91Ó°ÊÓ

Separation of variables is a common and powerful method 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 opposite side. This allows for easier integration and simplification. In our example, the original equation was given as \( \frac{1}{z} \frac{dz}{dt} = 5 \). To start the process, you switch it to \( \frac{dz}{dt} = 5z \).

By dividing both sides by \( z \), and multiplying by \( dt \), we arrive at \( \frac{dz}{z} = 5 \, dt \). This format is ideal for solving because each side is now isolated to involve only one variable, allowing us to integrate each side independently.

An initial condition is an essential part of finding a specific solution to a differential equation. It gives us the particular value that the function must satisfy at a specific point. For our example, the initial condition is \( z(1) = 5 \).

Applying this condition involves substituting the known values back into the general solution to pinpoint any constants or unknowns. It narrows down the infinite number of possible curves described by the differential equation to just one, ensuring the solution meets the specific scenario described in the problem.

Integration is a key step in solving differential equations, as it enables us to reverse the process of differentiation. After separating variables, each side of the equation is integrated independently. In our case, the integration results in \( \int \frac{1}{z} \, dz = \ln |z| \) on the left side, and \( \int 5 \, dt = 5t + C \) on the right side, where \( C \) is the constant of integration.

Finding these integrals gives us the general form of the solution, which includes the logarithmic expression transformed during the integration. Knowing \( C \) will be determined with the initial condition, this step sets up your equation for further simplification and solving.

Solving a differential equation means finding a function that satisfies the equation given the conditions imposed by the problem. In our situation, this involved rearranging, integrating, and simplifying the equation before applying the initial condition to find any constants.

In our solution, post integration and applying the condition \( z(1)=5 \), we discovered \( A = 5e^{-5} \). This gave us the particular solution \( z = 5e^{5(t-1)} \). Each step narrows down the possibilities until a unique solution that satisfies the original differential equation and the given conditions is reached. Understanding these steps is fundamental in handling such problems effectively.