Math Snap

STEP 1

1. We are given a first-order linear differential equation: $\frac{d y}{d t} = -k(y + A)$.
2. We need to find a function $y(t)$ that satisfies this differential equation.
3. $A$ and $k$ are positive constants, and $C$ is an arbitrary constant.

STEP 2

1. Rearrange the differential equation.
2. Solve the differential equation using separation of variables.
3. Identify the correct solution from the given options.

STEP 3

Rearrange the given differential equation $\frac{d y}{d t} = -k(y + A)$ to separate variables:
$$\frac{d y}{d t} = -k(y + A)$$ This can be rewritten as:
$$\frac{d y}{d t} = -ky - kA$$

STEP 4

Separate variables by moving all terms involving $y$ to one side and $t$ to the other side:
$$\frac{d y}{y + A} = -k \, dt$$ Integrate both sides:
$$\int \frac{1}{y + A} \, dy = \int -k \, dt$$ The left side integrates to:
$$\ln|y + A|$$ The right side integrates to:
$$-kt + C_1$$ Thus, we have:
$$\ln|y + A| = -kt + C_1$$

STEP 5

Solve for $y$ by exponentiating both sides to remove the natural logarithm:
$$|y + A| = e^{-kt + C_1}$$ This can be rewritten as:
$$y + A = \pm e^{C_1} e^{-kt}$$ Let $C = \pm e^{C_1}$, which is an arbitrary constant. Therefore, we have:
$$y + A = C e^{-kt}$$ Solve for $y$:
$$y = -A + C e^{-kt}$$

Identify the correct solution from the given options. The derived solution is:
$$y = -A + C e^{-kt}$$ The correct option is:
$$y = -A + C e^{-kt}$$ The solution to the differential equation is:
$$\boxed{y = -A + C e^{-kt}}$$