This problem involves analyzing the properties of exponential differential equation solutions. When we have $\frac{dP}{dt}=kP$ with $P(0)>0$, the population changes at a rate proportional to its current size. This proportional relationship always produces exponential solutions of the form $P(t) = Ce^{kt}$. Since $P(0) > 0$, we have $P(0) = Ce^{k \cdot 0} = C > 0$, so the constant $C$ is positive. Therefore, $P(t) = Ce^{kt}$ with $C > 0$ is the correct form for all solutions. Choice A would represent linear growth, which contradicts the proportional rate relationship. To verify exponential solutions: the form $Ce^{kt}$ emerges naturally from the proportional rate condition $\frac{dy}{dt} = ky$.

This problem involves solving an exponential differential equation where the rate of change is proportional to the current value. When we have $\frac{dP}{dt}=0.18P$, this means the population grows at a rate proportional to its current size with proportionality constant 0.18. This type of differential equation has solutions of the form $P(t) = Ce^{kt}$ where $k$ is the proportionality constant. Since our constant is positive (0.18), we get exponential growth with $P(t) = Ce^{0.18t}$. Choice A represents linear growth, not exponential growth as required by the proportional relationship. To recognize exponential models: positive constants in $\frac{dy}{dt} = ky$ give growth ($Ce^{kt}$), negative constants give decay ($Ce^{-kt}$).

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 question tests exponential decay in the context of radioactive decay, a classic application of exponential differential equations. The equation dM/dt = -λM tells us the decay rate is proportional to the current amount, with λ > 0 ensuring decay (negative rate). Solving this differential equation gives M(t) = Ce^(-λt), where C represents the initial amount of radioactive material. The exponential function with negative exponent captures how the material decreases over time, never quite reaching zero but approaching it asymptotically. Choice A (M₀ - λt) would represent linear decay with constant rate λ, not the proportional decay that characterizes radioactive materials. For any differential equation dy/dt = -ky with k > 0, the solution is y = Ce^(-kt), representing exponential decay.

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 involves solving an exponential differential equation where the investment growth rate is proportional to the current value. When we have $\frac{dV}{dt}=0.04V$, the investment grows at a rate proportional to its current amount with proportionality constant 0.04. This represents 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 $k = 0.04$, we get $V(t) = Ce^{0.04t}$. Choice B would represent simple interest (linear growth), but compound interest requires the exponential relationship. To identify compound vs simple interest: proportional growth rates ($\frac{dV}{dt} = rV$) indicate compound interest with exponential solutions.

This problem involves solving an exponential differential equation representing exponential decay. When we have $\frac{dD}{dt}=-0.08D$, the temperature deviation decreases at a rate proportional to its current value with constant -0.08. This negative proportionality constant indicates exponential decay toward equilibrium temperature. The general solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$, so with $k = -0.08$, we get $D(t) = Ce^{-0.08t}$. Choice A represents linear decay, but temperature deviations typically follow Newton's law of cooling, which requires exponential decay. To identify temperature deviation models: negative constants in proportional rate equations always produce exponential approaches to equilibrium.

This problem involves solving an exponential differential equation representing radioactive decay. When we have $\frac{dM}{dt}=-kM$ with $k>0$, the mass decreases at a rate proportional to the current amount. The negative sign with positive $k$ indicates decay, as radioactive materials lose mass proportionally to how much remains. This proportional decay relationship always produces exponential solutions of the form $M(t) = Ce^{-kt}$. Choice A would represent growth rather than decay since it has a positive exponent. To distinguish growth from decay in exponential models: negative coefficients in $\frac{dy}{dt} = ky$ produce decay solutions with negative exponents ($Ce^{-kt}$).

This problem involves solving an exponential differential equation where the rate of change is proportional to the current quantity. When we have $\frac{dy}{dx}=-\frac{3}{2}y$, the quantity decreases at a rate proportional to its current value with constant $-\frac{3}{2}$. The negative fractional constant indicates exponential decay at a rate of 1.5 times the current value. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$, so with $k = -\frac{3}{2}$, we get $y = Ce^{-(3/2)x}$. Choice C represents linear decay, but the proportional relationship requires exponential rather than linear behavior. To handle fractional decay constants: negative fractions in proportional rate equations still produce exponential decay solutions with the same fractional exponent.

This problem involves solving an exponential differential equation where the growth rate is proportional to the current population size. When we have $\frac{dN}{dt}=kN$ with $k>0$, the bacteria culture grows at a rate directly proportional to its current size. This proportional relationship creates exponential growth because larger populations produce proportionally more offspring. The standard form for solving $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$, so our solution becomes $N(t) = Ce^{kt}$. Choice A represents linear growth, which would occur if the growth rate were constant rather than proportional to size. To distinguish exponential from linear models: proportional rates ($\frac{dy}{dt} = ky$) always yield exponential solutions ($Ce^{kt}$), while constant rates yield linear solutions.