Problem 25 - Advanced Engineering Mathematics

Question

Verify that the piecewise-defined function $$ y= \begin{cases}-x^{2}, & x<0 \\ x^{2}, & x \geq 0\end{cases} $$ is a solution of the differential equation \(x y^{\prime}-2 y=0\) on the interval \((-\infty, \infty)\).

Step-by-Step Solution

Answer

The function satisfies the differential equation on \((-\infty, \infty)\).

Step 1: Write Down the Differential Equation

The given differential equation is \(x y' - 2y = 0\). We need to verify that the piecewise function satisfies this equation on the interval \((-\infty, \infty)\).

Step 2: Consider the Case When \(x < 0\)

In this case, the function is defined as \(y = -x^2\). The derivative is obtained by differentiating this expression with respect to \(x\):\[y' = \frac{d}{dx}(-x^2) = -2x.\]

Step 3: Verify the Equation for \(x < 0\)

Substitute into the differential equation: \(x(-2x) - 2(-x^2) = 0\).Simplify it: \(-2x^2 + 2x^2 = 0\). This is true, so the function satisfies the equation for \(x < 0\).

Step 4: Consider the Case When \(x \geq 0\)

In this case, the function is defined as \(y = x^2\). Differentiate with respect to \(x\): \[y' = \frac{d}{dx}(x^2) = 2x.\]

Step 5: Verify the Equation for \(x \geq 0\)

Substitute into the differential equation: \(x(2x) - 2(x^2) = 0\).Simplify it: \(2x^2 - 2x^2 = 0\). This simplifies to true, so the function satisfies the equation for \(x \geq 0\).

Step 6: Conclude for the Whole Interval

The function satisfies the differential equation for both \(x < 0\) and \(x \geq 0\), so it holds for the entire interval \((-\infty, \infty)\).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In simpler terms, they describe how a quantity changes over time. In the case of the exercise, we have the differential equation \(x y' - 2y = 0\). Here, \(y\) is the function of \(x\) that is being sought, and \(y'\) is its derivative with respect to \(x\).

This equation tells us how the function \(y\) and its rate of change \(y'\) must behave in relation to each other and the variable \(x\). Solving such equations often involves finding the unknown function \(y\) that satisfies the relationship for all values in a given interval.

To verify that a function satisfies a differential equation, like our piecewise function, means checking that the equation holds true on the specified interval when the function and its derivative are substituted into the equation.

Derivatives

Understanding derivatives is crucial since they represent the rate of change or the slope of the function at a given point. In the context of our exercise, we have two expressions due to the piecewise nature of the function.

For \(x < 0\), we have \(y = -x^2\), and its derivative \(y' = -2x\). For \(x \geq 0\), \(y = x^2\), and \(y' = 2x\). These derivatives tell us how rapidly \(y\) is changing for each segment of the piecewise function.

When verifying whether the function solves the differential equation, substituting these derivatives into the original equation produced a true statement, indicating that the function’s rate of change perfectly aligns with the behavior needed to satisfy the equation. This process of differentiation is a foundational tool used across calculus and differential equations.

Interval Notation

Interval notation is used to specify a range of values, and it is particularly useful when working with functions defined over certain domains. In our exercise, the interval given is \((-ity, ity)\), which means the function covers all real numbers.

When dealing with piecewise functions, intervals help us define which expression of the function applies. Here, \(y = -x^2\) is used for \(x < 0\), whereas \(y = x^2\) takes over for \(x \geq 0\).

Using interval notation, we can succinctly define these ranges and clearly communicate which formula should be applied to which section of the \(x\)-axis. Mastery of interval notation allows solving and understanding piecewise-defined functions comprehensively.