This problem requires solving a differential equation using separation of variables. For dy/dx = x y, separate as dy/y = x dx, assuming y ≠ 0. Integrate: ln|y| = $(x^2$)/2 + C. Exponentiate to y = C $e^{x^2/2}$. The distractor y = $e^{x^2/2}$ + C fails by adding C outside, not as a multiplier. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

This problem requires solving a differential equation using separation of variables. With dy/dx = y / $x^2$ for x ≠ 0, separate as dy/y = dx / $x^2$, assuming y ≠ 0. Integrate: ln|y| = -1/x + C. Exponentiate to y = C $e^{-1/x}$. The option y = $e^{-1/x}$ + C is wrong as it adds C instead of multiplying. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

This differential equation dy/dx = y/(1+x) is solved using separation of variables. Separating gives us dy/y = dx/(1+x), with y terms on the left and x terms on the right. Integrating both sides yields ln|y| = ln|1+x| + C₁, where we recognize that the integral of 1/(1+x) is ln|1+x|. Using logarithm properties, we can write ln|y| = ln|1+x| + ln|C| = ln|C(1+x)|, where C = $e^C$₁. Exponentiating gives y = C(1+x), which is the general solution. Choice B incorrectly suggests y = ln(1+x) + C, but this would give dy/dx = 1/(1+x), not y/(1+x) as required. Remember that when integrating gives matching logarithmic forms, the solution often involves the arguments of those logarithms in a product.

This Newton's Law of Cooling equation dT/dt = 4(T-10) requires separation of variables after recognizing the shifted form. We can rewrite this as dT/dt = 4T - 40, and to separate variables, we need dT/(T-10) = 4dt. Integrating both sides gives ln|T-10| = 4t + C₁. Exponentiating yields |T-10| = e^(4t+C₁) = Ce^(4t), where C = $±e^C$₁. Solving for T gives T = 10 + Ce^(4t), which is the general solution. Choice B incorrectly places the 10 as -10 outside the exponential term, which would give dT/dt = 4Ce^(4t) ≠ 4(T-10). When dealing with shifted exponential models like dT/dt = k(T-a), the solution is always T = a + Ce^(kt), not Ce^(kt) - a.

This differential equation $\frac{dT}{dt} = -2T$ is a classic candidate for separation of variables. We separate by writing $\frac{dT}{T} = -2dt$, isolating T terms on the left and t terms on the right. Integrating both sides gives us $\ln|T| = -2t + C_1$, where $C_1$ is an integration constant. Exponentiating both sides yields $T = e^{-2t + C_1} = e^{C_1} \cdot e^{-2t}$. Since $e^{C_1}$ is just another arbitrary constant, we write the general solution as $T = Ce^{-2t}$. Choice B incorrectly treats this as if T were absent from the right side, leading to a linear rather than exponential solution. The telltale sign for using separation of variables is when the derivative of a function equals that function multiplied by something that depends only on the independent variable.

This problem requires solving a differential equation using separation of variables. For dy/dx = 2 x y, separate as dy/y = 2 x dx, assuming y ≠ 0. Integrate: ln|y| = $x^2$ + C. Exponentiate to y = C $e^{x^2}$. The distractor y = $e^{x^2}$ + C fails by not using a multiplicative constant. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

This problem requires solving a differential equation using separation of variables. For dy/dx = y sin x, separate as dy/y = sin x dx, assuming y ≠ 0. Integrate: ln|y| = -cos x + C. Exponentiate to y = C $e^{-cos x}$. The distractor y = $e^{-cos x}$ + C fails by adding the constant outside the exponential form. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

This problem requires solving a differential equation using separation of variables. To solve dP/dt = 3P, separate variables by writing dP/P = 3 dt, assuming P ≠ 0. Integrate both sides: the left gives ln|P|, and the right gives 3t + C. Exponentiate to solve for P, yielding P = $Ce^{3t}$, where C absorbs the exponential constant. A tempting distractor like P = $3e^t$ + C fails because it suggests linear growth rather than exponential, which would apply if the equation were dP/dt = 3 instead. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

This problem requires solving a differential equation using separation of variables. For dT/dt = -2T, separate by writing dT/T = -2 dt, assuming T ≠ 0. Integrate: ln|T| on the left and -2t + C on the right. Solve for T by exponentiating, getting T = $Ce^{-2t}$. The distractor T = -2t + C fails as it implies linear decay, not exponential, which would fit dT/dt = -2. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

This problem requires solving a differential equation using separation of variables. Given dy/dx = -(3/x) y for x ≠ 0, separate as dy/y = -(3/x) dx, assuming y ≠ 0. Integrate to ln|y| = -3 ln|x| + C. This simplifies to y = C $x^{-3}$. The choice y = -3 ln|x| + C is incorrect as it stops at the integral without solving for y exponentially. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.