This problem involves solving an exponential differential equation representing compound interest growth. When we have $\frac{dB}{dt}=0.07B$, the savings balance grows at a rate proportional to its current amount with rate 0.07. This represents continuous compound interest where larger balances earn proportionally more interest. The differential equation $\frac{dy}{dt} = ky$ has the general solution $y = Ce^{kt}$, so with $k = 0.07$, we get $B(t) = Ce^{0.07t}$. Choice A would represent simple interest (linear growth), but continuous compounding requires exponential growth. To identify compound interest: when growth rate is proportional to current balance ($\frac{dB}{dt} = rB$), the solution is always exponential ($Ce^{rt}$).

This problem involves solving an exponential differential equation representing proportional decay. When we have $\frac{dy}{dt}=-ky$ with $k>0$, the quantity decreases at a rate proportional to its current amount. The negative sign with positive $k$ ensures decay behavior, as larger quantities experience proportionally larger decreases. The general solution to $\frac{dy}{dt} = -ky$ is $y = Ce^{-kt}$, representing exponential decay. Choice A would represent exponential growth since it lacks the negative exponent required for decay. To identify exponential decay: negative coefficients in proportional rate equations ($\frac{dy}{dt} = -ky$) always produce decay solutions with negative exponents ($Ce^{-kt}$).

This problem involves solving an exponential differential equation representing compound interest growth. When we have $\frac{dy}{dx}=0.5y$, the balance grows at a rate proportional to its current value with rate 0.5. This represents continuous compound interest where larger balances earn proportionally more interest. The differential equation $\frac{dy}{dx} = ky$ has the general solution $y = Ce^{kx}$, so with $k = 0.5$, we get $y = Ce^{0.5x}$. Choice A represents simple interest (linear growth), but continuous compounding requires the exponential relationship. To distinguish compound from simple interest: proportional growth rates ($\frac{dy}{dx} = ry$) always indicate compound interest with exponential solutions ($Ce^{rx}$).

This problem involves solving an exponential differential equation representing exponential decay with rate ln(2). When we have $\frac{dy}{dx}=-\ln(2)y$, the quantity decreases at a rate proportional to its current value with constant $-\ln(2)$. The negative constant indicates exponential decay, and the ln(2) factor relates to half-life calculations in exponential decay. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = -\ln(2)$, we get $y = Ce^{-(\ln 2)x}$. Choice D represents linear decay, but the proportional relationship requires exponential behavior. To handle negative logarithmic constants: $-\ln(2)$ produces exponential decay solutions with the corresponding negative exponent.

This problem involves solving an exponential differential equation where the rate of change is proportional to the current quantity. When we have $\frac{dy}{dx}=0.02y$, the quantity grows at a rate proportional to its current value with proportionality constant 0.02. This positive constant indicates exponential growth, though at a slow rate of 2% of the current value. The standard solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = 0.02$, we get $y = Ce^{0.02x}$. Choice C represents linear growth, which would occur if the rate of change were constant rather than proportional to the current quantity. To recognize small exponential growth: even small positive constants in proportional rate equations produce exponential growth, not linear growth.

This problem involves solving an exponential differential equation representing exponential decay. When we have $\frac{dy}{dx}=-0.12y$, the quantity decreases at a rate proportional to its current value with constant -0.12. The negative proportionality constant indicates exponential decay at 12% of the current value per unit change. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = -0.12$, we get $y = Ce^{-0.12x}$. Choice C represents linear decay, but the proportional relationship requires exponential rather than linear behavior. To recognize percentage-based exponential decay: negative decimal constants like -0.12 produce exponential decay solutions, representing 12% proportional decrease per unit.

This question addresses exponential decay in temperature, though the notation might seem counterintuitive at first. The equation dT/dt = kT with k < 0 means temperature changes at a rate proportional to its current value, but the negative k ensures cooling. The general solution is T(t) = Ce^(kt), and with initial condition T(0) = T₀, we get T(t) = T₀e^(kt). Since k is negative, we can write this as T₀e^(-|k|t), showing explicit decay. Choice C (T₀e^(-kt)) would actually represent growth since k < 0 makes -k positive, contradicting the cooling behavior. The key insight: in dy/dt = ky, the sign of k determines growth (k > 0) or decay (k < 0), but the solution form y = Ce^(kt) remains the same.

This problem involves solving an exponential differential equation where the rate of change is proportional to the current quantity. When we have $\frac{dy}{dx}=3y$, the quantity $y$ changes at a rate proportional to its current value with proportionality constant 3. This positive constant indicates exponential growth, as larger values of $y$ produce proportionally larger rates of increase. The standard solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = 3$, we get $y = Ce^{3x}$. Choice A represents linear growth, which would occur if the rate were constant rather than proportional to the current value. To recognize exponential vs linear behavior: proportional rates always yield exponential solutions, while constant rates yield linear solutions.

This problem presents the fundamental exponential growth differential equation dN/dt = kN with k > 0. When the growth rate is proportional to the current population size, we get exponential behavior described by N(t) = Ce^(kt). The positive constant k ensures the population grows over time, as e^(kt) increases when k > 0 and t increases. This proportional relationship is what distinguishes exponential growth from other types—the larger the population, the faster it grows. Choice E (N(t) = $Ck^t$) might seem plausible but represents a different exponential base (k instead of e), which doesn't solve our differential equation. Remember: differential equations of the form dy/dt = ky always have solutions y = Ce^(kt), where e is the natural exponential base.

This problem explicitly states that growth rate is a percentage of current size, the defining characteristic of exponential growth. The equation dP/dt = 0.07P means the population grows at 7% of its current value per unit time. The general solution to this exponential differential equation is P(t) = Ce^(0.07t), where C represents the initial population size. This captures how larger populations grow faster in absolute terms while maintaining the same relative growth rate. Choice A (0.07t + C) would mean the population grows by a constant 0.07 units per time period, not 7% of its current size. The key recognition pattern: when a rate is described as a percentage or proportion of the current value, you have an exponential differential equation.