x}=1$.
No, because $y'=3x$ but $\dfrac{y}{x}=3$.
Explanation
To verify a solution, substitute both the function and its derivative into the differential equation. For y = 3x, we compute y' = 3. The differential equation dy/dx = y/x becomes 3 = (3x)/x = 3, which is true for all x ≠ 0. Choice B incorrectly states y' = 3x instead of y' = 3. Choice E incorrectly claims y' = 0, which would only be true for a constant function. The systematic approach is to differentiate the proposed solution, substitute both y and y' into the equation, and verify the identity holds on the specified domain.
For $\dfrac{dy}{dx}=\dfrac{y}{x}+x$, is $y=x^2$ a solution on $x\ne 0$?
Yes, because $y'=2$.
No, because $y'=x$.
No, because $\dfrac{y}{x}+x=\dfrac{x^2}{x^2}+x=1+x$.
No, because $\dfrac{y}{x}+x=\dfrac{x}{x}+x=1+x$.
Yes, because $y'=2x$ and $\dfrac{y}{x}+x=\dfrac{x^2}{x}+x=2x$.
Explanation
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.
A curve is defined by $\dfrac{dy}{dx}=2xy$. Is $y=e^{x^2}$ a solution for all real $x$?
Yes, because $y'=2xe^{x^2}$ and $2xy=2xe^{x^2}$.
No, because $2xy=2x^2e^{x^2}$, which does not equal $y'$.
No, because $y'=2x$ but $2xy=2xe^{x^2}$.
Yes, because $y'=e^{2x}$ and $2xy=e^{2x}$.
No, because $y'=e^{x^2}$ but $2xy=2xe^{x^2}$.
Explanation
Verifying whether a proposed function satisfies a differential equation is a key skill in differential equations. To check if y = $e^{x^2}$ solves dy/dx = 2xy, compute y' using the chain rule, resulting in 2x $e^{x^2}$. Next, calculate the right-hand side 2xy, which is 2x times $e^{x^2}$, or 2x $e^{x^2}$. Since y' equals 2xy for all real x, it is a solution. A distractor like choice A fails by claiming y' = 2x, forgetting the chain rule that multiplies by $e^{x^2}$. Always substitute the proposed solution into both sides of the differential equation and simplify to check for equality across the domain.
Does $y=\tan x$ satisfy $\dfrac{dy}{dx}=1+y^2$ where defined?
No, because $y'=\sec x\tan x$.
Yes, because $y'=\sec^2 x$ and $1+\tan^2 x=\sec^2 x$.
Yes, because $y'=\cos^2 x$.
No, because $y'=1-\tan^2 x$.
No, because $1+y^2=1+\sec^2 x$.
Explanation
Verifying solutions for trigonometric differential equations requires applying derivatives correctly and using trigonometric identities. For y = tan x, we compute y' = sec²x. The right side 1 + y² becomes 1 + tan²x. Using the trigonometric identity 1 + tan²x = sec²x, both sides are equal, confirming the solution where tan x is defined. Choice B gives an incorrect derivative (sec x tan x is the derivative of sec x, not tan x). Choice C makes an error in the identity. Always use standard trigonometric derivatives and identities when verifying solutions involving trigonometric functions.
For $\dfrac{dy}{dx}=4x^3$, is $y=x^4+7$ a solution for all $x$?
Yes, because $y'=4x^3+7$.
No, because $y'=4x^4$.
No, because the constant $7$ changes $\dfrac{dy}{dx}$.
No, because $y'=x^3$.
Yes, because $y'=4x^3$ equals the right-hand side.
Explanation
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.