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Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable. $$x^{2} y^{\prime}(x)=y^{2}$$

Short Answer

Expert verified
Answer: The general solution is \(y(x)=\frac{x}{1-Cx}\), where C is the constant of integration.

Step by step solution

01

Identify the type of differential equation

The given equation is of the form: $$x^{2}\cdot \frac{dy}{dx} = y^{2}$$ It is a first-order separable differential equation because we can separate the variables y and x and rewrite it in the form: $$\frac{dy}{y^2}=\frac{dx}{x^2}$$
02

Rewrite the equation by separating variables

Divide both sides by \(y^2\) and \(x^2\) to separate the variables: $$\frac{dy}{y^2}=\frac{dx}{x^2}$$
03

Integrate both sides

Integrate both sides with respect to their respective variables: $$\int \frac{dy}{y^2}=\int \frac{dx}{x^2}$$ Now perform the integration: $$-\frac{1}{y}=-\frac{1}{x}+C$$ Where C is the constant of integration.
04

Solve for y as a function of x

To express the solution explicitly as a function of the independent variable x, we can take the reciprocal of both sides: $$\frac{1}{y}=\frac{1}{x}-C$$ Now, you can rewrite the equation as $$y(x)=\frac{1}{\frac{1}{x}-C}=\frac{x}{1-Cx}$$ Thus, the general solution of the given differential equation is: $$y(x)=\frac{x}{1-Cx}$$

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Key Concepts

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

First-order differential equations
First-order differential equations are equations involving the first derivative of a function. These types of equations are fundamental in mathematical analysis because they describe how a function changes intuitively. Typically, a first-order differential equation appears in the form \(\frac{dy}{dx} = f(x, y)\), where \(y\) and \(x\) are variables, and \(f(x, y)\) is a given function describing the relationship between them.
For many real-world situations, such as in physics or population dynamics, these equations model changing rates. They provide a way to predict future behavior of systems by looking at the current rate of change. Understanding these equations is crucial for interpreting various natural and social phenomena.
Integration
Integration is a mathematical process used to find a function from its derivative, which in the context of differential equations, is crucial for solving them. When we integrate a differential equation like \(\frac{dy}{y^2} = \int \frac{dx}{x^2}\), we essentially reverse the differentiation to find the original function, \(y(x)\).
This process provides a way to determine cumulative quantities, such as area under a curve or the total change across an interval for a given rate of change. Integration requires applying rules and techniques, like substitution or partial fractions, to evaluate integrals correctly. Understanding integration opens doors to solving more complex equations and predicting system behaviors over time.
Variable separation
Variable separation is a technique used to solve differential equations by rearranging them so that each variable is on its own side of the equation. This is particularly applicable to separable differential equations, where you can rearrange terms to isolate \(y\) and its derivatives from \(x\) and its derivatives.
For instance, in the equation \(\frac{dy}{y^2} = \frac{dx}{x^2}\), we separate the variables to integrate them easily. This requires rearranging terms, allowing us to solve the equation by integrating each part independently. Mastering this method simplifies solving differential equations and makes complex problems more manageable.
General solution
The general solution of a differential equation provides a family of functions that satisfy the equation. It includes an arbitrary constant, often denoted as \(C\), which can be adjusted based on initial or boundary conditions specific to a problem.
In the example, after integration, the general solution expressed is \(y(x) = \frac{x}{1 - Cx}\). This equation describes all possible curves that solve the differential equation. The solution becomes specific when a value for \(C\) is known from given conditions. Understanding the general solution is essential for applications in engineering, physics, or any field requiring differential equation modeling.

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Most popular questions from this chapter

Solve the differential equation for Newton's Law of Cooling to find the temperature function in the following cases. Then answer any additional questions. A cup of coffee has a temperature of \(90^{\circ} \mathrm{C}\) when it is poured and allowed to cool in a room with a temperature of \(25^{\circ} \mathrm{C}\). One minute after the coffee is poured, its temperature is \(85^{\circ} \mathrm{C}\). How long must you wait until the coffee is cool enough to drink, say \(30^{\circ} \mathrm{C} ?\)

Analyzing models The following models were discussed in Section 9.1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. Chemical rate equations Consider the chemical rate equations \(y^{\prime}(t)=-k y(t)\) and \(y^{\prime}(t)=-k y^{2}(t),\) where \(y(t)\) is the concentration of the compound for \(t \geq 0,\) and \(k>0\) is a constant that determines the speed of the reaction. Assume the initial concentration of the compound is \(y(0)=y_{0}>0\). a. Let \(k=0.3\) and make a sketch of the direction fields for both equations. What is the equilibrium solution in both cases? b. According to the direction fields, which reaction approaches its equilibrium solution faster?

Solve the equation \(y^{\prime}(t)=k y+b\) in the case that \(k y+b<0\) and verify that the general solution is \(y(t)=C e^{k t}-\frac{b}{k}\)

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=r B-m,\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(r>0\) reflects the annual interest rate, \(m>0\) is the annual rate of withdrawal, \(B_{0}\) is the initial balance in the account, and \(t\) is measured in years. a. Solve the initial value problem with \(r=0.05, m=\$ 1000 /\) year, and \(B_{0}=\$ 15,000 .\) Does the balance in the account increase or decrease? b. If \(r=0.05\) and \(B_{0}=\$ 50,000,\) what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?

Growth rate functions a. Show that the logistic growth rate function \(f(P)=r P\left(1-\frac{P}{K}\right)\) has a maximum value of \(\frac{r K}{4}\) at the point \(P=\frac{K}{2}\) b. Show that the Gompertz growth rate function \(f(M)=-r M \ln \left(\frac{M}{K}\right)\) has a maximum value of \(\frac{r K}{e}\) at the point \(M=\frac{K}{e}\)

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