Differential Equations Quiz: Implicit Vs Explicit Solutions

What this quiz covers

This quiz focuses on Implicit Vs Explicit Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $y' = \frac{x-y}{x+y}$
2. $y' = \frac{y-x}{x+y}$
3. $y' = \frac{x+y}{x-y}$ (correct answer)
4. $y' = \frac{x+2y}{x}$

Explanation: We perform implicit differentiation on the given relation with respect to $x$. First, rewrite the relation as $\arctan(y/x) - \frac{1}{2}\ln(x^2+y^2) = C$. Differentiating term by term: $\frac{d}{dx}(\arctan(y/x)) = \frac{1}{1+(y/x)^2} \cdot \frac{x y' - y}{x^2} = \frac{x^2}{x^2+y^2} \cdot \frac{x y' - y}{x^2} = \frac{x y' - y}{x^2+y^2}$. And $\frac{d}{dx}(\frac{1}{2}\ln(x^2+y^2)) = \frac{1}{2} \cdot \frac{1}{x^2+y^2} \cdot (2x+2yy') = \frac{x+yy'}{x^2+y^2}$. The derivative of the constant $C$ is $0$. So the equation becomes $\frac{x y' - y}{x^2+y^2} - \frac{x+yy'}{x^2+y^2} = 0$. Multiplying by $x^2+y^2$ gives $x y' - y - (x+yy') = 0$. Rearranging to solve for $y'$: $x y' - yy' = x+y$, so $y'(x-y) = x+y$. Finally, $y' = \frac{x+y}{x-y}$.

Question 2

1. There are two distinct explicit solutions, both of which are defined and differentiable for all real numbers $x$.
2. The implicit relation defines two explicit solutions, but they are valid only on intervals where $\sin(x) \ge 0$.
3. There are two explicit solutions, but they are only valid as solutions to the differential equation on intervals where $\sin(x) \neq -1$. (correct answer)
4. Only one continuous explicit solution can be defined from the relation, which must be $y(x) = \sqrt{\sin(x)+1}$.

Explanation: The implicit relation $y^2 = \sin(x) + 1$ yields two explicit functions: $y_1(x) = \sqrt{\sin(x)+1}$ and $y_2(x) = -\sqrt{\sin(x)+1}$. Since $-1 \le \sin(x) \le 1$, the term $\sin(x)+1$ is always non-negative, so both functions are defined for all real $x$. To be a solution to the differential equation $2yy' = \cos(x)$, or $y' = \frac{\cos(x)}{2y}$, the function must be differentiable and satisfy the equation. The derivative is undefined when $y=0$. This occurs when $\sin(x)+1=0$, i.e., $\sin(x)=-1$. This happens at $x = \frac{3\pi}{2} + 2k\pi$ for any integer $k$. At these points, the graphs of $y_1(x)$ and $y_2(x)$ have cusps, and the functions are not differentiable. Therefore, they can only be considered solutions on open intervals that do not contain these points.

Question 3

1. $y(x) = 2\sqrt{1-x^2}$ on the interval $(-1, 1)$ (correct answer)
2. $y(x) = -2\sqrt{1-x^2}$ on the interval $(-1, 1)$
3. $y(x) = 2\sqrt{1-x^2}$ on the interval $(-\infty, \infty)$
4. $y(x) = \sqrt{4-x^2}$ on the interval $(-2, 2)$

Explanation: First, solve the implicit relation for $y$: $\frac{y^2}{4} = 1 - x^2 \implies y^2 = 4(1-x^2) \implies y = \pm 2\sqrt{1-x^2}$. This gives two possible explicit solutions. The initial condition is $y(1/2) = \sqrt{3}$. Since $\sqrt{3}$ is positive, we must select the positive branch, $y(x) = 2\sqrt{1-x^2}$. Let's verify the initial condition: $y(1/2) = 2\sqrt{1-(1/2)^2} = 2\sqrt{3/4} = 2(\frac{\sqrt{3}}{2}) = \sqrt{3}$. The solution is valid. The interval of existence requires the argument of the square root to be positive, $1-x^2 > 0$ (not equal to zero, because $y$ is in the denominator of $y'$), which means $x^2 < 1$, or $-1 < x < 1$. So the interval of existence is $(-1, 1)$.

Question 4

1. For any constant $C$, the relation $F(x,y)=C$ defines a single, unique explicit solution $y=g(x)$.
2. If an explicit solution $y=g(x)$ is derived from $F(x,y)=C$, its interval of existence is the set of all $x$ for which $g(x)$ is a real number.
3. Any explicit solution $y=g(x)$ derived from $F(x,y)=C$ must satisfy the identity $F(x, g(x)) = C$ for all $x$ in its interval of existence. (correct answer)
4. The implicit solution $F(x,y)=C$ can always be obtained by separating variables and integrating the differential equation.

Explanation: By definition, an explicit solution $y=g(x)$ derived from an implicit solution $F(x,y)=C$ is a function that, when substituted back into the implicit relation, satisfies the relation for all $x$ in its domain (interval of existence). Thus, $F(x, g(x)) = C$ must hold true. Choice A is false; an implicit relation can define multiple explicit solutions (e.g., $x^2+y^2=C$ gives $y=\pm\sqrt{C-x^2}$) or no real solutions. Choice B is false; the interval of existence also requires that the solution be differentiable and that the original DE be defined, which may impose stricter constraints than just the function being real (e.g., denominators cannot be zero). Choice D is false; many differential equations are not separable, and their implicit solutions must be found by other methods like using integrating factors for exact equations.

Question 5

1. $-1/2$
2. $\pi+1$
3. $\pi-1$
4. $-(\pi+1)$ (correct answer)

Explanation: When you encounter an implicit solution to a differential equation, you need to use implicit differentiation to find the slope at any given point. The key insight is that both the function and its derivative are embedded in the implicit equation. Starting with $\sin(x+y) = xy + C$, first find the constant $C$ using the given point $(0, \pi)$. Substituting: $\sin(0+\pi) = 0 \cdot \pi + C$, so $\sin(\pi) = 0 = C$. Thus our specific solution is $\sin(x+y) = xy$. To find $y'$, differentiate both sides with respect to $x$. The left side gives $\cos(x+y) \cdot (1 + y')$ using the chain rule. The right side gives $y + xy'$ using the product rule. This yields: $\cos(x+y)(1 + y') = y + xy'$ At point $(0, \pi)$, we have $\cos(0+\pi)(1 + y') = \pi + 0 \cdot y'$, which simplifies to $-1(1 + y') = \pi$. Solving: $-1 - y' = \pi$, so $y' = -1 - \pi = -(\pi + 1)$. Choice A ($-1/2$) likely comes from incorrect differentiation or arithmetic errors. Choice B ($\pi + 1$) represents forgetting the negative sign from $\cos(\pi) = -1$. Choice C ($\pi - 1$) involves sign errors in the algebraic manipulation. Remember: with implicit differentiation problems, always substitute the given point values after differentiating, and be extra careful with signs when evaluating trigonometric functions at special angles.

Question 6

1. $e^y = 2x - 2 + e^{-x-1}$
2. $e^y = 2x - 2 + e^{-x}$
3. $e^y = 2x - 2$
4. $e^y = 2x - 2 + e^{1-x}$ (correct answer)

Explanation: When you encounter a problem asking for a specific solution to a differential equation, you need to apply the given initial condition to the general solution to find the particular constant value. Given the general solution $e^y = 2x - 2 + Ce^{-x}$ and the initial condition $y(1) = 0$, you substitute these values to find $C$. When $x = 1$ and $y = 0$: $e^0 = 2(1) - 2 + Ce^{-1}$ $1 = 0 + Ce^{-1}$ $1 = \frac{C}{e}$ $C = e$ Therefore, the specific solution is $e^y = 2x - 2 + ee^{-x} = 2x - 2 + e^{1-x}$, which matches answer choice D. Let's examine why the other options are incorrect: A) $e^y = 2x - 2 + e^{-x-1}$ represents $C = e^{-1} = \frac{1}{e}$, which would satisfy $e^0 = 0 + \frac{1}{e} \cdot e^{-1} = \frac{1}{e^2} \neq 1$. B) $e^y = 2x - 2 + e^{-x}$ corresponds to $C = 1$, giving us $e^0 = 0 + 1 \cdot e^{-1} = \frac{1}{e} \neq 1$. C) $e^y = 2x - 2$ implies $C = 0$, which would mean $e^0 = 0 \neq 1$. Study tip: Always substitute your initial conditions directly into the general solution to solve for the arbitrary constant. Double-check your algebra, especially when working with exponential expressions, as small errors in exponent manipulation lead to incorrect constants.

Question 7

1. The relation is not a solution because implicit differentiation yields $y' = \frac{-x}{y-C/2}$.
2. The relation is a correct family of solutions, as can be verified by implicit differentiation and substitution. (correct answer)
3. The relation is not a solution because the correct differential equation it satisfies is $y' = \frac{2xy}{y^2-x^2}$.
4. The relation is only a solution for $C=1$, as this is the only value allowing for a non-trivial solution.

Explanation: To verify if $x^2+y^2 = Cy$ is a solution, we differentiate it implicitly with respect to $x$: $2x + 2yy' = Cy'$. Our goal is to obtain an expression for $y'$ in terms of only $x$ and $y$. First, solve for $y'$: $2x = y'(C - 2y)$, so $y' = \frac{2x}{C-2y}$. This expression contains $C$. We can eliminate $C$ by using the original relation, $C = \frac{x^2+y^2}{y}$. Substituting this into the expression for $y'$: $y' = \frac{2x}{(\frac{x^2+y^2}{y}) - 2y}$. To simplify the denominator, find a common denominator: $y' = \frac{2x}{\frac{x^2+y^2-2y^2}{y}} = \frac{2x}{\frac{x^2-y^2}{y}}$. Finally, inverting and multiplying gives $y' = \frac{2xy}{x^2-y^2}$. This matches the given differential equation, so the proposal is correct.

Question 8

1. $\frac{\pi^3}{2}$ (correct answer)
2. $\frac{\pi^3}{2} + 1$
3. $\pi^2(\frac{\pi}{2} - 1)$
4. $\pi^2$

Explanation: Since the equation is exact, there is a potential function $F(x,y)$ such that $\frac{\partial F}{\partial x} = M = 2xy + \cos(y)$ and $\frac{\partial F}{\partial y} = N = x^2 - x\sin(y)$. Integrating $M$ with respect to $x$ gives $F(x,y) = \int (2xy + \cos(y)) dx = x^2y + x\cos(y) + g(y)$. To find $g(y)$, we differentiate this expression with respect to $y$ and set it equal to $N$: $\frac{\partial F}{\partial y} = x^2 - x\sin(y) + g'(y)$. Comparing this with $N = x^2 - x\sin(y)$, we see that $g'(y) = 0$, so $g(y)$ is a constant. We can incorporate this into the constant of integration $C$. The family of implicit solutions is $x^2y + x\cos(y) = C$. To find the value of $C$ for the curve passing through $(\pi, \pi/2)$, we substitute these values: $C = (\pi)^2(\frac{\pi}{2}) + \pi\cos(\frac{\pi}{2}) = \frac{\pi^3}{2} + \pi(0) = \frac{\pi^3}{2}$.

Question 9

1. $y(x) = \sqrt{x^2-5}$ on the interval $(\sqrt{5}, \infty)$ (correct answer)
2. $y(x) = -\sqrt{x^2-5}$ on the interval $(\sqrt{5}, \infty)$
3. $y(x) = \sqrt{x^2-5}$ on the interval $(-\infty, -\sqrt{5}) \cup (\sqrt{5}, \infty)$
4. $y(x) = \sqrt{x^2-5}$ on the interval $(-\infty, \infty)$

Explanation: First, use the initial condition $y(3)=2$ to find the constant $C$. Substitute $x=3$ and $y=2$ into the implicit solution: $2^2 - 3^2 = C$, which gives $4 - 9 = -5 = C$. The particular implicit solution is $y^2 - x^2 = -5$, or $y^2 = x^2 - 5$. Solving for $y$ yields two possible explicit solutions: $y = \pm\sqrt{x^2-5}$. The initial condition $y(3)=2$ has a positive $y$-value, so we must choose the positive branch: $y(x) = \sqrt{x^2-5}$. For this solution to be defined and real, the argument of the square root must be non-negative, $x^2-5 \ge 0$, which means $|x| \ge \sqrt{5}$. Additionally, the original differential equation, which can be found by differentiating the implicit solution ($2yy' - 2x = 0 \implies y' = x/y$), is undefined when $y=0$. This occurs when $x^2-5=0$, or $x=\pm\sqrt{5}$. Thus, the solution is valid for $|x|>\sqrt{5}$. The interval of existence for a solution to an IVP must be a single continuous interval containing the initial point $x=3$. The interval containing $x=3$ is $(\sqrt{5}, \infty)$.

Question 10

1. The explicit solution $y=g(x)$ cannot be found, therefore the relation cannot be a solution to any differential equation.
2. An explicit solution $y=g(x)$ exists for any initial condition, but finding it requires solving a non-separable differential equation.
3. An explicit solution $y=g(x)$ cannot be expressed in terms of elementary functions, but its derivative $y'$ can be found in terms of $x$ and $y$. (correct answer)
4. The explicit solution can be written as $y(x) = \sqrt[5]{\cos(x) - x - y}$, which shows it is implicitly defined.

Explanation: The given relation is a quintic polynomial in $y$. According to the Abel-Ruffini theorem, there is no general algebraic formula for the roots of a polynomial of degree five or higher in terms of its coefficients. This means we cannot isolate $y$ to find an explicit solution $y=g(x)$ using elementary functions (algebraic operations, exponentials, logarithms, trigonometric functions). However, we can still find the differential equation it solves using implicit differentiation: $5y^4 y' + y' + 1 + \sin(x) = 0$. Solving for $y'$ gives $y'(5y^4+1) = -1-\sin(x)$, so $y' = \frac{-1-\sin(x)}{5y^4+1}$. This shows that while the explicit solution $y(x)$ is not expressible in a simple closed form, its derivative $y'$ can be expressed in terms of $x$ and $y$.

Question 11

1. $y(x) = 1 + \sqrt{4 - 2\ln(x)}$ on the interval $(0, \infty)$
2. $y(x) = 1 - \sqrt{4 - 2\ln(x)}$ on the interval $(0, e^2)$
3. $y(x) = 1 + \sqrt{4 - 2\ln(x)}$ on the interval $(-e^2, 0) \cup (0, e^2)$
4. $y(x) = 1 + \sqrt{4 - 2\ln(x)}$ on the interval $(0, e^2)$ (correct answer)

Explanation: When you encounter a differential equation like $y' = \frac{1}{x(1-y)}$, recognize this as a separable equation. You can rearrange it so all terms involving $y$ are on one side and all terms involving $x$ are on the other. Separating variables gives you $(1-y)dy = \frac{1}{x}dx$. Integrating both sides: $\int(1-y)dy = \int\frac{1}{x}dx$, which yields $y - \frac{y^2}{2} = \ln|x| + C$. Using the initial condition $y(1) = 3$: $3 - \frac{9}{2} = \ln(1) + C$, so $C = -\frac{3}{2}$. This gives us $y - \frac{y^2}{2} = \ln(x) - \frac{3}{2}$. Rearranging: $y^2 - 2y + (2\ln(x) - 3) = 0$. Using the quadratic formula: $y = \frac{2 \pm \sqrt{4 - 4(2\ln(x) - 3)}}{2} = 1 \pm \sqrt{4 - 2\ln(x)}$. Since $y(1) = 3$ and $\sqrt{4 - 2\ln(1)} = 2$, we need $y = 1 + \sqrt{4 - 2\ln(x)}$. For the domain, the expression under the square root must be non-negative: $4 - 2\ln(x) \geq 0$, which means $\ln(x) \leq 2$, so $x \leq e^2$. Combined with the requirement that $x > 0$ (from the original equation), the domain is $(0, e^2)$. Choice A incorrectly extends the interval to infinity. Choice B has the wrong sign (minus instead of plus). Choice C includes negative values where $\ln(x)$ is undefined. Study tip: Always check your solution against the initial condition to determine the correct sign, and carefully analyze domain restrictions from both the original equation and your final solution.

Question 12

1. The exponential term prevents any algebraic manipulation to isolate $y$
2. The initial condition $y(0) = 2$ leads to an inconsistent system with no solution
3. The equation is not actually a valid implicit solution to any first-order differential equation
4. The mixed $xy$ term in the exponent creates a transcendental equation that generally requires numerical methods (correct answer)

Explanation: When working with implicit solutions to differential equations, you'll often encounter situations where converting to explicit form becomes mathematically challenging or impossible using elementary algebraic methods. Let's examine what happens when we try to solve $e^{xy} - x^2 + y = K$ for $y$ explicitly. Using the initial condition $y(0) = 2$, we first find $K$: $e^{0 \cdot 2} - 0^2 + 2 = 1 + 2 = 3$. So our equation becomes $e^{xy} - x^2 + y = 3$, or $e^{xy} + y = x^2 + 3$. The fundamental problem is that $y$ appears both inside the exponential function (as part of $xy$) and as a separate linear term. This creates a transcendental equation — one that mixes algebraic and transcendental (exponential) functions in a way that cannot be solved using standard algebraic operations. The $xy$ term in the exponent is the culprit, making this equation impossible to solve analytically for $y$ in terms of $x$. This confirms answer D. Looking at the incorrect options: A is wrong because exponential terms can often be manipulated algebraically (using logarithms, for instance) — the issue is specifically the mixed $xy$ term. B is incorrect since substituting $y(0) = 2$ gives us a consistent value $K = 3$. C is false because this type of implicit relation commonly arises from first-order differential equations. Study tip: When you see $y$ appearing both inside a transcendental function and elsewhere in the equation, expect that numerical methods will be needed for explicit solutions.

Question 13

1. Direct algebraic solving for $y$ since this is a quadratic equation in $y$
2. Using implicit differentiation to find $\frac{dy}{dx}$, then integrating to find $y(x)$
3. Recognizing this as a cubic equation in $y$ and applying the cubic formula or numerical methods (correct answer)
4. Parametric representation since explicit form may not exist in elementary functions

Explanation: The equation $x^3 + 3xy^2 - y^3 = 8$ can be rearranged as $-y^3 + 3xy^2 + (x^3 - 8) = 0$, or $y^3 - 3xy^2 - (x^3 - 8) = 0$. This is indeed a cubic equation in $y$ for any given value of $x$. While cubic equations can be solved using the cubic formula, the expressions are often complex. Choice A is wrong because this is cubic, not quadratic in $y$. Choice B is incorrect because implicit differentiation gives us the derivative, not the original function. Choice D is overly pessimistic - while the cubic formula yields complex expressions, explicit solutions do exist in elementary functions.

Question 14

1. An explicit solution $y = f(x)$ exists and can be found using inverse trigonometric functions
2. An explicit solution exists locally by the Implicit Function Theorem since $\frac{\partial F}{\partial y} \neq 0$ at the given point (correct answer)
3. No explicit solution exists because the equation involves both trigonometric and polynomial terms
4. The explicit solution requires solving a transcendental equation that has no closed form

Explanation: Let $F(x,y) = \sin(xy) + x^2y - C$. At the point $(\frac{\pi}{2}, 1)$, we have $\frac{\partial F}{\partial y} = x\cos(xy) + x^2 = \frac{\pi}{2}\cos(\frac{\pi}{2}) + (\frac{\pi}{2})^2 = 0 + \frac{\pi^2}{4} \neq 0$. Since the partial derivative with respect to $y$ is non-zero, the Implicit Function Theorem guarantees that locally around this point, $y$ can be expressed as an explicit function of $x$. Choice A is wrong because inverse trig functions alone won't solve this mixed equation. Choice C is incorrect - the combination of function types doesn't prevent explicit solutions from existing. Choice D confuses the difficulty of finding a closed form with the existence of an explicit function.

Question 15

1. It represents two different arbitrary constants that should be written as $C_1$ and $C_2$
2. It indicates two distinct solution branches that cannot cross each other due to uniqueness theorem (correct answer)
3. It shows that the original differential equation has no unique solution for any initial condition
4. It represents mathematical artifact with no physical meaning in differential equation solutions

Explanation: The $\pm$ sign represents two distinct solution branches: $y = x + \sqrt{x^2 + C}$ and $y = x - \sqrt{x^2 + C}$. By the uniqueness theorem for first-order ODEs, if the right-hand side and its partial derivative are continuous, then solution curves cannot intersect (except possibly at singular points). Each branch represents a different family of solution curves that cannot cross each other. Choice A is wrong because $C$ is the same constant - the branches differ by which square root we choose. Choice C is incorrect because uniqueness still holds for each branch given an initial condition. Choice D is wrong because the branches have real geometric meaning as different solution families.

Question 16

1. $y = \tan\left(\arctan(x) + \frac{\pi}{4}\right) = \frac{1+x}{1-x}$ for $x < 1$ (correct answer)
2. $y = \tan\left(\arctan(x) + \frac{\pi}{4}\right) = \frac{x+1}{1-x}$ for $x < 1$
3. $y = x + \tan\left(\frac{\pi}{4}\right) = x + 1$ for all real $x$
4. $y = \frac{\sin(\arctan(x) + \frac{\pi}{4})}{\cos(\arctan(x) + \frac{\pi}{4})}$ which cannot be simplified further

Explanation: When you encounter an implicit solution involving inverse trigonometric functions, the key is systematically applying trigonometric identities to convert it to explicit form. Start by isolating one of the inverse trig terms. From $\arctan(y) - \arctan(x) = \frac{\pi}{4}$, we get $\arctan(y) = \arctan(x) + \frac{\pi}{4}$. Taking the tangent of both sides: $y = \tan\left(\arctan(x) + \frac{\pi}{4}\right)$. Now apply the tangent addition formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. Here, $\tan(\arctan(x)) = x$ and $\tan\left(\frac{\pi}{4}\right) = 1$, so: $y = \frac{x + 1}{1 - x \cdot 1} = \frac{1 + x}{1 - x}$ The domain restriction $x < 1$ ensures the denominator doesn't equal zero. Looking at the wrong answers: Option B has the numerator terms in the wrong order ($x + 1$ instead of $1 + x$), though mathematically these are equivalent - but the formatting doesn't match the systematic derivation. Option C incorrectly assumes you can separate the arctangent terms linearly, treating $\tan(\arctan(x) + \frac{\pi}{4})$ as $x + 1$, which ignores the tangent addition formula entirely. Option D stops at the sine/cosine form without recognizing this equals the tangent function, missing the simplification opportunity. Study tip: When converting implicit solutions with inverse trig functions, always use the appropriate addition formulas rather than trying to distribute operations linearly. The tangent addition formula is especially common in differential equations solutions.

Question 17

1. The algebraic factoring $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$ was applied incorrectly
2. The cube root function is not defined for negative values of $C$
3. The solution $y = \sqrt[3]{C} - x$ is incomplete because cube roots can have three complex values (correct answer)
4. The approach loses generality by treating $C$ as a specific value rather than an arbitrary constant

Explanation: The factoring is correct: $x^3 + 3x^2y + 3xy^2 + y^3 = (x+y)^3$. From $(x+y)^3 = C$, we get $x+y = \sqrt[3]{C}$, so $y = \sqrt[3]{C} - x$. However, the student has only considered the real cube root. In the complex plane, the equation $w^3 = C$ has three solutions: $w = \sqrt[3]{C}, \sqrt[3]{C}\omega, \sqrt[3]{C}\omega^2$ where $\omega = e^{2\pi i/3}$ is a primitive cube root of unity. This gives three solution branches: $y = \sqrt[3]{C} - x$, $y = \sqrt[3]{C}\omega - x$, and $y = \sqrt[3]{C}\omega^2 - x$. In real differential equations, we typically focus on real solutions, but the student's approach misses the fact that there could be multiple real branches depending on the nature of $C$. Choice A is wrong - the factoring is correct. Choice B is incorrect - cube roots of negative numbers are well-defined real numbers. Choice D is not the main issue - $C$ remains arbitrary in the final form.

Question 18

1. $y = Ce^t$ where $C = xe^K$ is redefined as the arbitrary constant (correct answer)
2. $y = xe^{t+K}$ where $x$ remains as a parameter in the solution
3. $y = \pm e^{t+K}$ accounting for the absolute value signs in the logarithms
4. $y = e^{t+K-\ln|x|}$ maintaining the logarithmic form for generality

Explanation: When you encounter implicit relations involving logarithms in differential equations, your goal is to isolate the dependent variable while properly handling constants and parameters. Starting with $\ln|y| - \ln|x| = t + K$, you can use the logarithm property $\ln a - \ln b = \ln(a/b)$ to rewrite this as $\ln\left|\frac{y}{x}\right| = t + K$. Taking the exponential of both sides gives $\left|\frac{y}{x}\right| = e^{t+K}$, which means $\frac{y}{x} = \pm e^{t+K}$. Therefore, $y = \pm x e^{t+K}$. Since $x$ and the $\pm$ sign are both constants (not functions of $t$), you can combine them into a single arbitrary constant. Let $C = \pm x e^K$, giving you $y = Ce^t$. This is answer choice A. Choice B incorrectly keeps $x$ as a separate parameter rather than absorbing it into the constant. In differential equations solutions, you typically consolidate all constant terms. Choice C attempts to handle the absolute value but fails to account for $x$ properly, and the exponential form doesn't match what you get from the algebraic manipulation. Choice D maintains an unnecessarily complex logarithmic form instead of simplifying to the standard exponential solution format. Study tip: In differential equations, always look to combine constants, parameters, and signs into a single arbitrary constant $C$ when expressing your final solution. This creates the cleanest, most standard form that's typically expected.