Question

In each exercise, the unique solution of the initial value problem \(y^{\prime}+y=g(t), y(0)=y_{0}\) is given. Determine the constant \(y_{0}\) and the function \(g(t)\). \(y(t)=-2 e^{-t}+e^{t}+\sin t\)

Short answer

Answer: The constant \(y_0\) is -1 and the function \(g(t)\) is \(2 e^t + \cos t + \sin t\).

Step-by-step solution

Finding \(y(0)\)

To find \(y_0\), we simply need to evaluate the given function \(y(t)\) at \(t=0\). So, we have: \(y(0) = -2 e^{-0} + e^0 + \sin{0}\) Using the identities \(e^0 = 1\) and \(\sin{0} = 0\), we can write: \(y(0) = -2(1) + 1 + 0 = -1\) Hence, the constant \(y_0\) is -1.

Differentiating \(y(t)\)

Now we need to differentiate the given function \(y(t)\) with respect to \(t\). Let's find the derivative \(y'(t)\): \(y'(t) = \frac{d}{dt} \left(-2 e^{-t} + e^{t} + \sin t\right)\) Using the differentiation rules, we get: \(y'(t) = 2 e^{-t} + e^{t} + \cos t\)

Finding \(g(t)\)

Now we will substitute the obtained values of \(y(t)\) and \(y'(t)\) in the initial value problem equation: \(y'(t) + y(t) = g(t)\) So, we have: \((2 e^{-t} + e^{t} + \cos t) + (-2 e^{-t} + e^t + \sin t) = g(t)\) Combining terms, we can write: \(g(t) = 2 e^t + \cos t + \sin t\) Hence, the constant \(y_0 = -1\) and the function \(g(t) = 2 e^t + \cos t + \sin t\).

Key concepts

Initial Value Problem

An initial value problem is a common concept in differential equations.
It involves finding a function that satisfies a given differential equation along with a specific initial condition.
In our exercise, the initial condition is written as \(y(0) = y_0\).

To solve an initial value problem:

• Start by identifying the given differential equation and initial condition.
• Find a general solution to the differential equation.
• Use the initial condition to find the particular solution, which provides a unique function satisfying the problem.

These steps ensure that the solution not only meets the equation's requirements but also adheres to the starting value, making it relevant to the specific scenario modeled by the problem.

Solution Method

The solution method for an initial value problem often requires evaluating the given functions at specific points and finding derivatives.
In this exercise, the function \(y(t) = -2e^{-t} + e^{t} + \sin t\) was provided.

First, find the initial value \(y_0\) by substituting \(t = 0\) into \(y(t)\), which gives us \(-1\).
Then, differentiate the function \(y(t)\) with respect to \(t\):

• \(\frac{d}{dt}(-2e^{-t}) = 2e^{-t}\)
• \(\frac{d}{dt}(e^{t}) = e^{t}\)
• \(\frac{d}{dt}(\sin t) = \cos t\)

Adding these derivatives provides \(y'(t) = 2e^{-t} + e^{t} + \cos t\).
Use \(y'(t)\) and \(y(t)\) to identify \(g(t)\) in the equation \(y'(t) + y(t) = g(t)\).

This method carefully handles each component to satisfy both the differential equation and the initial condition.

Functional Differentiation

Functional differentiation is essential when working with differential equations.
It involves finding how a function changes as its variable changes, usually denoted by the derivative \(y'(t)\).

In our exercise, the given function \(y(t)\) involves:

• An exponential decay term \(-2e^{-t}\)
• An exponential growth term \(e^{t}\)
• A sine term \(\sin t\)

For each, apply the respective differentiation rules:

• The derivative of \(e^t\) and \(e^{-t}\) are themselves, modified by the chain rule when necessary.
• The derivative of \(\sin t\) is \(\cos t\).

By carefully differentiating each term, the function's derivatives reveal their rates of change, which is crucial when analyzing and solving differential equations.

Trigonometric and Exponential Functions

Trigonometric and exponential functions often appear in differential equations.
They have unique and predictable behaviors that make them useful in modeling processes that oscillate or grow exponentially.

In the exercise function \(y(t) = -2e^{-t} + e^{t} + \sin t\), each part serves a distinct purpose:

• \(-2e^{-t}\) represents exponential decay, decreasing as \(t\) increases.
• \(e^{t}\) shows exponential growth, increasing rapidly with \(t\).
• \(\sin t\) captures oscillation, as it varies between -1 and 1.

Both trigonometric and exponential functions provide familiar patterns in solutions.
Understanding their behaviors and mathematical properties assists students in visualizing and predicting the behavior of their differential equations.
These functions also simplify the process of differentiation and integration essential in solving these types of problems.