Problem 9 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Is \(y=\frac{x e^{x}}{2}+\frac{e^{x}}{3 x}\) a solution to the differential equation \(x \frac{d y}{d x}+(1-x) y=x e^{x} ?\) Justify your answer.

Step-by-Step Solution

Answer

Yes, \(y=\frac{x e^{x}}{2}+\frac{e^{x}}{3 x}\) is a solution to the differential equation \(x \frac{d y}{d x}+(1-x) y=x e^{x}\).

Step 1: Find derivative

Calculate the derivative of \(y=\frac{x e^{x}}{2}+\frac{e^{x}}{3 x}\) using the chain rule and the product rule. This yields \(\frac{d y}{d x}=\frac{e^{x}(x+1)}{2}-\frac{e^{x}}{3 x^{2}}\).

Step 2: Substitute into the given differential equation

Substitute \(\frac{d y}{d x}\) and \(y\) into the original differentiation equation \(x \frac{d y}{d x}+(1-x) y=x e^{x}\). This yields \(x[\frac{e^{x}(x+1)}{2}-\frac{e^{x}}{3 x^{2}}]+(1-x)[\frac{x e^{x}}{2}+\frac{e^{x}}{3 x}]=x e^{x}\).

Step 3: Simplify

Simplify the equation. This will yield the expression \(x e^{x}=x e^{x}\), which is the identical to the right side of our original differential equation.

Step 4: Conclusion

Since the right side equals the left side of the original differential equation, therefore, \(y=\frac{x e^{x}}{2}+\frac{e^{x}}{3 x}\) is a solution to the differential equation \(x \frac{d y}{d x}+(1-x) y=x e^{x}\).

Key Concepts

Chain Rule

The chain rule is a fundamental principle in calculus used to find the derivative of a composite function. In essence, if you have a function that is the combination of two other functions, the chain rule helps you differentiate it. For instance, if you have a function like \( g(h(x)) \) where both \( g \) and \( h \) are functions of \( x \) with known derivatives, then the derivative of \( g(h(x)) \) is \( g'(h(x)) \cdot h'(x) \).

Applying the chain rule correctly is crucial for solving many differential equations since it enables the differentiation of complex expressions. This concept was used in the exercise to calculate the derivative of \( y=\frac{x e^{x}}{2}+\frac{e^{x}}{3 x} \) effectively. By recognizing the composite functions involved, such as \( x \cdot e^{x} \) and \( e^{x}/x \) and differentiating each part accordingly, students can unravel more complex differential equations with ease.

Product Rule

The product rule is another indispensable tool in calculus utilized when differentiating an equation that is the product of two functions. The product rule states that if you have two functions, \( u(x) \) and \( v(x) \) then the derivative of their product \( u(x) \cdot v(x) \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). This proves crucial when functions are not merely combined, but intertwined through multiplication.

In the provided exercise, the product rule is essential for deriving \( y \) since \( y \) is expressed as a product of \( x \) and \( e^{x} \) for the first term. Understanding when and how to apply the product rule ensures that students can tackle a wide range of problems in calculus, and in particular, correctly solve differential equations that involve products of functions.

Differential Equation Solutions

Solving differential equations is a core skill in various scientific disciplines. A differential equation is an equation that relates one or more functions and their derivatives. It defines a relationship between the rate of change of a quantity and the quantity itself. Solutions to these equations are functions that satisfy the given equation.

To solve the differential equation, as seen in the textbook problem \( x \frac{d y}{d x}+(1-x) y=x e^{x} \) step by step, we first find the derivative of \( y \) using rules such as the chain and product rules. We then substitute the function \( y \) and its derivative into the equation to check for consistency. If the original equation holds after substitution and simplification, as it does in this case with \( x e^{x}=x e^{x} \), we can conclude that the function \( y \) is indeed a solution to the differential equation. This process is instrumental in validating solutions and understanding the nature of differential equations.