91Ó°ÊÓ

When faced with a differential equation like \( \frac{dP}{dt} = 0.2t \), it's crucial to determine whether the equation is separable. A separable differential equation is one that can be rearranged to isolate the derivative on one side, separating the variables. In this case, we strive to express the equation

• in a form where all terms involving \( P \) are on one side
• and all terms involving \( t \) are on the other.

What makes separable differential equations special is that both sides of the equation can be integrated separately.
For our equation, \( \frac{dP}{dt} = 0.2t \) is already quite simple. By integrating both sides with respect to their respective variables, we set the stage for finding the solution. This technique often proves handy in recognizing the nature of the potential solution.

Verifying a solution involves substituting this solution back into the original equation to see if it holds true. This is a crucial step to validate whether a proposed function is indeed a solution. In the original exercise, the given solution was \( P = Be^{0.2t} \). This equation seems plausible due to the use of exponential terms which are common in the solutions to differential equations.Let's break down the verification process:

• Take the derivative of the proposed solution \( P = Be^{0.2t} \). With the chain rule, the derivative becomes \( \frac{dP}{dt} = 0.2Be^{0.2t} \).
• Compare this derivative with the original differential equation \( \frac{dP}{dt} = 0.2t \).

Since these expressions are not equal unless under special conditions, it signals that the initial solution proposal does not satisfy the differential equation. Thus, verifying solutions is a vital step in ensuring accuracy in solving differential equations.

When solving differential equations, especially separable ones, integration is a key tool. Integration helps us move from the rate of change (derivative) to the function itself. For the equation \( \frac{dP}{dt} = 0.2t \), we approach integration by tackling each side separately.Let's walk through this:

• On the left side, \( \int dP = P + C \) since integrating a basic differential \( dP \) yields the function \( P \) plus a constant \( C \).
• On the right side, we have \( \int 0.2t \, dt = 0.1t^2 + C \). This involves basic power rule integration where \( n \)-th power of \( t \) changes by adding 1 to the exponent and dividing by the new exponent.

Putting it together, the correct solution to the original equation comes out as \( P = 0.1t^2 + C \). This solution displays how powerful integration can be, transforming a dynamic equation into a clear expression of \( P \) in terms of \( t \).