91Ó°ÊÓ

A first-order linear differential equation is a type of differential equation that involves a function and its first derivative. These equations are expressible in the form \( \frac{dy}{dt} + P(t)y = Q(t) \). In most practical cases, such differential equations model real-world processes, like the cooling of an object to ambient temperature, which is our scenario here.
To solve a first-order linear differential equation, typical steps include:

• Rewriting the equation in its standard form.
• Using techniques such as separation of variables.

In our example, we started with the equation \( dT + 0.15(T - 10) dt = 0 \). By arranging this equation, we convert it into the form \( dT = -0.15(T - 10) dt \). This transformation allows further process steps like separation of variables.
Understanding these equations' structure helps in breaking them down logically to reach a solution.

Separation of variables is a method used to solve differential equations. It involves rearranging the equation to isolate each variable on opposite sides of the equation. This prepares the equation for integration. Let’s delve into how this works:
In the equation \( \frac{dT}{T - 10} = -0.15 dt \), variables \( T \) and \( t \) are separated. This separation allows us to integrate both sides independently. - **Left Side Integration:** Integrating \( \int \frac{1}{T-10} \, dT \), results in \( \ln |T-10| \). - **Right Side Integration:** Integrating \( \int -0.15 \, dt \), leads to \( -0.15t + C \), where \( C \) is a constant of integration.
With variables separated and integrated, you can solve for \( T \) by exponentiating both sides: \( |T-10| = e^{-0.15t + C} \). This method provides a straightforward way of working through the equation, enabling you to find \( T \) as a function of \( t \).

Initial conditions in differential equations are used to determine specific solutions from general solutions by providing specified values. These are crucial because they make an equation adaptable to real-world scenarios.
For our example, the initial condition is given as: when \( t=0 \), \( T=100 \). Applying this involves substituting these values into the equation to solve for any constants introduced during integration.

• Put \( t=0 \) and \( T=100 \) into \( C_1 e^0 = 90 \).
• This simplifies to \( C_1 = 90 \). Therefore, substituting back, we get \( T - 10 = 90 e^{-0.15t} \).

Initial conditions help create a meaningful specific solution from the general solution, as we solve for constants using the conditions given. This ensures the solution is uniquely aligned with the scenario we're modeling.