Problem 7 - Thomas Calculus

Question

In Exercises \(7-10\) , write an equivalent first-order differential equation and initial condition for \(y .\) $$ y=-1+\int_{1}^{x}(t-y(t)) d t $$

Step-by-Step Solution

Answer

The first-order differential equation is \( \frac{dy}{dx} = x - y \) with initial condition \( y(1) = -1 \).

Step 1: Understanding the Problem

We are given an equation that expresses \(y\) using an integral. We need to derive a first-order differential equation that corresponds to this integral equation. Also, we need to identify an initial condition for \(y\).

Step 2: Differentiate Both Sides

Differentiate both sides of the equation with respect to \(x\): \( \frac{dy}{dx} = \frac{d}{dx} \left(-1 + \int_{1}^{x} (t - y(t)) \, dt \right) \). According to the Fundamental Theorem of Calculus, the derivative of the integral from 1 to \(x\) of a function \(f(t)\) is \(f(x)\), so we have:\[ \frac{dy}{dx} = x - y \].

Step 3: Identify the Initial Condition

Since the integral starts from 1, plug in \(x = 1\) into the original equation to find an initial condition for \(y\). From the equation:\[ y = -1 + \int_{1}^{1} (t - y(t)) \, dt \],since the integral of a function over an interval of zero length is zero, we have:\( y(1) = -1 \).

Step 4: Write the First-Order Differential Equation and Initial Condition

From Step 2, the differential equation is \( \frac{dy}{dx} = x - y \) and the initial condition determined in Step 3 is \( y(1) = -1 \). Thus, the equivalent first-order differential equation with the initial condition is:\[ \frac{dy}{dx} = x - y, \quad y(1) = -1 \].

Key Concepts

Initial Conditions

Initial conditions in differential equations provide vital information about the value of the solution at a particular point. They help determine a unique solution to the differential equation by anchoring the curve to a specific starting point.

In our exercise, we used an initial condition to find the value of the function when the variable, often 'x', is set to a particular value. This means that we specify what 'y' is at a specific point, say at time zero or at some fixed 'x'.

• Initial Condition Example: For our problem, we found that when \( x = 1 \), \( y(1) = -1 \).
• This means at the starting point, the value of our function, \( y \), is -1.

The role of initial conditions is crucial because from a mathematical perspective, many differential equations can have an infinite number of solutions if no initial conditions are provided. They help us pin down exactly which solution fits our problem.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, showing that they are inverse operations. It states that if a function is continuous over an interval, the integral of its derivative over that interval gives the function itself, up to a constant.

In our exercise, we used the theorem to simplify the task of differentiating an integral expression.

• When we differentiated \( \int_{1}^{x} (t - y(t)) \, dt \), we directly found \( f(x) = x - y(x) \).
• This simplifies the process of finding a derivative for integral expressions.

Understanding this theorem is key because it helps us transition smoothly from integral expressions back to their original functions, allowing for straightforward computation of derivatives.

Differentiation Rules

Differentiation rules are the tools in calculus that allow us to find the derivative of various functions efficiently. They provide a systematic way to deal with sums, products, quotients, and compositions of functions.

When we differentiate, we are finding how a function changes as its input changes. In the step where we differentiated both sides, we were looking to find the rate of change of \( y \) with respect to \( x \).

Some key differentiation rules involved in solving differential equations include:

• Power Rule: Used when differentiating functions like \( x^n \).
• Chain Rule: Important for functions within other functions.
• Product and Quotient Rules: Useful when dealing with products or quotients of functions.

By applying these rules, we can dissect complex expressions into manageable parts, aiding in the derivation of meaningful and simplified derivatives. This is essential for solving and understanding first-order differential equations efficiently.