Question

To establish the uniqueness part of Theorem \(2.1\), assume \(y_{1}(t)\) and \(y_{2}(t)\) are two solutions of the initial value problem \(y^{\prime}+p(t) y=g(t), y\left(t_{0}\right)=y_{0} .\) Define the difference function \(w(t)=y_{1}(t)-y_{2}(t)\). (a) Show that \(w(t)\) is a solution of the homogeneous linear differential equation \(w^{\prime}+p(t) w=0\). (b) Multiply the differential equation \(w^{\prime}+p(t) w=0\) by the integrating factor \(e^{P(t)}\), where \(P(t)\) is defined in equation (11), and deduce that \(e^{P(t)} w(t)=C\), where \(C\) is a constant. (c) Evaluate the constant \(C\) in part (b) and show that \(w(t)=0\) on \((a, b)\). Therefore, \(y_{1}(t)=y_{2}(t)\) on \((a, b)\), establishing that the solution of the initial value problem is unique.

Short answer

Question: Prove the uniqueness of the solution to the given initial value problem. Answer: By defining the difference function \(w(t) = y_1(t) - y_2(t)\) and showing that \(w(t)\) is a solution to the homogeneous linear differential equation, we then multiplied by the integrating factor and deduced that \(e^{P(t)}w(t) = C\). Evaluating the constant \(C\) using the initial conditions and showing that \(w(t) = 0\), we concluded that the initial value problem has a unique solution, as \(y_1(t) = y_2(t)\) on \((a, b)\).

Step-by-step solution

Show that \(w(t)\) is a solution of the homogeneous linear differential equation

Consider the difference function, \(w(t) = y_1(t) - y_2(t)\). Differentiate \(w(t)\) with respect to \(t\) to obtain its derivative \(w'(t)\): $$w'(t) = y_1'(t) - y_2'(t).$$ Since both \(y_1(t)\) and \(y_2(t)\) are solutions of the initial value problem, they satisfy the given differential equation: $$y_i'(t) + p(t) y_i(t) = g(t),$$ where \(i = 1, 2\). Subtract the two equations: $$(y_1'(t) + p(t) y_1(t)) - (y_2'(t) + p(t) y_2(t)) = g(t) - g(t).$$ Simplifying, we get $$w'(t) + p(t) w(t) = 0.$$ Therefore, \(w(t)\) is a solution of the homogeneous linear differential equation \(w'(t) + p(t)w(t) = 0\).

Multiply by the integrating factor and deduce that \(e^{P(t)}w(t) = C\)

We will multiply the differential equation \(w'(t) + p(t)w(t) = 0\) by the integrating factor \(e^{P(t)}\), where \(P(t)\) is an antiderivative of \(p(t)\). This means that \(P'(t) = p(t)\). When we multiply the equation, we have: $$e^{P(t)}(w'(t) + p(t)w(t)) = e^{P(t)}\cdot 0.$$ Distribute the \(e^{P(t)}\) term: $$e^{P(t)}w'(t) + e^{P(t)}p(t)w(t) = 0.$$ Notice that the left-hand side of the equation is the derivative of the product \(e^{P(t)}w(t)\) with respect to \(t\). So, we can write this as: $$\frac{d}{dt}(e^{P(t)}w(t)) = 0.$$ Integrate both sides with respect to \(t\): $$e^{P(t)}w(t) = C.$$ Here, \(C\) is the constant of integration.

Evaluate the constant \(C\) and show that \(w(t) = 0\)

Now, we need to find the value of the constant \(C\) using the initial conditions provided in the problem statement: \(y(t_0) = y_0\). Since \(y_1(t_0) = y_2(t_0) = y_0\), we have: $$w(t_0) = y_1(t_0) - y_2(t_0) = 0.$$ Using the expression for \(w(t)\) we found in Step 2: $$e^{P(t_0)}w(t_0) = C \Rightarrow e^{P(t_0)}\cdot 0 = C$$ So, we have \(C = 0\). Using this value of \(C\) in the equation \(e^{P(t)}w(t) = C\), we get: $$e^{P(t)}w(t) = 0 \Rightarrow w(t) = 0.$$ Since \(w(t) = y_1(t) - y_2(t) = 0\), we can conclude that $$y_1(t) = y_2(t) \text{ on } (a, b).$$ Thus, the solution of the initial value problem is unique, as required.

Key concepts

Uniqueness Theorem

The Uniqueness Theorem is crucial when solving differential equations because it guarantees that a solution to an initial value problem, if it exists, is uniquely determined by the initial conditions. In our scenario, we focus on a first-order linear differential equation:

• The problem involves finding functions \( y_1(t) \) and \( y_2(t) \) that both satisfy the differential equation and initial condition.
• By defining a difference function \( w(t) = y_1(t) - y_2(t) \), the theorem helps us explore whether the uniqueness can be established.

By showing that the difference \( w(t) \) is zero everywhere in the interval, it implies \( y_1(t) \) must equal \( y_2(t) \), thus eviscerating any multiplicity of solutions and establishing that the solution is unique for the initial conditions given.

Initial Value Problem

An Initial Value Problem (IVP) is a differential equation along with an initial condition which specifies the value of the solution at a specific point. For example, if you have:

• A differential equation like \( y'(t) + p(t)y = g(t) \),
• An initial condition such as \( y(t_0) = y_0 \).

This setup means you're looking for a function \( y(t) \) that not only solves the equation but also passes through the point \( (t_0, y_0) \). The initial value given crucially defines the start of the trajectory of the function, setting the unique path through the infinite possibilities of solutions that might exist without it.

Homogeneous Linear Differential Equation

A homogeneous linear differential equation is a special type of equation where the function and its derivative terms are proportional and add up to zero. Let's consider the equation:

• \( w'(t) + p(t)w(t) = 0 \) is an example of such an equation.
• Here, "homogeneous" means that the equation is equal to zero.

The zero on the right side indicates there's no external force or input driving the function, but rather it balances out to zero through its terms. This property simplifies the analysis because solutions often involve exponential functions related to the zeros naturally emerging from the characteristic equation.

Integrating Factor

When tackling a first-order linear differential equation, an integrating factor is a mathematical tool that simplifies the process of finding a solution. The integrating factor, often expressed as \( e^{P(t)} \), where \( P(t) \) is the integral of \( p(t) \) fulfills a key role:

• By multiplying through by \( e^{P(t)} \), the differential equation can be rewritten in a form that allows straightforward integration: \( \frac{d}{dt}(e^{P(t)} w(t)) = 0 \).
• This transformation permits us to integrate both sides easily, leading to the conclusion that \( e^{P(t)} w(t) = C \), where \(C\) is a constant.

This manipulation ultimately helps isolate the function and solve these problems by turning the differential equation into a simpler arithmetic problem of finding where \( C = 0 \). This approach greatly assists in handling problems where the equation involves both the function and its derivative.