Problem 3 - Thomas Calculus in SI Units

Question

Show that each function \(y=f(x)\) is a solution of the accompanying differential equation. $$y=\frac{1}{x} \int_{1}^{x} \frac{e^{t}}{t} d t, \quad x^{2} y^{\prime}+x y=e^{x}$$

Step-by-Step Solution

Answer

The function is a solution to the differential equation because substituting \(y\) and \(y'\) simplifies to \(e^x\), satisfying the equation.

Step 1: Define the Given Function

The given function is \(y=f(x) = \frac{1}{x} \int_{1}^{x} \frac{e^{t}}{t} \, dt\). We need to find its derivative \(y'\) to show it satisfies the differential equation.

Step 2: Differentiate the Function

To find \(y'\), differentiate \(y=f(x)\) using the Leibniz rule for differentiation under the integral sign. Differentiate with respect to \(x\): \[ y = \frac{1}{x} \left( \int_1^x \frac{e^t}{t}\, dt \right) = \frac{d}{dx} \left[ \frac{1}{x} \int_1^x \frac{e^t}{t}\, dt \right] \]Apply the Leibniz integral rule and product rule for differentiation of \(\frac{1}{x}\): \[ y' = \frac{-1}{x^2} \int_1^x \frac{e^t}{t} \, dt + \frac{1}{x} \cdot \frac{e^x}{x} = \frac{-1}{x^2} \int_1^x \frac{e^t}{t} \, dt + \frac{e^x}{x^2} \]Simplifying this further gives:\[ y' = \frac{e^x}{x^2} - \frac{1}{x^2} \int_1^x \frac{e^t}{t} \, dt \]

Step 3: Substitute into the Differential Equation

Substitute \(y\) and \(y'\) into the differential equation \(x^2 y' + x y = e^x\). Substitute \(y'\): \[ x^2 \left( \frac{e^x}{x^2} - \frac{1}{x^2} \int_1^x \frac{e^t}{t}\,dt \right) + x \cdot \left( \frac{1}{x} \int_1^x \frac{e^t}{t}\,dt \right) \]Simplify by distribution:\[ \left( x^2 \cdot \frac{e^x}{x^2} \right) - x^2 \cdot \frac{1}{x^2} \int_1^x \frac{e^t}{t}\,dt + \int_1^x \frac{e^t}{t}\,dt \] Notice that the second term and third term cancel out, leaving:\[ e^x \]

Step 4: Verify Solution

After substituting and simplifying, the expression \(x^2 y' + x y = e^x\) holds true, which verifies that the given function \(y = \frac{1}{x} \int_1^x \frac{e^t}{t} dt\) is indeed a solution to the differential equation.

Key Concepts

Understanding the Leibniz Rule

When you are faced with differentiating an integral whose limits are functions of the variable of differentiation, the Leibniz rule is your friend. It simplifies the process of finding derivatives under the integral sign.
The Leibniz rule states that if you have an integral \( I(x) = \int_{a(x)}^{b(x)} f(t, x) \, dt \), then its derivative is:\[\frac{d}{dx} I(x) = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt\]In simpler terms:- Evaluate the integrand at each integral's bound, then multiply by the derivative of each bound.- Add the integral of the partial derivative of the integrand with respect to \(x\).
In the original exercise, although the bound \(b(x)\) is the variable \(x\), and \(a(x) = 1\) (a constant), it showcases applying the rule effectively. This powerful tool allows us to navigate through more complex calculus problems.

The Role of Integration

Integration is one of the fundamental concepts in calculus, providing a way to find areas under curves and solve differential equations.
In the given exercise function, integration helps in expressing "accumulated change." Specifically, it calculates an integral from 1 to \(x\) of \(\frac{e^t}{t}\).
Here’s why integration matters:

• It calculates total accumulation, whether that's area, volume, or, in this case, a functional form.
• Integral bounds can depend on other variables, which ties into the use of the Leibniz rule for differentiation.
• It smooths out variable expressions to find encompassing solutions to differential equations.

When viewing calculus as a broader subject, integration complements differentiation: one unravels rates of change, while the other collects and sums infinitesimal quantities over a domain.

Exploring Derivatives

The concept of derivatives in calculus reveals how a function changes as its input changes.
In the given problem, differentiating the function correctly helps determine if it fulfills the given differential equation.
Why derivatives matter:

• They measure the rate of change. Understanding how a function grows or shrinks is central in calculus.
• In solving differential equations, deriving correctly enables you to validate whether a function is a solution.
• The derivative can reveal more about the behavior of the function, enabling deeper analysis and understanding.

In the solution to the exercise, differentiating involved using both the Leibniz rule and general product differentiation to gauge \(y'\), leading to finding whether it fits into the original differential equation contextually.

Bridging Concepts with Calculus

Calculus is the mathematics of change and brings together concepts like derivatives and integration to solve complex problems, such as differential equations.
Here's how calculus interlinks with our problem:

• It equips us with methods to both differentiate and integrate, allowing for the expression and manipulation of variables over defined intervals.
• Transforms problems into environmentally solvable forms – breaking down an equation like \(x^2 y' + x y = e^x\) allows us to see the constituent parts more clearly and check solutions thoroughly.
• It encompasses stark, practical methods (e.g., Leibniz rule) to tie together seemingly distinct concepts.

Through this process, calculus demonstrates its strength in providing robust frameworks for approaching and solving mathematical challenges.