This equation $\frac{dy}{dt}=\frac{1+y^2}{1+t^2}$ separates to $\frac{dy}{1+y^2}=\frac{dt}{1+t^2}$. Both sides have the form $\frac{du}{1+u^2}$, which integrates to $\arctan(u)$. Therefore, we get $\arctan(y)=\arctan(t)+C$. Choice C incorrectly attempts to use logarithms, but $\int\frac{du}{1+u^2}=\arctan(u)$, not $\frac{1}{2}\ln(1+u^2)$ (which would be for $\int\frac{2u,du}{1+u^2}$). When solving separable equations: (1) separate variables, (2) recognize standard integral forms, (3) apply the correct antiderivative formulas, and (4) add the constant of integration.

This differential equation $\frac{dB}{dt}=\frac{B}{1+t^2}$ requires separation of variables. We separate to get $\frac{dB}{B}=\frac{dt}{1+t^2}$. The left side integrates to $\ln|B|$, while the right side is the standard integral $\arctan(t)$, giving us $\ln|B|=\arctan(t)+C$. Choice C incorrectly treats $\frac{1}{1+t^2}$ as the antiderivative rather than recognizing it as the derivative of $\arctan(t)$. When solving separable equations: (1) separate variables, (2) recognize standard integral forms, (3) integrate both sides, and (4) include the constant of integration.

Separation of variables is the method to find the general solution for this equation. Rearrange as $ \frac{dy}{y} = (1 + x^2) , dx $. Integrate to get $ \ln |y| = x + \frac{x^3}{3} + K $. Exponentiating yields $ y = C e^{x + \frac{x^3}{3}} $, where $ C = \pm e^K $. Choice D, $ y = C (x + \frac{x^3}{3}) $, is a tempting distractor that could result from forgetting to exponentiate after integrating and instead treating it linearly. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.

This population model requires separation of variables to find the general solution. Rewrite as $ \frac{dP}{P(4 - P)} = dt $, separating P and t. Use partial fractions: $ \frac{1}{P(4 - P)} = \frac{1}{4} \left( \frac{1}{P} + \frac{1}{4 - P} \right) $, then integrate to $ \ln|P| - \ln|4 - P| = 4 t + K $, simplifying to $ \ln\left| \frac{P}{4 - P} \right| = 4 t + K $. Exponentiating and solving gives $ P(t) = \frac{4}{1 + C e^{-4 t}} $. Choice B is a tempting distractor with a positive exponent, which could stem from an error in the sign during integration or exponentiation, resulting in unstable growth. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.

This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dy/dx = y / (1 + $x^2$), separate variables by writing dy / y = dx / (1 + $x^2$). Integrate both sides: ∫ (1/y) dy = ∫ 1/(1 + $x^2$) dx, yielding ln|y| = arctan x + C. This is the general solution. A tempting distractor is choice C, ln|y| = 1/(1 + $x^2$) + C, which mistakenly integrates to the integrand itself instead of arctan x. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.

This problem uses separation of variables for the equation $\frac{dS}{dt}=\frac{3t}{S}$. We separate by multiplying both sides by $S$ and $dt$ to get $S,dS=3t,dt$. Integrating both sides yields $\frac{1}{2}S^2=\frac{3}{2}t^2+C$, where we use the power rule on both sides. Choice A incorrectly omits the coefficient $\frac{1}{2}$ from integrating $S,dS$, while choice E incorrectly integrates $3t$ as $3t$ instead of $\frac{3}{2}t^2$. For separable equations: (1) separate variables completely, (2) integrate each side independently, (3) include all coefficients from integration, and (4) add the constant of integration.

This equation $\frac{dv}{dt}=5t\sqrt{v}$ requires separation of variables. We rewrite $\sqrt{v}$ as $v^{1/2}$ and separate to get $v^{-1/2}dv=5t,dt$. Integrating the left side using the power rule gives $2v^{1/2}=2\sqrt{v}$, while the right side gives $\frac{5}{2}t^2$. Therefore, $2\sqrt{v}=\frac{5}{2}t^2+C$. Choice B incorrectly omits the coefficient 2 from integrating $v^{-1/2}$, while choice E incorrectly integrates $5t$ as $5t$ instead of $\frac{5}{2}t^2$. When solving: (1) convert radicals to powers, (2) separate variables, (3) apply power rule carefully, and (4) simplify the final form.

This is Newton's law of cooling with $\frac{dT}{dt}=-4(T-20)$, solved using separation of variables. We separate to get $\frac{dT}{T-20}=-4dt$. Integrating both sides yields $\ln|T-20|=-4t+C$, where we use the substitution $u=T-20$ on the left side. Choice E incorrectly has a positive $4t$ instead of negative, likely from mishandling the negative sign in the original equation. When solving separable equations: (1) separate variables preserving all signs, (2) use substitution when needed, (3) integrate both sides, and (4) check that signs match the original equation.

For $\frac{dy}{dx}=(x^2+1)e^{y}$, we separate variables to get $e^{-y}dy=(x^2+1)dx$. Integrating the left side gives $-e^{-y}$ (since $\int e^{-y}dy=-e^{-y}$), while the right side gives $\frac{1}{3}x^3+x$. Therefore, $-e^{-y}=\frac{1}{3}x^3+x+C$. Choice A incorrectly omits the negative sign when integrating $e^{-y}dy$, a common error with exponential functions. For separable equations: (1) separate variables, (2) rewrite to standard forms if needed, (3) integrate carefully with correct signs, and (4) verify your antiderivatives.

For $\frac{dy}{dx}=\frac{x}{y+2}$, we use separation of variables by writing $(y+2)dy=x,dx$. Integrating the left side requires the substitution $u=y+2$, giving $\int(y+2)dy=\frac{1}{2}(y+2)^2$. The right side integrates to $\frac{1}{2}x^2$, so we have $\frac{1}{2}(y+2)^2=\frac{1}{2}x^2+C$. Choice E incorrectly uses $y$ instead of $(y+2)$ in the integration, missing the shift in the denominator. For separable equations: (1) separate variables, (2) identify needed substitutions, (3) integrate both sides carefully, and (4) include all terms from the original equation.