91Ó°ÊÓ

A differential equation is separable if it can be written such that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the other side. In the original equation \( \frac{d y}{d t} = \frac{t+1}{t y} \), we separate the variables by multiplying both sides by \( y \, dt \):

- We get: \( y \, dy = \frac{t+1}{t} \, dt \).

At this point, each side depends on only one variable, allowing us to integrate. Understanding this concept is crucial for solving these types of equations, as it simplifies the process and makes it possible to integrate both sides independently.

Initial conditions provide specific values for the variables in a differential equation, typically given as \( y(t_0) = y_0 \). They are used to find the exact solution out of all possible solutions to a differential equation. In our exercise, the initial condition is \( y(1) = -3 \).

- This means when \( t = 1 \), \( y = -3 \).

Applying this, we substitute \( t = 1 \) and \( y = -3 \) into our integrated equation \( \frac{y^2}{2} = t + \ln|t| + C \). This helps us solve for the constant \( C \). Without the initial condition, we wouldn't have a specific solution, just a family of potential solutions.

Integration is the process of finding the antiderivative of a function. This step is crucial in solving separable differential equations. After separating variables and setting up the equation \( \int y \, dy = \int \left( 1 + \frac{1}{t} \right) \, dt \), we integrate both sides:

- For \( \int y \, dy \), we use the power rule: \( \frac{y^2}{2} \).
- On the right: \( \int 1 \, dt = t \) and \( \int \frac{1}{t} \, dt = \ln|t| \), giving the integral \( t + \ln|t| \).

Combining these, we get: \( \frac{y^2}{2} = t + \ln|t| + C \), where \( C \) is the constant of integration. Integration takes the rates of change given by the differential equation and expresses them as functions of \( t \).

After integrating, we generally face unknown constants that need to be determined based on initial conditions. In our example, we had: \( \frac{y^2}{2} = t + \ln|t| + C \). Using the initial condition \( y(1) = -3 \), we substitute \( t = 1 \) and \( y = -3 \) into this integrated equation:

- This gives us: \( \frac{(-3)^2}{2} = 1 + \ln|1| + C \), which simplifies to \( \frac{9}{2} = 1 + 0 + C \), hence \( C = \frac{7}{2} \).

We substitute \( C \) back into the general solution: \( \frac{y^2}{2} = t + \ln|t| + \frac{7}{2} \). Solving for \( y \) yields: \( y = \pm \sqrt{2t + 2\ln|t| + 7} \). Since the initial condition specifies \( y(1) = -3 \), we select the negative root: \( y = -\sqrt{2t + 2\ln|t| + 7} \). This is our final solution.