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.

Separation of variables is used to solve this differential equation for y in terms of x. Rearrange as $y^3$ dy = 3 $x^2$ dx. Integrate to obtain (1/4) $y^4$ = $x^3$ + K. Multiply by 4 to get $y^4$ = 4 $x^3$ + C, where C = 4K. Choice B, $y^4$ = 3 $x^3$ + C, is a tempting distractor that might occur from forgetting to multiply the right side by 4 after integrating, thus mismatching the coefficients. 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 using separation of variables to solve the differential equation for the bacteria population. Start by rewriting the equation as dP / [P(1 - P)] = 0.5 dt, separating the variables P and t. Use partial fractions to express 1/[P(1 - P)] as 1/P + 1/(1 - P), then integrate both sides to get ln|P| - ln|1 - P| = 0.5 t + K, which simplifies to ln|P / (1 - P)| = 0.5 t + K. Exponentiating yields |P / (1 - P)| = $e^K$ $e^{0.5 t}$, and solving for P gives the logistic form P(t) = 1 / (1 + C $e^{-0.5 t}$), where C accounts for the constant and sign. A tempting distractor like choice E presents an equivalent form but with a different parameterization of the constant, which might confuse students but represents the same family of solutions. 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 requires the skill of finding general solutions to differential equations using separation of variables. To solve dy/dx = x / (1 + $y^2$), separate variables by writing (1 + $y^2$) dy = x dx. Integrate both sides: ∫ (1 + $y^2$) dy = ∫ x dx, yielding y + (1/3) $y^3$ = (1/2) $x^2$ + C. This is the implicit general solution. A tempting distractor is choice C, ln|1 + $y^2$| = (1/2) $x^2$ + C, which incorrectly integrates the left side as a log instead of a polynomial. 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 requires the skill of finding general solutions to differential equations using separation of variables. To solve dy/dx = x sqrt(y), separate variables by writing dy / sqrt(y) = x dx. Integrate both sides: ∫ $y^{-1/2}$ dy = ∫ x dx, yielding 2 sqrt(y) = (1/2) $x^2$ + C. Solve for y: sqrt(y) = (1/4) $x^2$ + K, so y = ((1/4) $x^2$ + $K)^2$. A tempting distractor is choice C, sqrt(y) = (1/2) $x^2$ + C, which forgets to divide by 2 after integrating the left side. 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 requires the skill of finding general solutions to differential equations using separation of variables. To solve dT/dt = -k (T - 20), separate variables by writing dT / (T - 20) = -k dt. Integrate both sides: ∫ 1/(T - 20) dT = ∫ -k dt, yielding ln|T - 20| = -k t + C. Exponentiate to get T - 20 = A $e^{-k t}$, or T = 20 + A $e^{-k t}$. A tempting distractor is choice B, T(t) = 20 + C $e^{k t}$, which has a positive exponent and would model heating away from equilibrium instead of approaching it. 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.