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.

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 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.

This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: dy/(y + 1) = dx/x, then integrate both sides to get ln|y + 1| = ln|x| + C. Exponentiate to solve for y: y + 1 = $e^C$ x, or y = K x - 1. Use the initial condition y(1) = 1: 1 = K * 1 - 1, so K = 2, yielding y = 2x - 1. A tempting distractor is y = 2x + 1, which might result from incorrect sign when solving for y. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.

This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: dy/(y - 2) = dx/x, then integrate both sides to get ln|y - 2| = ln|x| + C. Exponentiate to solve for y: y - 2 = $e^C$ x, or y = K x + 2. Use the initial condition y(1) = 5: 5 = K * 1 + 2, so K = 3, yielding y = 3x + 2. A tempting distractor is y = 2 + 3 ln x, which might result from incorrect integration. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.

The skill here involves applying an initial condition to find the particular solution of a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution V(t) = C $e^{-0.2 t}$. Plugging in the initial condition V(0) = 10 yields 10 = C $e^{0}$, which simplifies to C = 10. Therefore, the particular solution is V(t) = 10 $e^{-0.2 t}$. A tempting distractor like V(t) = C $e^{-0.2 t}$ fails because it remains the general solution without specifying the constant using the initial condition. Always substitute the initial values into the general solution to determine the constant and obtain the unique particular solution.

This problem requires applying an initial condition to find the particular solution of a differential equation. The general solution to 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·1 = C. Therefore, C = 50 and 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 differential equations with initial conditions, always substitute the initial values into the general solution to determine the specific constant.

This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: y dy = $x^2$ dx, then integrate both sides to get (1/2) $y^2$ = (1/3) $x^3$ + C. Solve for y: $y^2$ = (2/3) $x^3$ + 2C, or y = sqrt((2/3) $x^3$ + D). Use the initial condition y(0) = 4: 16 = D, so D = 16, yielding y = sqrt((2/3) $x^3$ + 16). A tempting distractor is y = sqrt((1/3) $x^3$ + 16), which might result from incorrect coefficient in integration. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.

This problem requires applying an initial condition to find the particular solution for exponential decay. The general solution to dV/dt = -0.1V is V(t) = Ce^(-0.1t), representing exponential decay with rate 0.1. Using the initial condition V(0) = 30: 30 = Ce^(-0.1·0) = $Ce^0$ = C·1 = C. Therefore C = 30, giving the particular solution V(t) = 30e^(-0.1t). Choice C incorrectly uses positive 0.1 in the exponent, which would represent growth rather than drainage. When the differential equation has a negative coefficient, the solution must have a negative exponent to model decay correctly.

This problem requires applying an initial condition to a linear differential equation with variable coefficient. Separating dy/dx = (1+x)y gives dy/y = (1+x)dx, which integrates to ln|y| = x + x²/2 + C. The general solution is y = Ae^(x+x²/2) where A = $e^C$. Using the initial condition y(0) = 2: 2 = Ae^(0+0²/2) = $Ae^0$ = A·1 = A. Thus A = 2 and the particular solution is y = 2e^(x+x²/2). Choice C incorrectly drops the linear term x from the exponent, showing only x². Always integrate each term carefully and use the initial condition to find the multiplicative constant.