Question

Differential Equations In Exercises \(125-128\) , verify that the function satisfies the differential equation. $$ \text{Function} \quad \text{Differential Equation} $$ $$ y=\frac{1}{x}, x>0 \quad x^{3} y^{\prime \prime}+2 x^{2} y^{\prime}=0 $$

Short answer

The function \(y=1/x\) does indeed satisfy the differential equation \(x^3 y'' + 2x^2 y' = 0\), as demonstrated by the step-by-step solution.

Step-by-step solution

Finding the first derivative

To calculate the first derivative of the function \(y=1/x\), the rule for the derivative of \(1/x\) or \(x^{-1}\) can be used, which is \(-1/x^2\) or \(-x^{-2}\). Therefore, the first derivative \(y'\) is \(-1/x^2\).

Determining the second derivative

Next, the second derivative of the function \(y\), or \(y''\), is needed. This is the derivative of \(y'\). Again, the power rule for derivatives is used, taking the derivative of \(-1/x^2 = -x^{-2}\) as \(2x^{-3}\). Therefore, the second derivative \(y''\) is \(2/x^3\).

Verification

To verify that these functions satisfy the differential equation \(x^3 y'' + 2x^2 y' = 0\), substitute \(y', y''\) into the differential equation and check out if the equation balances. That is, if we substitute \(y' = -1/x^2\) and \(y'' = 2/x^3\) into the differential equation, the equation becomes \(-2 + 2 = 0\), which is in fact a true statement.

Key concepts

Derivatives

Understanding derivatives is crucial in the study of calculus, as they represent the rate at which a function is changing at any point. In other words, a derivative describes how a function's output value changes in response to changes in its input value. For example, in physics, the derivative of the position function with respect to time gives the velocity, which tells us how position changes over time.

When working with derivatives, particularly in differential equations, we use them to find the behaviors of functions, such as growth rates in biology or changing trends in economics. In the example of the function \(y = \frac{1}{x}\), calculating the first derivative \(y'\) gives us a sense of how \(y\) reacts as \(x\) varies. We see that the derivative \(-\frac{1}{x^2}\) becomes smaller as \(x\) increases, indicating that the function is decreasing at a slower rate.

Second Derivative

The second derivative of a function gives us insight into the curvature of the function's graph. Essentially, it informs us about the rate at which the first derivative is changing. This can be thought of as an acceleration in the context of motion—how the velocity of an object changes over time.

Understanding the second derivative is important when dealing with the concavity of functions and can help identify points of inflection, where the concavity changes. In our example, the second derivative \(y'' = \frac{2}{x^3}\) of \(y = \frac{1}{x}\) shows that as \(x\) increases, the rate at which the slope of the tangent line changes is slowing down, since the second derivative is positive but getting smaller. In the context of the differential equation provided, calculating the second derivative with accuracy is pivotal for verifying the equation.

Power Rule

The power rule is a basic, yet powerful, tool in differentiation. It applies to functions of the form \(f(x) = x^n\), where \(n\) is any real number. The rule states that the derivative of such a function is \(nx^{n-1}\). The power rule simplifies the process of finding derivatives by reducing it to a multiplication of the power by the coefficient and a decrement of the power by one.

For our function \(y = x^{-1}\), applying the power rule to find the first derivative gives us \(y' = -x^{-2}\), and to find the second derivative of the function we apply the power rule again to obtain \(y'' = 2x^{-3}\). These simplified expressions show the importance of mastering the power rule in order to quickly and efficiently derive more complicated derivatives that are essential in verifying and solving differential equations.