Question

Finding a Particular Solution In Exercises \(25-28\) , find an equation of the graph that passes through the point and has the given slope. $$ (8,2), \quad y^{\prime}=\frac{2 y}{3 x} $$

Short answer

The equation for the graph that passes through the point (8,2) and has the given slope is \(y = \frac{x^{2/3}}{2}\)

Step-by-step solution

Find the general solution of the differential equation

This differential equation is a type of first order linear differential equation called a separable differential equation because we can rewrite the equation in the form of \( \frac{dy}{dx} = f(x)g(y) \). To solve it, we separate the variables y and x on each side of the equation and then integrate each side with respect to the appropriate variable. So, starting from \(y'(x) = \frac{2y}{3x}\) we can rewrite this as \(\frac{dy}{y} = \frac{2dx}{3x}\). Then we can integrate each side, giving us the equation \(\ln|y| = \frac{2}{3} \ln|x| + C\), where C is the constant of integration.

Represent the solution in an exponential form

The natural logarithm and the exponential function are inverses of each other, so we can cancel them out using the rules of logarithms to rewrite the solution as an exponential equation. This gives us the equation \(|y| = e^{C}x^{2/3}\), or equivalently \(y = \pm e^{C}x^{2/3}\). This is the general solution to the differential equation.

Find the particular solution

To find the particular solution that passes through the point (8,2), we substitute these values into the general solution and solve for C. This gives us \(2 = \pm e^{C}(8^{2/3})\), or \(C = \ln \left(\pm \frac{2}{8^{2/3}} \right)\). Because y is positive when x is positive, we take the positive square root to give \(C = \ln \left(\frac{1}{2} \right) = -\ln 2\). This gives the particular solution as \(y = e^{-\ln 2}x^{2/3} = \frac{x^{2/3}}{2}\)

Key concepts

First Order Linear Differential Equation

A first order linear differential equation is a type of equation involving derivatives that expresses dynamics in a relationship between a function and its rates of change. Specifically, it involves first order derivatives, meaning it includes terms like \( \frac{dy}{dx} \). These equations are "linear" because none of the variables or their derivatives are raised to a power other than one. In the context of our problem, the equation \( y^{\prime}=\frac{2 y}{3 x} \) is a first order linear differential equation because it includes the first derivative of \( y \) with respect to \( x \).

First order linear equations are prevalent in modeling real-world situations such as population growth, radioactive decay, or cooling of objects. Understanding how to solve them involves recognizing their structure and being able to separate variables when possible. In many cases, as shown in this problem, these equations can be tractively rewritten into separable forms to allow for further solving.

Particular Solution

A particular solution to a differential equation is a solution that satisfies not only the equation itself but also meets a set of initial conditions (such as a specific point in the function's domain). In our original problem, we are tasked with finding a particular solution for the equation that passes through the point \((8, 2)\). This means we start with the general solution that describes a family of solutions and pinpoint the one that goes through this exact point.

This is done by substituting the given values of \( x \) and \( y \) into the general solution, which allows us to solve for any constants and refine the solution to fit the specific conditions. Essentially, while the general solution offers a wide view of possibilities, the particular solution narrows it down to the specific curve or path that matches the given criteria. For our equation \[ y = e^{-\ln 2} x^{2/3} = \frac{x^{2/3}}{2} \] by using the initial point, we ascertain that this unique form passes through \((8, 2)\).

General Solution

The general solution of a differential equation includes all possible solutions of the equation, encompassing an entire family of functions that could potentially satisfy the differential equation. Determining the general solution involves solving the differential equation but leaving the constant(s) of integration undetermined. These constants reflect all potential vertical shifts on the graph and ensure the completeness of the solution set.
In our example, once the equation is separated and integrated, we arrived at \[ \ln|y| = \frac{2}{3} \ln|x| + C \] which, when expressed in exponential form, results in \[ y = \pm e^{C} x^{2/3} \] Here, \( C \) is the constant of integration, and the \( \pm \) symbol represents that the function can be either positively or negatively oriented; these variations encapsulate the entirety of possible solutions for the original differential equation.

Only when a specific condition is given, like a point that the function must pass through, do we arrive at a particular solution, as mentioned earlier.

Integration

Integration is a fundamental technique used to solve differential equations. It is the inverse operation of differentiation and is used here to "undo" the derivatives present in differential equations. To solve a differential equation by separation of variables, we often separate the variables and then integrate each side independently.

In our exercise, we had the equation \( \frac{dy}{y} = \frac{2dx}{3x} \) To integrate, we need to handle each variable separately. For this, we integrated \( \int \frac{dy}{y} \) resulting in \( \ln|y| \), and \( \int \frac{2}{3}\frac{dx}{x} \) which leads to \( \frac{2}{3} \ln|x| \).

This crucial step transforms a challenging derivative problem into a more manageable algebraic form, solidifying the connection between the variables in terms of their primitive functions. Through integration, we thus derive expressions that aid us in determining both the general and particular solutions of differential equations.