This logistic differential equation requires applying an initial condition to determine the particular solution. The general solution to $\frac{dP}{dt}=\frac{1}{100}P(100-P)$ is $P(t)=\frac{100}{1+Ce^{-t}}$ for some constant $C$. Substituting the initial condition $P(0)=20$: $20=\frac{100}{1+Ce^0}=\frac{100}{1+C}$. Solving: $20(1+C)=100$, so $20+20C=100$, giving $C=4$. Therefore, $P(t)=\frac{100}{1+4e^{-t}}$. Choice C shows the general form without determining $C$, which is incomplete. For logistic equations, always use the initial population to find the specific constant in the denominator.

This problem requires applying an initial condition to find the particular solution of a differential equation. The general solution to $\frac{dP}{dt}=0.2P$ is $P(t)=Ce^{0.2t}$, where $C$ is an arbitrary constant. To find $C$, we substitute the initial condition $P(0)=50$: $50=Ce^{0.2(0)}=Ce^0=C$, so $C=50$. Therefore, the particular solution is $P(t)=50e^{0.2t}$. Choice B shows the general solution without applying the initial condition, which is a common error. When solving initial value problems, always substitute the initial condition into the general solution to determine the specific constant value.

This problem requires applying an initial condition after solving a separable differential equation. Separating variables in $\frac{dy}{dx}=xy$ gives $\frac{dy}{y}=x,dx$. Integrating both sides yields $\ln|y|=\frac{x^2}{2}+C$, so $y=Ae^{x^2/2}$ where $A=e^C$. Using the initial condition $y(0)=3$: $3=Ae^{0^2/2}=Ae^0=A$, so $A=3$. Therefore, the particular solution is $y=3e^{x^2/2}$. Choice C incorrectly integrates to get $e^{x^2}$ instead of $e^{x^2/2}$, forgetting the factor of $\frac{1}{2}$ from integration. When solving separable equations, carefully integrate both sides and then apply the initial condition to find the constant.

This exponential growth problem requires applying an initial condition to determine the particular solution. The general solution to $\frac{dS}{dt}=3S$ is $S(t)=Ce^{3t}$, representing exponential growth with rate 3. Applying the initial condition $S(0)=2$: $2=Ce^{3(0)}=Ce^0=C$, so $C=2$. Thus, the particular solution is $S(t)=2e^{3t}$. Choice D reverses the roles of the initial value and growth rate, giving $S(t)=3e^{2t}$, which satisfies neither the differential equation nor the initial condition. For exponential growth/decay, the coefficient in the differential equation becomes the exponent's coefficient, while the initial value becomes the multiplicative constant.

This homogeneous differential equation requires applying an initial condition at a non-zero point. The general solution to $\frac{dy}{dx}=\frac{y}{x}$ is $y=Cx$ (found by separating variables: $\frac{dy}{y}=\frac{dx}{x}$, giving $\ln|y|=\ln|x|+\ln|C|$). Using the initial condition $y(2)=7$: $7=C(2)$, so $C=\frac{7}{2}$. Therefore, the particular solution is $y=\frac{7}{2}x$. Choice B incorrectly assumes $C=7$, forgetting to solve for $C$ using the given point $(2,7)$. When the initial condition is given at a point other than $x=0$, substitute both coordinates to find the constant.

This problem involves applying an initial condition to Newton's law of cooling. The differential equation $\frac{dT}{dt}=-\frac{1}{5}(T-20)$ has general solution $T(t)=20+Ce^{-t/5}$, where 20 is the ambient temperature. Using the initial condition $T(0)=80$, we get $80=20+Ce^0=20+C$, which gives $C=60$. Thus, the particular solution is $T(t)=20+60e^{-t/5}$. Choice C represents the general solution without determining $C$, missing the crucial step of applying the initial condition. For cooling problems, always identify the equilibrium temperature and use the initial condition to find the specific constant.

The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dv/dt = $-v^3$ is v(t) = 1 / sqrt(2 t + C), derived by separating variables, integrating with power rules, and solving for v (taking the positive root since v(0) > 0). Substituting v(0) = 2 gives 2 = 1 / sqrt(C), so C = 1/4. Thus, the particular solution is v(t) = 2 / sqrt(1 + 8 t) after simplifying. A tempting distractor is v(t) = 2 / sqrt(1 - 8 t), which uses a negative sign, but it fails because the differential equation's negative rate leads to a positive term inside the square root for decreasing v. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.

The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dT/dt = - (1/10) (T - 20) is T(t) = 20 + C $e^{-t/10}$, obtained by rewriting the equation, separating variables, and integrating. Using the initial condition T(0) = 80 gives 80 = 20 + C, so C = 60. Therefore, the particular solution is T(t) = 20 + 60 $e^{-t/10}$. A tempting distractor is T(t) = 20 + 60 $e^{t/10}$, which assumes heating instead of cooling, but it fails because the negative rate constant indicates exponential decay toward the ambient temperature. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.

The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to $\frac{dy}{dx} = \frac{2x}{1 + x^2} y$ is $y(x) = C (1 + x^2)$, derived by separating variables, recognizing the integral as $\ln(1 + x^2)$, and exponentiating. Applying the initial condition $y(0) = 3$ yields $3 = C (1)$, so $C = 3$. Thus, the particular solution is $y(x) = 3 (1 + x^2)$. A tempting distractor is $y(x) = 3 \sqrt{1 + x^2}$, which might arise from mishandling the integral, but it fails because the correct integration leads to a linear power, not a square root. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.

The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dP/dt = P(3-P) is P(t) = 3 / (1 + C $e^{-3t}$), derived from separating variables, integrating with partial fractions, and solving for P. Applying the initial condition P(0) = 1 gives 1 = 3 / (1 + C), so C = 2. Therefore, the particular solution is P(t) = 3 / (1 + 2 $e^{-3t}$). A tempting distractor is P(t) = 3 / (1 + 2 $e^{3t}$), which uses a positive exponent, but it fails because the logistic growth requires a negative exponent for convergence to the carrying capacity. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.