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A first-order differential equation involves the first derivative of an unknown function, usually with respect to a variable. In this particular example, the given equation is \( \frac{dy}{dt} = 6y \). This equation is of first-order because it involves the first derivative of \( y \) with respect to \( t \). The equation is also linear because it can be expressed in the standard form \( \frac{dy}{dt} = ky \), where \( k \) is a constant, in this case, 6. First-order linear differential equations have the general property that they can be solved, among other methods, by using separation of variables. The aim is to find a function \( y(t) \) that satisfies this differential equation. These types of equations often illustrate exponential growth or decay, depending on whether \( k \) is positive or negative. The process of solving these by hand provides a foundational understanding applicable in various scientific fields.

Separation of variables is a common method used to solve first-order differential equations like the one given \( \frac{dy}{dt} = 6y \). The idea is to rearrange the equation so that each variable appears on one side of the equation only. This process starts by rewriting the differential equation as \( \frac{1}{y}dy = 6dt \). Next, each side of this equation is integrated separately. This gives us \( \int \frac{1}{y} \,dy = \int 6 \,dt \), leading to \( \ln|y| = 6t + C \). The constant \( C \) is introduced as part of the integration process. By solving these integral equations, the separation of variables effectively brings about a solution to the original differential equation by allowing direct integration and then finding a general solution that can be simplified further.

An initial value problem involves finding a specific solution to a differential equation that also satisfies an initial condition. This is crucial for differentiating among the infinitely many solutions a differential equation might have. Consider the solution \( y(t) = C'e^{6t} \) obtained from the integration step. The initial value problem is addressed by using the condition \( y(0) = 1 \). Applying this condition helps determine the value of the constant \( C' \). Substitute \( t = 0 \) into the equation: \( 1 = C' e^0 \), simplifying to \( C' = 1 \). Thus, the specific solution to the initial value problem is \( y(t) = e^{6t} \). This definitive solution respects both the differential equation and the given initial condition, ensuring that the entire problem is solved accurately.