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Magazine Chapter 1: Problem 38
\(d y / d x=x \ln x\)
Short Answer
Expert verified
y = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C.
Step by step solution
01
Understand the Differential Equation
The differential equation given is \(\frac{dy}{dx} = x \ln x\). This is a first-order differential equation.
02
Separate the Variables
Rewrite the equation to separate the variables y and x. This gives \(\frac{dy}{dx} = x \ln x\). Move dx to the right: \(\frac{dy}{dx} = x \ln x\) becomes \(\frac{dy}{dx} = x \ln x \: dx\).
03
Integrate Both Sides
Integrate both sides of the equation. Move the dy to the left side to isolate: \[ \int dy = \int x \ln x \: dx \]
04
Use Integration by Parts for Right Side
To integrate \(x \ln x\), use the integration by parts formula: \[ \int u \ dv = uv - \int v \ du \]. Set \(u = \ln x\) and \(dv = x \: dx\). Then, \(du = \frac{1}{x} \: dx\) and \(v = \frac{x^2}{2}\).
05
Apply Integration by Parts
Substitute into the integration by parts formula: \[ \int x \ln x \: dx = \ln x \: \frac{x^2}{2} - \int \frac{x^2}{2} \times \frac{1}{x} \: dx \]. Simplify the integral: \[ = \frac{x^2}{2} \ln x - \int \frac{x}{2} \: dx \].
06
Solve the Integral
Solve the remaining integral: \[ \int \frac{x}{2} \: dx = \frac{1}{2} \int x \: dx = \frac{1}{2} \times \frac{x^2}{2} = \frac{x^2}{4} \].
07
Combine and Simplify
Combine and simplify the results from the integration by parts: \[ y = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C\], where C is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
separation of variables
Separation of variables is a method used to solve a first-order differential equation where we rewrite the equation so that each variable appears on a different side of the equation. This makes it easier to integrate and thus solve for the function. In the equation \(\frac{dy}{dx} = x \ln x\), we start by rearranging terms to get the differential elements \(dy\) and \(dx\) on opposite sides. This results in \(\frac{dy}{dx} = x \ln x\) becoming \(dy = x \ln x \ dx\). With variables separated, we can then integrate both sides independently: \(\int dy = \int x \ln x \ dx\). The goal is to isolate \(y\) to solve the differential equation.
integration by parts
Integration by parts is a technique aimed at integrating products of functions. It uses the formula: \[\int u \ dv = uv - \int v \ du\]. This approach helps when straightforward integration isn't easy. In our case, we apply integration by parts to the integral \(\int x \ln x \ dx\). Here, we let \(u = \ln x\) and \(dv = x \ dx\). By differentiating and integrating, we get \(du = \frac{1}{x} \ dx\) and \(v = \frac{x^2}{2}\). Plugging these into our formula results in: \[\int x \ln x \ dx = \ln x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \ dx\],dramatically simplifying the process of finding our integral.
constant of integration
Whenever we integrate a function, we need to include a constant of integration \(C\). This constant represents an unknown value that arises because indefinite integration can have multiple solutions differing by a constant. When we solve an integral, such as: \[\int dy = y + C\], the \(C\) is crucial. It ensures our solution covers all possible anti-derivatives of the integrand. In our specific differential equation, after performing all steps, we write the final solution as: \( y = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C\), showing the general solution with the integration constant.
differential equations
Differential equations are equations that relate a function to its derivatives. They play a crucial role in mathematical modeling of real-world phenomena. A first-order differential equation, which involves the first derivative of the unknown function but no higher derivatives, can be written generally as \( \frac{dy}{dx} = f(x,y) \). In our example, \( \frac{dy}{dx} = x \ln x \) is such an equation. Solving differential equations often involves techniques like separation of variables or integrating factors to find the function \(y = y(x)\) that satisfies the equation. These techniques help transform complex real-world problems into manageable mathematical ones.
