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Finding a Particular Solution Curve In Exercises 29-32, find an equation of the curve that passes through the point and has the given slope. $$(8,2), \quad y^{\prime}=\frac{2 y}{3 x}$$

Short Answer

Expert verified
The solution curve that passes through the point (8,2) and satisfies the differential equation \(y' = \frac{2y}{3x}\) is \(y = \frac{2}{\exp(\frac{16}{3})} \times e^{\frac{2x}{3}}\).

Step by step solution

01

Rewrite the differential equation

Rewrite the differential equation in separating variables form. For the given equation \(y' = \frac{2y}{3x}\), we can rewrite this as \(\frac{3x}{2y}\cdot y' = 1\).
02

Integrate both sides

By integrating both sides of the equation, we can find the general solution of the given differential equation. \(\int \frac{3x}{2y} \cdot dy = \int dx\). Performing the integration gives: \(\frac{3}{2}\cdot \ln |y| = x + C\), where C is the constant of integration.
03

Solve for y

Exponentiating both sides of this equation to solve for y will yield the general solution curve in terms of x. Multiply the constant C in the exponent, resulting in: \(y = A \times e^{\frac{2x}{3}}\), where \(A = e^{C}\) is the new constant of integration.
04

Apply the given point

Substitute the given point (8,2) into the equation to solve for the constant A. \(2 = A \times \exp(\frac{16}{3})\). Solving for A gives \(A = 2/\exp(\frac{16}{3})\)
05

Write the final solution

Substitute A into the equation from Step 3 to obtain the particular solution curve that passes through the given point (8,2). The solution is \(y = \frac{2}{\exp(\frac{16}{3})} \times e^{\frac{2x}{3}}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equation
A differential equation is a mathematical statement that equates a function with its derivatives. In simple words, it relates the change in one variable to another. For example, in physics, these equations can represent how a changing velocity affects the position of an object over time.

In our textbook exercise, the equation presented is a first-order differential equation, which is one of the most basic types of ordinary differential equations. The equation is given as \(y' = \frac{2y}{3x}\), where \(y'\) represents the derivative of \(y\) with respect to \(x\), implying that the rate of change of \(y\) is proportional to \(\frac{y}{x}\).

Differential equations can be classified in various ways, such as by order (first, second, etc.), linearity (linear vs. nonlinear), and whether they involve one variable (ordinary) or more (partial). The equation we're considering is first-order because it involves only the first derivative, and it is also linear because the derivative of \(y\) is not raised to any power or multiplied by itself.
Integrating Factors
An integrating factor is a mathematical tool used to solve certain types of differential equations, particularly those that are not readily separable. The purpose of an integrating factor is to multiply both sides of a differential equation by a function that will make the equation easier to solve — typically transforming it into a perfect differential, allowing for direct integration.

This technique was not needed for the example given in the exercise since the variables could be separated easily. However, it is important to understand that integrating factors can be immensely helpful when dealing with more complex differential equations where separation of variables is not possible. For instance, if we had a non-separable linear first-order equation such as \(y' + P(x)y = Q(x)\), an integrating factor could be \(e^{\int P(x)dx}\), which would allow us to multiply through the equation and rearrange it into a form where the left-hand side is the derivative of a product of functions.
Particular Solution
The particular solution of a differential equation is a specific solution that satisfies not only the differential equation itself but also an additional condition, such as an initial value or boundary condition. This additional requirement ensures that out of the infinite number of possible solutions to the differential equation (the general solution), we pinpoint the one that relates to a specific physical or geometrical situation.

In the exercise, after finding the general solution \(y = A \times e^{\frac{2x}{3}}\), we use the given point (8,2) as our specific condition to find the exact value of the constant \(A\). By plugging these x and y values into the general solution, we are able to solve for \(A\) and thus obtain the unique curve that not only satisfies the differential equation but also passes through the point (8,2), rendering it the particular solution.
Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations involving functions of only one independent variable and their derivatives. They are 'ordinary' as they deal with functions of one variable, in contrast to 'partial' differential equations, which involve partial derivatives of functions of multiple variables. ODEs are a fundamental tool in expressing the dynamics of various systems across mathematics, physics, engineering, biology, and beyond.

ODEs vary in complexity from simple separable equations, like the one in our exercise, to much more complex nonlinear equations that might not have an explicit solution. The first step in solving an ODE, like the one we encountered (\(y' = \frac{2y}{3x}\)), usually involves trying to separate the variables and integrate, or looking for an integrating factor if separation isn't possible. The primary goal in solving ODEs is often to find the general solution, which encompasses an entire family of solutions, and as needed, a particular solution that fulfills given specific conditions.

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Most popular questions from this chapter

Solving a Homogeneous Differential Equation In Exercises \(77-82,\) solve the homogeneous differential equation in terms of \(x\) and \(y .\) A homogeneous differential equation is an equation of the form $$M(x, y) d x+N(x, y) d y=0$$ where \(M\) and \(N\) are homogeneous functions of the same degree. To solve an equation of this form by the method of separation of variables, use the substitutions \(y=v x\) and \(d y=x d v+v d x\) . $$\left(x^{3}+y^{3}\right) d x-x y^{2} d y=0$$

Using a Logistic Differential Equation In Exercises 55 and 56, the logistic differential equation models the growth rate of a population. Use the equation to (a) find the value of \(k,(\) b) find the carrying capacity, (c) graph a slope field using a computer algebra system, and (d) determine the value of \(P\) at which the population growth rate is the greatest. $$\frac{d P}{d t}=0.1 P-0.0004 P^{2}$$

In Exercises \(49-56\) , find the general solution of the first-order differential equation for \(x>0\) by any appropriate method. $$x d x+\left(y+e^{y}\right)\left(x^{2}+1\right) d y=0$$

True or False? In Exercises 83-85, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The function \(y=0\) is always a solution of a differential equation that can be solved by separation of variables.

Modeling Data The table shows the cost of tuition and fees \(M\) (in dollars) at public four-year universities for selected years. (Source: The College Board) $$\begin{array}{|c|c|c|c|c|}\hline \text { Year } & {1980} & {1985} & {1990} & {1995} \\ \hline \text { Cost,M } & {2320} & {2918} & {3492} & {4399} \\\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|c|}\hline \text { Year } & {2000} & {2005} & {2010} & {2015} \\ \hline \text { cost,M } & {4845} & {6708} & {8351} & {9410} \\\ \hline\end{array}$$ (a) Use a graphing utility to find an exponential model \(M_{1}\) for the data. Let \(t=0\) represent 1980. (b) Use a graphing utility to find a linear model \(M_{2}\) for the data. Let \(t=0\) represent 1980 . (c) Which model fits the data better? Explain. (d) Use the exponential model to predict when the cost of tuition and fees will be \(\$ 15,000 .\) Does the result seem reasonable? Explain.

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