Question

For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. $$y^{\prime}(t)=\frac{t^{2}}{y^{2}+1} ; y(-1)=1, y(0)=0, y(-1)=-1$$

Short answer

#Answer# a. The general solution of the equation is \(y = \tan(-\frac{1}{t} + C)\). b. For the initial conditions: - \(y(-1) = 1\): The arbitrary constant is \(C_1 = 1 + \frac{\pi}{4}\). - \(y(0) = 0\): There is no valid solution. - \(y(-1) = -1\): The arbitrary constant is \(C_3 = 1 - \frac{\pi}{4}\). c. Sketch the solution curves on the provided graph using the following equations: - \(y = \tan(-\frac{1}{t} + 1 + \frac{\pi}{4})\) for the initial condition \(y(-1) = 1\). - \(y = \tan(-\frac{1}{t} + 1 - \frac{\pi}{4})\) for the initial condition \(y(-1) = -1\). Note: The actual sketching of the solution curves depends on the provided graph in your textbook or exercise sheet.

Step-by-step solution

Separate the Variables

To separate the variables in the equation \(y^{\prime}(t) = \frac{t^2}{y^2 + 1}\), we will divide both sides by \(t^2\), and then multiply both sides by \(y^2 + 1\). So, we get: $$\frac{y^{\prime}(t)}{t^2} = \frac{1}{y^2 + 1}$$ Now, we can write this as: $$\frac{dy}{dt}\frac{1}{t^2} = \frac{1}{y^2+1}$$

Integrate Both Sides

Now that we have separated the variables, we can integrate both sides with respect to their respective variables: $$\int \frac{dy}{y^2+1} = \int \frac{dt}{t^2}$$ The left-hand side is a common integral, which is the arctangent function, and the integral on the right-hand side is the power rule for integration: $$\arctan{y} = -\frac{1}{t} + C$$

Solve for the General Solution

In order to find the general solution, we need to solve for y: $$y = \tan{(-\frac{1}{t} + C)}$$ This is our general solution.

Solve for the Arbitrary Constant for Each Initial Condition

We will use the three initial conditions: \(y(-1) = 1, y(0) = 0,\) and \(y(-1) = -1\), and find the corresponding arbitrary constants. 1. Initial condition \(y(-1)=1\): $$1 = \tan(-\frac{1}{-1} + C_1)$$ $$\arctan{1} = C_1 - 1$$ $$\frac{\pi}{4} = C_1 - 1$$ $$C_1 = 1 + \frac{\pi}{4}$$ 2. Initial condition \(y(0)=0\): $$0 = \tan(-\frac{1}{0} + C_2)$$ Since the denominator is zero, this initial condition will not have a valid solution. Therefore, we cannot obtain an arbitrary constant \(C_2\). 3. Initial condition \(y(-1)=-1\): $$-1 = \tan(-\frac{1}{-1} + C_3)$$ $$\arctan{-1} = C_3 - 1$$ $$-\frac{\pi}{4} = C_3 - 1$$ $$C_3 = 1 - \frac{\pi}{4}$$

Sketch the Solution Curves

Using the graph of the general solution provided, we can sketch the solution curve for each initial condition except the second one where we couldn't obtain the arbitrary constant: 1. First initial condition \(y(-1) = 1\) → equation is \(y = \tan(-\frac{1}{t} + 1 + \frac{\pi}{4})\). Plot this equation on the graph. 2. Third initial condition \(y(-1) = -1\) → equation is \(y = \tan(-\frac{1}{t} + 1 - \frac{\pi}{4})\). Plot this equation on the graph. Note: Since we were not given an actual graph, the actual sketching of the solution curves cannot be demonstrated here. Please use the provided graph in your textbook or exercise sheet to sketch the solutions.

Key concepts

General Solution of Differential Equations

Understanding the general solution to a differential equation is crucial for students delving into calculus and differential equations. A differential equation represents a relationship between functions and their derivatives. The general solution refers to the solution that contains all possible specific solutions of the differential equation and typically includes one or more arbitrary constants, representing an infinite family of curves.

In the case of separable differential equations like the one in our exercise, the process involves rearranging the equation to isolate the derivatives of the function on one side and the function itself on the other side. This is achieved before integration. The aim is to express the solution in terms of the independent variable but with an arbitrary constant, which will be denoted as 'C' in the solution. This constant 'C' is what makes the solution 'general', as it can be adjusted to fit any particular, or 'specific', initial condition presented alongside the differential equation.

The step-by-step solution provided in the exercise walks through the separation of variables and subsequent integration to arrive at the general solution. It emphasizes the importance of being comfortable with integration techniques, as these are the tools that will allow us to solve for the general solution in most cases.

Initial Value Problem

An initial value problem is a specific type of differential equation problem that not only asks for the general solution but also necessitates the determination of the arbitrary constant based on given initial conditions. These initial conditions specify the value of the function, and sometimes its derivatives, at a particular point. This enables us to find the unique solution that passes through a given point on the curve, effectively 'pinpointing' one particular solution out of the infinite set described by the general solution.

For the exercise at hand, after finding the general solution, the next step is typically to plug in the initial conditions to solve for the constant 'C'. This constant, once determined, provides us with the exact solution that will satisfy both the differential equation and the initial condition. It is important to note that not all initial conditions may lead to a valid solution, as illustrated in the provided step-by-step solution, where one of the initial conditions led to a division by zero, indicating no solution in that particular case.

Arctangent Function Integration

Integration is a fundamental concept in calculus, essential for solving differential equations. In our exercise, integrating the left-hand side results in the arctangent function, expressed as \(\arctan{y}\). The arctangent function, also denoted as \(\tan^{-1}{y}\), is the inverse of the tangent function. It is helpful to remember that the integral of \(\frac{1}{y^2+1}\) with respect to \(y\) is \(\arctan{y}\), a fact that is commonly used in calculus.

The arctangent function emerges frequently when dealing with integrals that result from separable differential equations involving expressions like \(y^2 + 1\) in the denominator. It is important for students to become comfortable with this integration as it simplifies the process of finding solutions to a wide range of differential equations. Mastery of these integral functions, including arctangent, assures that students can tackle more complex problems efficiently.

Tangent Function Solutions

The tangent function, denoted \(\tan{y}\), plays a central role in the solutions to certain trigonometric integrals and differential equations. After integrating to obtain the arctangent function and finding the arbitrary constants using initial conditions, as we've done in our exercise, we then express \(y\) in terms of the \(\tan\) function to get the specific solutions.

Using the inverse properties of the arctangent and tangent functions allows us to solve for the dependent variable \(y\) in terms of the independent variable \(t\) and the constant derived from initial conditions. It is key to understand how these trigonometric functions behave and their relationship to one another, particularly as they describe waveforms and oscillatory behavior common in many physical and natural phenomena. Knowing how to manipulate and solve for these functions makes handling a wide range of mathematical problems possible, from the simple to the complex. Students should practice these solutions to gain fluency in their application to various scenarios.