Question

The solution of the differential equation satisfying initial condition \(y(0)=1\) is given. \(y^{\prime}=-y+2 ; \quad y(t)=2-e^{-t}\)

Short answer

Answer: Yes, the given solution satisfies both the differential equation and the initial condition, confirming that it is the correct solution for the initial value problem.

Step-by-step solution

Find the derivative of the given solution

Differentiate the given function, \(y(t) = 2-e^{-t}\), with respect to \(t\). Using the chain rule for the differentiation of the exponential function, we get: \(y'(t) = \frac{d}{dt}(2-e^{-t}) = 0 - \frac{d}{dt}(e^{-t}) = 0 + e^{-t}\), as the derivative of the exponential function \(e^{at}\) with respect to \(t\) is \(ae^{at}\).

Substitute the given solution and its derivative into the differential equation

Insert \(y(t) = 2 - e^{-t}\) and \(y'(t) = e^{-t}\) into the given differential equation, \(y' = -y + 2\): \(e^{-t} = -(2 - e^{-t}) + 2\)

Check if the differential equation is satisfied

Simplify the equation in step 2 to see if both sides are equal: \(e^{-t} = -2 + e^{-t} +2\) \(e^{-t} = e^{-t}\) The given solution does satisfy the differential equation.

Check if the initial condition is satisfied

Now, plug in the initial condition, \(y(0) = 1\), into the given solution, \(y(t)=2-e^{-t}\) and check if it's true: \(y(0) = 2 - e^{-0} = 2 - 1 = 1\) The given solution does satisfy the initial condition. Since the given solution satisfies both the differential equation and the initial condition, it is the correct solution for the given initial value problem.

Key concepts

Initial Value Problems

An Initial Value Problem in the context of differential equations involves finding a function that not only satisfies a given differential equation but also meets a specific initial condition. The initial condition provides a starting value, typically expressed as the value of the function at a particular point.Initial value problems are essential for determining a unique solution to a differential equation. Without an initial condition, a differential equation might have infinitely many solutions. The initial condition "anchors" the function, ensuring that the solution curve passes through a specified point, making the solution applicable to real-world scenarios.For example, in the exercise above, the differential equation is \( y' = -y + 2 \) with the initial condition \( y(0) = 1 \). Solving this problem requires finding a function \( y(t) \) that satisfies both the equation and the condition. From the steps shown, we see that the function \( y(t) = 2 - e^{-t} \) satisfies the differential equation when substituted back into the equation and meets the initial condition of \( y(0) = 1 \). This demonstrates how initial conditions are vital for defining a specific solution from a family of potential solutions.

Exponential Functions

Exponential functions have the form \( y = Ce^{kt} \), where \( C \) is a constant, \( e \) is the base of the natural logarithm, and \( k \) is a constant that impacts the growth or decay rate of the function. Exponential functions are significant in many areas of mathematics because they model growth and decay processes, such as population growth, radioactive decay, and continuous compounding interest.In the given exercise, the solution \( y(t) = 2 - e^{-t} \) includes an exponential function \( e^{-t} \). The exponential component \( e^{-t} \) describes decay over time, as the negative sign in the exponent causes the value to decrease as \( t \) increases.The behavior of exponential functions is particularly useful in solving differential equations, as they frequently arise when the rate of change of a quantity is proportional to the quantity itself. Understanding the properties of exponential functions is crucial for tackling problems like the one in this exercise, where the solution involves exponential decay.

Chain Rule in Calculus

The chain rule is a powerful tool in calculus used to differentiate composite functions. When you have a function within another function, the chain rule makes it possible to find the derivative of this complex structure.The mathematical representation of the chain rule is given by:- If \( y = f(g(t)) \), then the derivative of \( y \) with respect to \( t \) is \( y' = f'(g(t)) \, g'(t) \).In the exercise's solution, the chain rule is applied when differentiating the function \( y(t) = 2 - e^{-t} \). The exponential component \( e^{-t} \) is a composite function comprising the outer function \( e^u \) and the inner function \( u = -t \). Applying the chain rule, the derivative of \( y(t) \) with respect to \( t \) is calculated by:- First differentiating the outer function, \( rac{d}{du}(e^u) = e^u \), evaluates to \( e^{-t} \) when substituting back \( u = -t \).- The derivative of the inner function, \( u = -t \), is \( rac{d}{dt}(-t) = -1 \).- Therefore, the derivative is \( e^{-t} \cdot (-1) = -e^{-t} \) which simplifies to \( e^{-t} \) due to the subtraction involved in the original differentiating step.The chain rule thus ensures accuracy when derivatives involve multiple layers of functions, as shown in this solution process.