Question

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: Determine the general solution of the first-order differential equation: \(y^{\prime}(t)=3+e^{-2t}\) Answer: The general solution of the given differential equation is: \(y(t) = 3t -\frac{1}{2}e^{-2t} + C\)

Step-by-step solution

Identify the given differential equation

The given first-order differential equation is: $$y^{\prime}(t)=3+e^{-2 t}$$

Integrate both sides with respect to \(t\)

To find the general solution, we will integrate both sides: $$\int y^{\prime}(t) dt = \int (3 + e^{-2t}) dt$$

Perform integration

Integration of the left side, we have: $$y(t) = \int(3 + e^{-2t}) dt$$ Now, we can integrate the right side by splitting the integral: $$y(t) = \int 3 dt + \int e^{-2t} dt$$ Integrate 3 with respect to t: $$\int 3 dt = 3t + C_1$$ Integrate \(e^{-2t}\) with respect to t by using substitution method (let \(u= -2t\) and \(dt = -\frac{1}{2}du\)). $$\int e^{-2t} dt = -\frac{1}{2}\int e^u du=-\frac{1}{2}e^{u}=-\frac{1}{2}e^{-2t}+C_2$$

Combine the integrals and add the arbitrary constants

Add the integrals and constants together to find the general solution: $$y(t) = 3t + C_1 -\frac{1}{2}e^{-2t}+C_2$$ Combine arbitrary constants \(C_1\) and \(C_2\) into a single arbitrary constant \(C\): $$y(t) = 3t -\frac{1}{2}e^{-2t} + C$$

Write the final general solution

The general solution of the given differential equation is: $$y(t) = 3t -\frac{1}{2}e^{-2t} + C$$

Key concepts

Integrals

The concept of integrals is fundamental in calculus and is the inverse process of differentiation. When we talk about integrating a function, we are looking to find a function whose derivative is the given function. This process is known as anti-differentiation.
In the exercise above, integration is used to solve the differential equation \(y'(t) = 3 + e^{-2t}\). We integrate both sides with respect to \(t\), which involves finding an antiderivative for each term on the right-hand side.

• The integral of \(3\) with respect to \(t\) is \(3t + C_1\), where \(C_1\) is an arbitrary constant.
• For \(e^{-2t}\), a substitution method is used. Substituting \(u = -2t\), the integral becomes \(-\frac{1}{2}e^{-2t} + C_2\), where \(C_2\) is another arbitrary constant.

Combining these integrals gives you the antiderivative needed to solve the differential equation, resulting in the general solution. Each integral reflects a summation aspect of calculus, indicating accumulated change or total quantity.

First-Order Differential Equations

First-order differential equations involve derivatives of the first degree—the simplest type of differential equations. These equations form the foundation upon which more complex equations are built.
In the given exercise, the differential equation \(y'(t) = 3 + e^{-2t}\) can be classified as a first-order differential equation because it includes the first derivative of \(y(t)\), that is, \(y'(t)\).

• The form of a typical first-order differential equation is \( \frac{dy}{dt} = f(t, y) \). However, in this exercise, \(y'(t) = 3 + e^{-2t} \) is a simpler representation because \(f(t,y)\) depends only on \(t\) and not on \(y\).
• Solving such equations often requires integration, as in this example, where integrating both sides with respect to \(t\) helps isolate \(y(t)\).

Such equations are crucial for modeling natural phenomena, where a rate of change in one variable is proportional to or influenced by another.

General Solution

The concept of a general solution is essential in the study of differential equations. It represents not just a single answer, but a family of solutions comprising all possible particular solutions. Each particular solution is obtained by assigning specific values to the arbitrary constants.
In our exercise, after integrating to solve the differential equation \(y'(t) = 3 + e^{-2t}\), we derive the general solution \(y(t) = 3t - \frac{1}{2}e^{-2t} + C\). Here, \(C\) is an arbitrary constant encompassing the constants \(C_1\) and \(C_2\) initially introduced during integration.

• The general solution highlights the infinite possibilities of particular solutions that depend on different initial conditions or additional information.
• Representing solutions in this flexible format aids in understanding how a system may behave under various circumstances.

The general solution emphasizes the broad applicability and adaptability of mathematical models in different contexts and scenarios.