91Ó°ÊÓ

The technique of separation of variables is commonly used to solve differential equations. It involves rearranging the equation so that all terms involving one variable are on one side, and all terms involving the other variable are on the other side. This is an essential step because it allows us to integrate each side separately.

In the exercise, we begin with the differential equation \( \frac{dP}{dt} = \sqrt{Pt} \). To separate the variables, we rearrange it to \( \frac{dP}{\sqrt{P}} = \sqrt{t} \ dt \). Here's why this works:

• We isolate \( dP \) and terms involving \( P \) on one side.
• We isolate \( dt \) and terms involving \( t \) on the other side.

Now, each side of the equation is ready for integration. This technique is useful for solving many types of differential equations and is a fundamental concept to master in calculus.

Initial conditions are specific values provided for a function and its variables, usually at a starting point, that help find a unique solution to a differential equation. They are critical because, without initial conditions, a differential equation could have infinitely many solutions.

In our example, the initial condition is \( P(1) = 2 \). This means that when \( t = 1 \), the value of \( P \) must be 2. This condition helps determine the constant of integration that appears after we integrate the separated equation.

In Step 4 of the solution, we used this initial condition to solve for \( C \). We substitute \( t = 1 \) and \( P = 2 \) into the equation to find \( C \), ensuring our solution matches the given initial scenario. This specification is vital in real-world problems where specific starting values guide the expected result.

Integration is the process of finding the antiderivative of a function. This is a crucial step when solving differential equations, especially after separating variables. Integration turns the differential equation into an algebraic equation that reveals the relationship between the variables.

In our solution, we need to integrate both sides of the separated equation \( \int \frac{1}{\sqrt{P}} \, dP = \int \sqrt{t} \, dt \). The integration yields:

• The left side becomes \( 2\sqrt{P} + C_1 \).
• The right side becomes \( \frac{2}{3}t^{3/2} + C_2 \).

The constants \( C_1 \) and \( C_2 \) are the constants of integration, which are crucial because they account for the indefinite nature of antiderivatives. By applying the initial condition, we solve for the combined constant \( C = C_2 - C_1 \), making it possible to find the specific solution that satisfies our original problem.