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Chapter 1: Problem 13

Solve the given differential equation. $$y^{\prime}-x^{-1} y=2 x^{2} \ln x$$

Short Answer

Expert verified
The given differential equation is a first-order linear differential equation, which can be solved using an integrating factor. The integrating factor is \(IF = x^{-1}\). After multiplying the equation by the integrating factor and integrating both sides, we find the general solution for \(y(x)\) to be \(y(x) = x^3\ln x - \frac{1}{2}x^3\).

Step by step solution

01

Identify the given differential equation

The differential equation is given as: \[y' - x^{-1}y = 2x^2 \ln x\] It is a first-order linear differential equation of the form: \[\frac{dy}{dx} - P(x)y = Q(x)\]
02

Find the integrating factor

To find the integrating factor (IF), we have: \[IF = e^{\int P(x) dx}\] For our given equation, \(P(x) = -x^{-1}\), so we have: \[IF = e^{\int -x^{-1} dx}\] \[IF = e^{-\ln x} = x^{-1}\]
03

Multiply the equation by the integrating factor

We multiply both sides of the equation by the integrating factor: \[x^{-1}(y' - x^{-1}y) = x^{-1}(2x^2 \ln x)\] Now, the left side of the equation is the derivative of \((x^{-1}y)\) with respect to x, so we have: \[\frac{d}{dx}(x^{-1}y) = 2x\ln x\]
04

Integrate both sides with respect to x

We will now integrate both sides of the equation with respect to x: \[\int \frac{d}{dx}(x^{-1}y) dx = \int 2x\ln x\, dx\] On the left side, the integral of the derivative is just the function itself: \[x^{-1}y = \int 2x\ln x\, dx\]
05

Solve the integral on the right side

To solve the integral on the right side, we can use integration by parts: Let \(u = \ln x\), so \(du = x^{-1}dx\), and \(dv = 2x dx\), so \(v = x^2\) Using integration by parts, we get: \[\int u\, dv = uv - \int v\, du\] \[\int 2x\ln x\, dx = x^2 \ln x - \int x^2 \cdot x^{-1}dx\] \[\int 2x\ln x\, dx = x^2 \ln x - \int x\, dx\] \[\int 2x\ln x\, dx = x^2 \ln x - \frac{1}{2}x^2\]
06

Solve for y(x)

Now we have: \[x^{-1}y = x^2 \ln x - \frac{1}{2}x^2\] Multiplying both sides by x, we get the solution for y(x): \[y(x) = x^3\ln x - \frac{1}{2}x^3\] This is the general solution to the given differential equation.

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Key Concepts

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

Integrating Factor
To solve first-order linear differential equations, one powerful method is using an integrating factor. An integrating factor is a function that we multiply through the differential equation to easily solve it.

In the context of the equation \( y' - x^{-1} y = 2x^2 \ln x \), the integrating factor is determined by the coefficient of \( y \), which is \( P(x) = -x^{-1} \). The integrating factor (IF) is then calculated as follows:
  • We compute \( IF = e^{\int P(x) \, dx} \).
  • Here, \( IF = e^{\int -x^{-1} \, dx} = e^{-\ln x} \).
  • By simplifying, we find \( IF = x^{-1} \).
The beauty of this method lies in transforming the left-hand side of the differential equation into the derivative of a product, simplifying further integration and solution of the equation.
Integration by Parts
Integration by parts is a technique used in calculus to integrate products of functions. It's especially useful when direct integration is not feasible.

It is based on the formula:
  • \( \int u \, dv = uv - \int v \, du \)
In our problem, we apply this method to integrate \( 2x \ln x \) where:
  • Let \( u = \ln x \) so \( du = x^{-1} dx \).
  • Let \( dv = 2x \, dx \) giving \( v = x^2 \).
Applying integration by parts:
  • \( \int 2x \ln x \, dx = x^2 \ln x - \int x^2 \cdot x^{-1} \text{ } dx \)
  • This simplifies to \( x^2 \ln x - \int x \, dx = x^2 \ln x - \frac{1}{2}x^2 \).
This step is crucial for solving parts of differential equations where products of functions appear.
Differential Equations Solution
Solving differential equations involves transforming and integrating to find an unknown function. The key is getting the equation into a recognizable form to apply known strategies.

In the case of linear differential equations such as \( y' - x^{-1} y = 2x^2 \ln x \), after applying an integrating factor, the equation simplifies.

Once both sides of the equation are multiplied by the integrating factor, the left side becomes the derivative of a product: \( \frac{d}{dx}(x^{-1}y) \).
  • This equality allows us to integrate both sides: \( \int \frac{d}{dx}(x^{-1}y) \, dx = \int 2x \ln x \, dx \).
  • The integration of the left side simply gives back \( x^{-1}y \).
By solving the integral on the right through methods like integration by parts, we get closer to the explicit function \( y(x) \), providing us the solution to the differential equation.
Linear Differential Equations
Linear differential equations are those where the unknown function and its derivative appear linearly, making them often more straightforward to solve.

A general form of first-order linear differential equations is:
  • \( \frac{dy}{dx} + P(x)y = Q(x) \)
The problem \( y' - x^{-1}y = 2x^2 \ln x \) fits this form, where:
  • \( P(x) = -x^{-1} \)
  • \( Q(x) = 2x^2 \ln x \)
This classification helps in systematically applying methods like the integrating factor. The goal is to simplify the equation by rewriting it in such a way that standard integration techniques can be applied, leading to a workable solution for \( y(x) \). This methodical approach is key to tackling linear differential equations efficiently.

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