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, you usually begin by finding a general solution of the differential equation.

Short answer

Question: Determine whether the following statements are true or false. Provide an explanation for your answer. a. The general solution of the differential equation y'(t) = 1 is y(t) = C. b. The differential equation y''(t) - y(t)y'(t) = 0 is a linear differential equation. c. To find the solution to an initial value problem, we first find the general solution and then use the initial conditions to determine the constant values.

Step-by-step solution

a. Statement a: Determine truth value and reasoning

The given differential equation is \(y'(t) = 1\). To find the general solution, we can integrate both sides of the equation to get \(y(t) = t + C\), where C is an integration constant. So, the statement is false; the correct general solution should be \(y(t)=t+C\).

b. Statement b: Determine truth value and reasoning

The given differential equation is \(y''(t) - y(t)y'(t) = 0\). Recall that a linear differential equation must have linear terms (with respect to dependent variable 'y' and its derivatives 'y''', 'y"", ...). Since the given equation contains a term \(y(t) y'(t)\) which is a product of 'y' and its first derivative, this equation is not linear. Therefore, the statement is false.

c. Statement c: Determine truth value and reasoning

To find the solution to an initial value problem, we usually start by finding the general solution to the differential equation. Then, we use the given initial conditions to determine the values of the integration constants. So, the statement is true and accurate.

Key concepts

General Solution

A general solution of a differential equation contains all possible solutions, taking into account constant integration values. This is seen in the exercise from the equation \( y'(t) = 1 \). By integrating, you derive \( y(t) = t + C \), where \( C \) is an arbitrary constant.

Here's why this is important:

• Integrating introduces a constant because differentiation of a constant yields zero, and thus does not affect derivative terms.
• This constant \( C \) represents any horizontal shift in the solution, providing a broad family of solutions.
• Hence, a general solution is necessary to encompass all specific solutions that might satisfy varied initial conditions.

It’s crucial to note that without the constant \( C \), you’d miss out on valid solutions. In this example, \( y(t) = t \) alone isn’t sufficient as a general solution since it doesn’t account for this constant shift or the full set of possible solutions.

Initial Value Problem

An initial value problem (IVP) in differential equations consists of finding a specific solution to a differential equation that not only satisfies the equation but also fulfills predetermined initial conditions. The typical process begins with finding the general solution. In our exercise, starting with the general solution \( y(t) = t + C \) is the first step.

Here's how the process unfolds:

• Identify the differential equation, complete with any initial conditions given, like \( y(0) = y_0 \).
• Find the general solution as discussed previously.
• Substitute the initial conditions into the general solution to solve for the constant value(s).
• The result is the specific or particular solution to the IVP.

The general solution provides a framework. However, the initial condition ensures the solution fits a specific scenario. Without this step, one cannot anchor the solution to conduct meaningful evaluations or predictions.

Linear Differential Equation

A linear differential equation is characterized by terms that are linear in the dependent variable and its derivatives. It's essential that the equation reads as a sum of terms, each involving the dependent variable or its derivatives raised to the first power, and possibly multiplied by a function of the independent variable.

Consider our exercise scenario: the equation \( y''(t) - y(t)y'(t) = 0 \) was evaluated for linearity. Here’s what this assessment reveals:

• In a linear differential equation, mixed products like \( y(t)y'(t) \) are not allowed because they are nonlinear. Each term should contain only one dependent variable or its derivatives.
• An equation like \( y''(t) - y(t)y'(t) = 0 \) doesn't satisfy the linearity requirement due to the multiplicative term \( y(t)y'(t) \).
• Such an equation is classified as nonlinear because this product introduces complexity in the relationship between \( y \) and its derivatives.

Recognizing an equation's linearity is crucial because it determines the methods available for finding solutions. Linear differential equations, given their simpler structure, often have more straightforward solution techniques.