Question

Determine whether the following statements are true and give an explanation or counterexample. a. The general solution of the differential equation \(y^{\prime}(t)=1\) is \(y(t)=t\) b. The differential equation \(y^{\prime \prime}(t)-y(t) y^{\prime}(t)=0\) is second order and linear. c. To find the solution of an initial value problem, we usually begin by finding a general solution of the differential equation.

Short answer

#Answer# a. False b. False c. True

Step-by-step solution

Part a: Determine if the general solution is \(y(t) = t\)

Firstly, we need to integrate the given differential equation with respect to t: $$\int y'(t) dt = \int 1 dt$$ $$\implies y(t) = t + C$$ Here, C is any constant of integration. Therefore, the general solution is \(y(t) = t + C\). As we can see, the equation \(y(t)=t\) is part of the general solution but not the complete solution. Thus, statement "a" is false.

Part b: Determine if the given differential equation is second order and linear

The given differential equation is: $$y^{\prime \prime}(t)-y(t) y^{\prime}(t)=0$$ Clearly, it is a second-order differential equation, as the highest order of derivative is 2. Now we have to check if it is linear. The equation is linear if the coefficients of y and its derivatives are constants or functions independent of y. However, in this case, the term \(y(t)y'(t)\) shows that the coefficient of \(y'(t)\) is a function of y. Therefore, the given differential equation is not linear. Thus, statement "b" is false.

Part c: Check if finding the general solution is the initial step in solving an initial value problem

To solve an initial value problem, one usually follows these steps: 1. Find the general solution of the given differential equation. 2. Use the initial conditions to determine the constants in the general solution. 3. Write the particular solution that satisfies the initial conditions. As it is clear that finding the general solution of the differential equation is indeed the first step in solving an initial value problem, statement "c" is true.

Key concepts

General Solution of Differential Equation

When encountering a differential equation, the ultimate goal is to find a function that satisfies it. A general solution of a differential equation is an expression containing one or more constants whose specific values have not yet been determined. These constants, often represented as C, which appear due to the process of integration, can take on any real value and allow the general solution to represent a family of functions.

In the context of the exercise provided, for the first statement, the differential equation is very simple: y'(t)=1, indicating that the slope of the function y(t) is always 1, regardless of t. Integrating this equation gives us $$y(t)= \begin{cases} \text{General solution:} & t+C \text{Exercise's assertion:} & t\text{not accounting for } C \text{(false statement)} \text{.}\text{Proof by contradiction to improve understanding:} \text{Assuming} & y(t)=t \text{ is the complete solution,} \text{ contradicts the} \text{ existence of the constant of integration } C \text{ that arises from indefinite integration.}\text{ In learning about general solutions}, it is essential to acknowledge that incorporation of the constant C is non-negotiable, unless further conditions are specified that allow for its value to be determined.

When students attempt to understand this concept, they should verbally re-articulate that the general solution encompasses all potential specific solutions, and the constant C provides the necessary flexibility to adapt to various scenarios or initial conditions.

Second Order Differential Equation

Differential equations are classified based on various attributes, including their order, which is determined by the highest derivative present in the equation. A differential equation is said to be of second order if the highest derivative that occurs in the equation is the second derivative of the unknown function. These types of equations often describe systems with acceleration, like the motion of springs or electrical circuits.

The example given in the exercise, $$y^{\text{''}}(t) - y(t)y'(t) = 0$$is, without a doubt, a second order differential equation because the second derivative, y''(t), is present. However, the catch is that it's not linear because of the term y(t)y'(t) which introduces a product of the unknown function y(t) and its first derivative y'(t), making it non-linear. A linear second order differential equation would have the form $$y^{\text{''}}(t) + p(t)y'(t) + q(t)y(t) = g(t)$$where p(t), q(t), and g(t) are functions of t, or constants, but do not include the function y(t) itself or its derivatives.
When focusing on solving these equations, it's central to differentiate between linear and non-linear equations, as the methods to solve each type vary greatly. Students should practice identifying the characteristics - such as linearity and order - to better approach solving these equations with appropriate methods.

Initial Value Problem

In the realm of differential equations, an initial value problem is a differential equation accompanied by a condition that specifies the value of the unknown function and its derivatives at a given point. This point is usually given in terms of one of the independent variables, most commonly time, and these conditions allow us to pinpoint a specific solution from the general solution.Although finding the general solution is usually the first step, as step three from the exercise solution correctly states, it's critical to remember that this is only an intermediate part of the process. The complete steps include using the initial condition to determine the values of the constants present in the general solution. Specifically, the exercise proves the process as follows: 1. General solution found. 2. Initial conditions applied. 3. Constants determined for the particular solution that aligns with initial conditions.

For a successful execution, students must understand that these steps are sequential and each step relies on the completion of the one prior. Furthermore, practice in solving such problems enhances comprehension of the underlying physical or geometric phenomena described by the differential equations, and reinforces the significance of the initial values.