Question

Consider the differential equation \(y^{\prime}=|y|\). (a) Is this differential equation linear or nonlinear? Is the differential equation separable? (b) A student solves the two initial value problems \(y^{\prime}=|y|, y(0)=1\) and \(y^{\prime}=y\), \(y(0)=1\) and then graphs the two solution curves on the interval \(-1 \leq t \leq 1\). Sketch what she observes. (c) She next solves both problems with initial condition \(y(0)=-1\). Sketch what she observes in this case.

Short answer

Answer: The given differential equation is nonlinear and separable. Question: What are the solutions to the initial value problems IVP1, IVP2, IVP3, and IVP4? Answer: The solutions to the initial value problems are as follows: - IVP1: \(y(t) = e^{t}\) - IVP2: \(y(t) = e^{t}\) - IVP3: \(y(t) = -e^{-t}\) - IVP4: \(y(t) = -e^{t}\)

Step-by-step solution

(a) Linear or Nonlinear and Separable or Nonseparable

To determine if the differential equation is linear or nonlinear, we look at the differential equation \(y^{\prime}=|y|\). As the absolute value of \(y\) introduces a nonlinearity (due it's piecewise definition), this differential equation is nonlinear. Now let's check if the differential equation is separable, i.e. if we can write it in the form \(\frac{dy}{dt} = f(y)g(t)\). In this case, we have \(\frac{dy}{dt} = |y|\) and \(|y|\) only depends on \(y\) and not both \(y\) and \(t\). Thus, we can say that the differential equation is separable.

(b) IVP1: \(y^{\prime}=|y|\), \(y(0)=1\) and IVP2: \(y^{\prime}=y\), \(y(0)=1\)

Initially, we will solve both initial value problems: 1. IVP1: For \(y^{\prime}=|y|\), \(y(0)=1\), we integrate by separation of variables and depending on the value of \(y\). Case 1 (y>=0): \(\frac{dy}{dt}=y\), integrating we obtain: \(y(t) = ke^{t}\), where k is a constant, and using initial condition \(y(0) = 1\), we have \(k = 1\), hence \(y(t) = e^{t}\) Case 2 (y<=0): \(\frac{dy}{dt}=-y\), integrating we obtain: \(y(t) = ke^{-t}\), but since the function has to be continuous and initial condition is \(y(0) = 1\), this case will not apply. Here is a rough sketch of IVP1; the curve starts at the initial value 1, and increases exponentially with the function \(y(t) = e^{t}\) 2. IVP2: For \(y^{\prime}=y\), \(y(0)=1\), this one is simpler: \(y(t) = e^{t}\), and it is exactly the same as case 1 from IVP1. These two solution curves will coincide because in both initial value problems, the solutions are actually the same on the interval \(-1 \leq t \leq 1\).

(c) IVP3: \(y^{\prime}=|y|\), \(y(0)=-1\) and IVP4: \(y^{\prime}=y\), \(y(0)=-1\)

Now, let us solve both problems with initial condition \(y(0)=-1\). 1. IVP3: For \(y^{\prime}=|y|\), \(y(0)=-1\), similarly, we integrate by separation of variables: Case 1 (y>=0): \(\frac{dy}{dt}=y\), integrating we obtain: \(y(t) = ke^{t}\), but since the function has to be continuous and initial condition is \(y(0) = -1\), this case will not apply. Case 2 (y<0): \(\frac{dy}{dt}=-y\), integrating we obtain: \(y(t) = ke^{-t}\), and using initial condition \(y(0) = -1\), we have \(k=-1\), hence \(y(t) = -e^{-t}\) Here is a rough sketch of IVP3; the curve starts at the initial value -1 and increases to 0 as the function \(y(t) = -e^{-t}\) 2. IVP4: For \(y^{\prime}=y\), \(y(0)=-1\), we have: \(y(t) = -e^{t}\) Here is a rough sketch of IVP4; the curve starts at the initial value -1 and decreases exponentially with the function \(y(t) = -e^{t}\) For the interval \(-1 \leq t \leq 1\), IVP3 shows a curve starting from \(y(-1) = -e\) and increasing towards y = 0. On the other hand, IVP4 shows a curve which starts at the same initial value \(y(-1)=-e\), however, it decreases exponentially as the function \(y(t) = -e^{t}\).

Key concepts

Nonlinear Differential Equations

Differential equations can either be linear or nonlinear, and understanding the difference is key to solving them. Linear differential equations have solutions that can be superimposed, meaning that if two solutions of the equation exist, their sum is also a solution. Nonlinear differential equations do not have this linear property and often involve terms like powers or products of the function and its derivatives.

In our example, the differential equation provided is \(y^{\prime} = |y|\). This is a nonlinear differential equation due to the presence of the absolute value function \(|y|\). Since the absolute value function modifies the graph's behavior depending on the sign of \(y\), it introduces a piecewise aspect to the solution. Hence, nonlinear differential equations can exhibit more complex and varied types of behaviors than linear ones.

Understanding the nature of nonlinear equations helps in choosing appropriate solution methods and anticipating potential complications.

Separable Differential Equations

A separable differential equation is a type of differential equation that can be rewritten such that all terms involving one variable and its derivative are on one side of the equation, and all terms involving the other variable are on the opposite side. This allows us to integrate each side independently.

In the example \(y^{\prime} = |y|\), we can separate the variables into \(\frac{dy}{|y|} = dt\). This separation is possible because the equation only dependent on \(y\), not on \(t\). With the separated terms, we can integrate both sides to find the solution to the equation.

Separable differential equations are often more straightforward to solve than other types, making them a common starting point for solving differential equations. Using the technique of separation of variables, even nonlinear differential equations can sometimes be simplified and solved.

Initial Value Problems

An Initial Value Problem (IVP) in the context of differential equations is a problem where the solution is required to satisfy a differential equation and an initial condition. The initial condition provides the specific value of the function at a particular point, usually located at the beginning of the interval of interest. This condition allows us to determine a specific solution out of the potentially infinite set of solutions that a differential equation could have.

For instance, given \(y^{\prime} = |y|\) and an initial condition like \(y(0) = 1\) or \(y(0) = -1\), we aim to find a solution that specifically passes through the point \((0, 1)\) or \((0, -1)\), respectively. The inclusion of initial conditions helps us solve for the constant \(k\) that arises during integration, thus allowing us to pinpoint the exact curve which represents the behavior of the function over time.

Initial value problems are foundational in the study of differential equations, particularly in physics and engineering, where they describe systems evolving over time with known starting states.