Problem 36 \((t+y) d t-t d y=0, y(1)=1\)... [FREE SOLUTION] (2024)

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First-order linear differential equations involve derivatives of the first order. The general form of such an equation is \({a(t) y' + b(t) y = c(t)}\). These equations are critical in various applications across physics, engineering, and other sciences. In our exercise, the equation given is \({(t + y) dt - t dy = 0}\), which can be simplified and rearranged to fit this form. Recognizing a first-order linear differential equation allows us to use specific techniques to solve it.

To identify this type of equation, ensure it has the derivative of the dependent variable (e.g., \({y'}\)) and can be written in the specified form. This classification is the first essential step for solving the problem systematically.

Separation of variables is a method used to solve differential equations. Here, we rearrange terms so variables are on different sides of the equation. This process makes it easier to integrate each side separately.

In our exercise, after rewriting \({(t + y) dt = t dy}\), we divide both sides by \({t (t + y)}\) to separate variables: \({\frac{dt}{t} = \frac{dy}{t + y}}\).

By isolating the variables, we can integrate each side with respect to its variable, moving us closer to the solution. This technique is extremely powerful for solving differential equations that can be separated in this manner.

Integration is the process of finding integrals, helping us solve differential equations. Once the variables are separated, we integrate each side regarding their respective variables.

For our problem, we integrate \({ \frac{1}{t} dt = \frac{1}{t + y} dy }\). These integrals yield \({ \text{ln} |t| = \text{ln} |t + y| + C }\), where \({ C }\) is the integration constant.

Integration simplifies the relationship between the variables, reducing the problem to algebraic techniques to solve for the function. Ensuring proper integration is essential for accurately solving differential equations.

Initial conditions specify the value of the solution at a particular point. They are crucial for finding the specific solution to a differential equation, given an infinite number of possible functions fitting the general solution.

In our exercise, we use the initial condition \({ y(1) = 1 }\). By substituting this into our general solution, we solve for the constant \({ k }\). Here, \({ 1 = \frac{1 \times (1 - k)}{k} }\) leads to \({ k = 2 }\).

Applying initial conditions ensures our solution fits the specific requirements of the problem's scenario. This step finalizes our solution: \({ y = -\frac{t}{2} + 1 }\). Properly applying initial conditions is the final crucial step in solving differential equations.