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Finding a General Solution Using Separation of Variables In Exercises 5-18, find the general solution of the differential equation. $$\frac{d y}{d x}=\frac{x}{y}$$

Short Answer

Expert verified
The general solution of the given differential equation is \(y = \pm \sqrt{x^2 + 2C}\)

Step by step solution

01

Identifying the Differential Equation

The given differential equation is \(\frac{d y}{d x}=\frac{x}{y}\). This equation could be written as \(y \frac{d y}{d x}=x\).
02

Separating variables

Next, separate the variables by moving any term involving y (along with differentials) on one side and all the terms involving x (along with differentials) on the other. The equation will then become \(y dy = x dx\).
03

Integration on both sides

Now, integrate both sides of the equation with respect to their variables. This gives \(\int y dy = \int x dx\).
04

Computing the integrals

The integral of \(y\) with respect to \(y\) is \(0.5 y^2\) and the integral of \(x\) with respect to \(x\) is \(0.5 x^2\). Including the constant of integration \(C\) on one side, the equation becomes \(0.5 y^2 = 0.5 x^2 + C\).
05

Simplification to express y

Simplify the equation to solve for \(y\) to get the general solution for the given differential equation. The general solution of the differential equation is \(y = \pm \sqrt{x^2 + 2C}\)

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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 equation that relates a function with its derivatives. In practice, it models how a certain quantity changes over time or space. The equation provided in the exercise, \( \frac{d y}{d x} = \frac{x}{y} \), is an example of a first-order differential equation. It links the derivative of the function \( y \) with respect to \( x \) to a ratio involving \( x \) and \( y \).

Understanding how to solve this type of equation is crucial for describing various physical phenomena, such as growth rates in biology, motion in physics, or changing rates in economics. The solution process often requires manipulating the equation to isolate the derivative on one side. In this case, through separation of variables, the equation was manipulated into a form where \( y \) and \( x \) are on opposite sides, facilitating the integration process that leads to the general solution.
Integrating Factors
Integrating factors are a technique used to solve certain differential equations that cannot be easily separated. While the equation from the exercise, \( \frac{d y}{d x} = \frac{x}{y} \), is solveable via separation of variables, not all differential equations are this straightforward. An integrating factor is a function that is multiplied by a given differential equation, making it possible to rewrite the equation as the derivative of a product of functions.

In general, for a linear first-order differential equation in the form \( \frac{d y}{d x} + P(x) y = Q(x) \), the integrating factor can be found by calculating the exponentiated integral of \( P(x) \) and can transform the equation into an exact equation. Although the integrating factor method was not required for the original equation, it’s a valuable strategy in a differential equations toolkit, especially for equations that do not separate naturally.
General Solution
Finally, the general solution of a differential equation is an expression that contains all possible solutions to the equation. It is called 'general' because it includes a constant of integration \( C \), which can take on any value. In our exercise, after integrating both sides of the separated equation, we arrive at \(0.5 y^2 = 0.5 x^2 + C\). Simplifying this, we achieve the general solution \( y = \pm \sqrt{x^2 + 2C} \).

The \( \pm \) sign in front of the square root acknowledges that the function \( y \) could be either positive or negative, allowing the solution to cover all possible cases of the square of a number. The constant \( C \) represents the indefinite integral's arbitrary constant, reflecting the infinite set of curves that satisfy the equation, each curve corresponding to a different initial condition or starting point. The general solution captures the essence of all these infinitely many specific solutions.

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Most popular questions from this chapter

Newton's Law of Cooling When an object is removed from a furnace and placed in an environment with a constant temperature of \(80^{\circ} \mathrm{F},\) its core temperature is \(1500^{\circ} \mathrm{F}\) . One hour after it is removed, the core temperature is \(1120^{\circ} \mathrm{F}\) . (a) Write an equation for the core temperature \(y\) of the object \(t\) thours after it is removed from the furnace. (b) What is the core temperature of the object 6 hours after it is removed from the furnace?

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