Question

In Problems \(1-4\), show that the indicated function is a solution of the given differential equation; that is, substitute the indicated function for \(y\) to see that it produces an equality. $$ \frac{d y}{d x}+\frac{x}{y}=0 ; y=\sqrt{1-x^{2}} $$

Short answer

The function \( y = \sqrt{1-x^2} \) satisfies the differential equation.

Step-by-step solution

Differentiate the Function

Given the function \( y = \sqrt{1-x^2} \). We need to find \( \frac{dy}{dx} \). Use the chain rule for differentiation: \( y' = \frac{d}{dx}(1-x^2)^{1/2} \). This gives \( y' = \frac{1}{2}(1-x^2)^{-1/2} \times (-2x) = \frac{-x}{\sqrt{1-x^2}} \).

Substitute into the Differential Equation

The given differential equation is \( \frac{dy}{dx} + \frac{x}{y} = 0 \). Substitute \( \frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}} \) and \( y = \sqrt{1-x^2} \) into the equation.

Simplify the Differential Equation

Insert the derivatives and values into the differential equation: \( \frac{-x}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}} \). This simplifies to \( \frac{-x + x}{\sqrt{1-x^2}} = \frac{0}{\sqrt{1-x^2}} = 0 \).

Verify the Equality

Observe that the simplified equation results in zero. Since the left-hand side equals the right-hand side, \( 0 = 0 \), it confirms that the function \( y = \sqrt{1-x^2} \) is indeed a solution to the differential equation \( \frac{dy}{dx} + \frac{x}{y} = 0 \).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes. This is particularly useful in finding the slope of a curve at any given point. The process of differentiation allows us to determine the derivative of a function, typically represented as \( \frac{dy}{dx} \), which reads as "the derivative of \( y \) with respect to \( x \)."
In the exercise above, the function given is \( y = \sqrt{1-x^2} \). Our task is to differentiate this function to solve the differential equation. By applying differentiation, we aim to find the expression for \( \frac{dy}{dx} \), which will be plugged into the original differential equation to verify the solution.

To differentiate \( y = \sqrt{1-x^2} \), you apply the power rule and the chain rule, which will be explained in the next section. Mastery of differentiation helps us in calculus to solve complex problems related to rates of change and motion.

Chain Rule

The chain rule is a crucial technique in differentiation, particularly when dealing with compositions of functions. It helps us find the derivative of a function that is composed of other functions. The core idea is to multiply the derivative of the outer function by the derivative of the inner function.
In our exercise, the function is \( y = (1-x^2)^{1/2} \). Here, the outer function is \( u^{1/2} \) and the inner function is \( u = 1-x^2 \).

• First, differentiate the outer function: \( \frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2} \).
• Next, differentiate the inner function: \( \frac{d}{dx}(1-x^2) = -2x \).

Using the chain rule, multiply these derivatives: \( y' = \frac{1}{2}(1-x^2)^{-1/2} \times (-2x) \).
This results in: \( y' = \frac{-x}{\sqrt{1-x^2}} \), which fits into the given differential equation, demonstrating how the chain rule facilitates solving such problems.

Solution Verification

Solution verification involves substituting the found derivative back into the original differential equation to ensure it satisfies the equation. This step is critical as it confirms whether our differentiated function was computed correctly.
In the problem, we are given the differential equation \( \frac{d y}{dx} + \frac{x}{y} = 0 \). After finding \( \frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}} \), we substitute both \( \frac{dy}{dx} \) and \( y = \sqrt{1-x^2} \) into this equation.

• The substitution step results in: \( \frac{-x}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}} \).
• Upon simplifying, the equation becomes: \( \frac{-x + x}{\sqrt{1-x^2}} = \frac{0}{\sqrt{1-x^2}} = 0 \).

This simplification shows that both sides of the equation equal zero, verifying that the function \( y = \sqrt{1-x^2} \) is indeed a solution of the differential equation. This step ensures our solution is both valid and complete, reinforcing the principles of differential equations.