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A differential equation is an equation that relates a function with its derivatives.
These types of equations are crucial in modeling real-world phenomena in fields like physics, engineering, and economics.
The given problem involves solving the differential equation: \[ y^{\text{'}} + y^{2} = y \]
where \( y(t) \) is the function we need to find.
In this problem, the function \( f(t) = \left(e^{-t} + 1 \right)^{-1} \) is given, and we are asked to show it satisfies the differential equation.To solve the differential equation, we calculated the derivative of \( f(t) \) and substituted it back into the equation.
The differential equation helps us understand how the function \( y(t) \) changes over time. This equation tells us that the change in \( y \) (represented by \( y^{\text{'}} \)) is related to the value of \( y \) and its square.

The chain rule is a fundamental technique in calculus for finding the derivative of composite functions.
When we have a function composed of other functions, we use the chain rule to differentiate it step-by-step.
In the solution process, we decomposed our function into simpler parts to apply the chain rule effectively.
Let \( u = e^{-t} + 1 \). Then, \( y = u^{-1} \), making the derivative \( y' \) the product of the derivatives of \( y \) concerning \( u \) and \( u \) concerning \( t \): \(y' = \frac{dy}{du} \cdot \frac{du}{dt} \).Breaking this process down makes the calculation manageable:

• First, we computed \( \frac{du}{dt} = -e^{-t} \).
• Then, we computed \( \frac{dy}{du} = -u^{-2} \).

Combining these results gave us
\(y' = -u^{-2} \cdot (-e^{-t}) = e^{-t} u^{-2} \), which was then simplified further.
The chain rule is like peeling an onion: it allows us to handle complex outer and inner layers one at a time.

Initial conditions specify the value of the function at a particular point, providing a starting point for solving differential equations.
They are essential because they help determine the unique solution to our problem.
In our exercise, the initial condition is given as \( y(0) = \frac{1}{2} \). This means that at time \( t = 0 \), our function \( y \) should equal \( \frac{1}{2} \).
When we check this against our solution
\( y(0) = \left(e^{0} + 1 \right)^{-1} = \frac{1}{2} \),
it confirms that our function satisfies the initial condition.
Verifying initial conditions is crucial:

• It ensures the solution is accurate.
• It provides a specific solution path.

Think of initial conditions as the coordinates of a starting point on a map: they tell us exactly where to begin our journey.

Derivatives measure how a function changes as its input changes.
They are a core concept in calculus, exploring rates of change and slopes of curves.
In our problem, finding the derivative of the function \( f(t) = \left(e^{-t} + 1 \right)^{-1} \) was essential to solve the differential equation.
Here are the key steps we followed:

• We used the chain rule to find the derivative step-by-step
• We computed \( \frac{du}{dt} \) and \( \frac{dy}{du} \)
• We substituted back into our original function

After computing, we obtained \( y' = e^{-t} (e^{-t} + 1)^{-2} \).
This derivative tells us how the function \( y \) changes with respect to \( t \), providing critical insights into the function's behavior.
Derivatives are like the speed of a car, helping us understand how quickly or slowly a function's value is changing at any given point.