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A separable differential equation is a type of differential equation that can be rearranged so that each variable and its derivative appears on opposite sides of the equation. In simpler terms, you can manipulate the equation so that one side contains only terms involving the dependent variable and its derivative, and the other side only involves the independent variable.

The key advantage here is that once the equation is separated, you can integrate both sides independently. This method is particularly appealing because it often reduces a complex problem into simpler, manageable parts. For example, in the given problem, we start with the differential equation \( w'(t) = a w(t)^{2/3} \).

By separating the variables, we rewrite it as \( \frac{dw}{w^{2/3}} = a \, dt \). This arrangement means we are set up to integrate both sides separately to gradually solve for the unknown function.

Integration techniques are strategies used to find the antiderivative or integral of a function. Depending on the type of function you're dealing with, different techniques may be applied, such as substitution, integration by parts, or recognizing simple power rules.

In our scenario, after separating the variables, we need to integrate \( \int w^{-2/3} \, dw \) and \( \int a \, dt \). Using a power rule for integration, we know the general form for integrating \( x^n \) is \( \frac{x^{n+1}}{n+1} \) plus a constant of integration.

For the left side, we get \( \frac{3}{1} w^{1/3} + C \), where \( C \) is the integration constant. The integration of the right side is straightforward as it involves the constant \( a \), resulting in \( at + C_2 \).

Using these techniques allows us to transition from a differential equation to an explicit formula, showing how the dependent variable changes over time.

Initial conditions are values given at the start of a problem to determine specific solutions of differential equations. They are crucial because they allow us to solve for the constant of integration, making the solution unique to a given scenario.

In this exercise, the initial condition is provided as \( w(0) = 1 \), which means the weight of the cell at time zero is 1 unit. After integrating, we had an expression involving an unknown constant: \( w^{1/3} = \frac{at}{3} + C \).

Substituting the initial condition into this equation gives us \( 1 = C^3 \), hence \( C = 1 \).

This step is crucial because it allows us to find the specific value of \( C \), ensuring that our final solution, \( w(t) = \left(\frac{at}{3} + 1\right)^3 \), accurately reflects the cell's weight over time given the starting condition.