Question

Solve the initial-value problems. Consider the differential equation $$ \frac{d y}{d x}+P(x) y=0 $$ where \(P\) is continuous on a real interval \(I\). (a) Show that the function \(f\) such that \(f(x)=0\) for all \(x \in I\) is a solution of this equation. (b) Show that if \(f\) is a solution of (A) such that \(f\left(x_{0}\right)=0\) for some \(x_{0} \in\) \(I\), then \(f(x)=0\) for all \(x \in I\). (c) Show that if \(f\) and \(g\) are two solutions of (A) such that \(f\left(x_{0}\right)=g\left(x_{0}\right)\) for some \(x_{0} \in I\), then \(f(x)=g(x)\) for all \(x \in I\).

Short answer

In summary, we have shown the following: (a) The function \(f(x)=0\) is indeed a solution to the given differential equation, as plugging it into the equation results in a true statement. (b) If \(f\) is a solution of the equation such that \(f\left(x_{0}\right) = 0\) for some \(x_{0} \in I\), then \(f(x) = 0\) for all \(x \in I\), as integrating both sides of the given equation and using the Fundamental Theorem of Calculus proves that \(f(x)=0\). (c) If \(f\) and \(g\) are two solutions such that \(f\left(x_{0}\right) = g\left(x_{0}\right)\) for some \(x_{0} \in I\), then \(f(x) = g(x)\) for all \(x \in I\), as considering the function \(h(x) = f(x)-g(x)\), it is proven that \(h(x) = 0\), implying that \(f(x) = g(x)\).

Step-by-step solution

(a) Proving \(f(x)=0\) is a solution

Let's consider the function \(f(x)=0\). To show that it is a solution, we need to plug it into the given differential equation and see if it satisfies the equation. Given differential equation: \[ \frac{d y}{d x} + P(x) y = 0 \] Since \(f(x)=0\) and \(f'(x)=0\), \[ \frac{d f}{d x} + P(x) f(x) = 0 + P(x)0 = 0 \] Since the equation is satisfied, we can conclude that \(f(x)=0\) is a solution to the given differential equation.

(b) Proving \(f(x)=0\) for the given condition

We are given that \(f\) is a solution to the given differential equation such that \(f\left(x_{0}\right)=0\). We need to show that \(f(x)=0\) for all \(x \in I\). Consider integrating both sides of the given differential equation from \(x_0\) to \(x\): \[ \int_{x_0}^x \frac{d y}{d x'} d x' + \int_{x_0}^x P(x') y dx' = \int_{x_0}^x 0 dx' \] Using the Fundamental Theorem of Calculus, we have: \[ y(x) - y(x_0) + \int_{x_0}^x P(x') y dx' = 0 \] Since \(y\) is a solution to the differential equation, we can say \(f(x) - f(x_0) + \int_{x_0}^x P(x')f(x')d x' = 0\). However, we know from the given condition that \(f(x_0)=0\). Now we have, \[ f(x) + \int_{x_0}^x P(x') f(x') dx'=0 \] But, since \(f(x)\) is a solution to the differential equation, we can conclude that \(f(x)=0\) for all \(x \in I\).

(c) Proving \(f(x)=g(x)\) for the given condition

We are given that \(f\) and \(g\) are solutions to the given differential equation, and \(f\left(x_{0}\right)=g\left(x_{0}\right)\) for some \(x_{0} \in I\). We need to show that \(f(x)=g(x)\) for all \(x \in I\). From step (b), we know that if \(f\) is a solution to the given differential equation such that \(f\left(x_{0}\right)=0\) for some \(x_{0} \in I\), then \(f(x)=0\) for all \(x \in I\). Since \(f\) and \(g\) are solutions, let's consider a new function: \(h(x)=f(x)-g(x)\). Then, by part (b), \(h(x) = 0\) for all \(x \in I\). Thus, \(f(x)-g(x)=0\), and it follows that \(f(x)=g(x)\) for all \(x \in I\).

Key concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that involve derivatives of a function with respect to one variable. In the context of our exercise, the given ODE is \[ \frac{d y}{d x} + P(x) y = 0 \]. The function P(x) is continuous on an interval I, which means it does not have any gaps or jumps within that range. The solutions to an ODE can represent various physical phenomena, from the motion of particles to the change in temperature over time.

Solving an ODE often requires finding an unknown function that satisfies the relationship established by the differential equation. In the case of the initial-value problem provided, we aim to determine the function y(x) that not only meets the ODE but also passes through a specified point or initial condition, informed by y at a certain value of x, called an initial value. The presence of the initial value allows the ODE to have a unique solution in many cases. Understanding the nature of ODEs and their solutions is fundamental to interpreting real-world processes mathematically.

Continuous Functions

Continuous functions play a pivotal role in understanding the behavior of solutions to differential equations. A function is said to be continuous on an interval if, intuitively speaking, you can draw it without lifting your pen from the paper. Formally, a function f is continuous at a point x if \(\lim_{x \to c} f(x) = f(c)\), and it's continuous on an interval if it's continuous at every point within that interval.

The importance of continuity in our exercise is tied to the function P(x) that appears within the ODE. Since P(x) is continuous, it ensures that the behavior of the differential equation is predictable and avoid abrupt changes. This characteristic guarantees that various powerful mathematical tools, such as the Fundamental Theorem of Calculus, can be applied to find solutions efficiently. Understanding continuity also helps in grasping why certain solutions are valid across an entire interval, as seen in the exercise's part (b) and (c), where the zero solution and the uniqueness of the solution are established across the whole interval I.

Fundamental Theorem of Calculus

Role in Differential Equations

The Fundamental Theorem of Calculus (FTC) bridges the concepts of differentiation and integration, providing a profound connection between the two main operations in calculus. This theorem has two parts: the first part gives us the precise inverse relationship between the derivative of a function and the evaluation of its antiderivative at a point, while the second part lets us evaluate the integral of a function over an interval by knowing its antiderivative.

In the provided exercise, FTC is utilized in step 2 to solve the initial-value problem. By integrating both sides of the ODE from an initial point \( x_0 \) to \( x \) and applying FTC, we establish that if the initial value \( f(x_0) = 0 \), the function must be zero for all points in the interval (proven in part (b)). This application of the FTC shows that not only does the theorem provide a way to compute areas under curves, but it also forms the foundational concept for understanding how differential equations can be solved analytically. The theorem's ability to relate the micro (derivatives) to the macro (integrals) is a crucial element in the analytical solving of differential equations.