Question

Finding general solutions Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots\) to denote arbitrary constants. y^{\prime}(t)=3+e^{-2 t}

Short answer

Question: Find the general solution of the given first-order linear differential equation y'(t) = 3 + e^{-2t}. Answer: The general solution of the given differential equation is y(t) = 3t - (1/2)e^{-2t} + C, where C is an arbitrary constant representing all possible solutions.

Step-by-step solution

Identify the given differential equation

We have a first-order linear differential equation of the form: y'(t) = 3 + e^{-2t}

Find the antiderivative of the right side function

To find the antiderivative of the right side function 3 + e^{-2t}, we will break it down into two parts (antiderivative of 3 and antiderivative of e^{-2t}) and then sum up both results. The antiderivative of 3 with respect to t is: 3t + C_1, where C_1 is an arbitrary constant. The antiderivative of e^{-2t} with respect to t is: \frac{-1}{2}e^{-2t} + C_2, where C_2 is an arbitrary constant. Now, we will sum up both antiderivatives to obtain the antiderivative of 3 + e^{-2t}: (3t + C_1) + (\frac{-1}{2}e^{-2t} + C_2)

Simplify the antiderivative and write the general solution

Combining the terms from Step 2, we get the general solution for the given differential equation as: y(t) = 3t - \frac{1}{2}e^{-2t} + C Here, C = C_1 + C_2 is an arbitrary constant that represents all possible solutions.

Key concepts

General Solution

A general solution of a differential equation is a solution that incorporates all possible solutions for a differential equation. When solving differential equations, you will often find arbitrary constants, such as \( C \) or \( C_1, C_2 \), which represent these possible variations.
The reason we include these constants is to account for the infinite number of solutions that can satisfy the equation, depending on initial conditions or specific parameters set by the problem.
When we found the general solution for the differential equation \( y'(t) = 3 + e^{-2t} \), we ended up with:

• \( y(t) = 3t - \frac{1}{2}e^{-2t} + C \)

This equation is the general solution because the term \( C \) allows it to adjust to any initial conditions, which would lead to the particular solution.

Antiderivative

The antiderivative, sometimes referred to as an indefinite integral, is the process of determining a function \( F(t) \) whose derivative is the given function \( f(t) \). This process is vital when solving differential equations because it helps to find the function \( y(t) \) that solves the equation.
For our problem, the right side of the equation, \( 3 + e^{-2t} \), was divided into separate functions:

• The antiderivative of \( 3 \) is \( 3t + C_1 \).
• The antiderivative of \( e^{-2t} \) is \( -\frac{1}{2}e^{-2t} + C_2 \).

After finding these antiderivatives, we summed them to form the overall antiderivative:

• \( (3t + C_1) + (-\frac{1}{2}e^{-2t} + C_2) \)

Finally, we simplified this expression into \( y(t) = 3t - \frac{1}{2}e^{-2t} + C \), where \( C = C_1 + C_2 \). Both constants combined into a single constant \( C \) represent any constant shift in the family's functions.

First-Order Linear Differential Equation

First-order linear differential equations are a specific type of differential equations characterized by the highest derivative being the first derivative. These equations generally have the form:

• \( y'(t) + p(t) y(t) = q(t) \)

In our example, \( y'(t) = 3 + e^{-2t} \) is a simple first-order linear differential equation where the \( y(t) \) term coincidentally does not explicitly appear. This makes it a straightforward case to solve.
Such equations are fundamental because they provide a bridge between pure theory and practical modeling in areas like physics and engineering, describing how systems evolve over time. In practice, solving first-order linear differential equations usually involves:

• Identifying the equation's form.
• Extracting or reorganizing terms to find the antiderivative.
• Arriving at a general solution that uses arbitrary constants.
• Optionally, applying initial conditions or constraints to identify a particular solution.