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Chapter 19: Problem 7

Solve \(x \frac{\mathrm{d} y}{\mathrm{~d} x}+y=x^{4}\).

Short Answer

Expert verified
The solution is \( y = \frac{x^4}{6} + \frac{C}{x^2} \).

Step by step solution

01

Recognize the Type of Differential Equation

The given equation \( x \frac{\mathrm{d} y}{\mathrm{~d} x} + y = x^4 \) is a first-order linear differential equation in the form \( P(x) \frac{\mathrm{d} y}{\mathrm{~d} x} + Q(x)y = R(x) \) where \( P(x) = x \), \( Q(x) = 1 \), and \( R(x) = x^4 \). This indicates we can use an integrating factor method to solve it.
02

Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by the formula \( e^{\int \frac{Q(x)}{P(x)} \, dx} \). Here, \( \frac{Q(x)}{P(x)} = \frac{1}{x} \). Thus, \( \mu(x) = e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = |x| \). We can choose \( \mu(x) = x \) because \( x \) is positive in this context.
03

Multiply through by the Integrating Factor

Multiply the entire differential equation \( x \frac{\mathrm{d} y}{\mathrm{~d} x} + y = x^4 \) by the integrating factor \( x \) to get: \[ x^2 \frac{\mathrm{d} y}{\mathrm{~d} x} + xy = x^5. \]
04

Recognize the Left-Hand Side as a Derivative

Notice that the expression \( x^2 \frac{\mathrm{d} y}{\mathrm{~d} x} + xy \) is the derivative of \( x^2y \) with respect to \( x \) according to the product rule. Thus, we can rewrite the equation as: \[ \frac{\mathrm{d}}{\mathrm{d} x}(x^2 y) = x^5. \]
05

Integrate Both Sides

Integrate both sides with respect to \( x \). The left side becomes: \[ \int \frac{\mathrm{d}}{\mathrm{d} x}(x^2 y) \, dx = x^2 y, \]and the right side becomes: \[ \int x^5 \, dx = \frac{x^6}{6} + C, \]where \( C \) is the constant of integration.
06

Solve for y

Equating the integrated expressions gives:\[ x^2 y = \frac{x^6}{6} + C. \]Solving for \( y \) yields:\[ y = \frac{x^6}{6x^2} + \frac{C}{x^2} = \frac{x^4}{6} + \frac{C}{x^2}. \]

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Key Concepts

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

First-Order Linear Differential Equations
First-order linear differential equations are an essential class of differential equations frequently encountered in mathematics and engineering. These equations are written in the standard form
  • \( P(x) \frac{\mathrm{d} y}{\mathrm{~d} x} + Q(x)y = R(x) \).
In this structure:
  • \(P(x)\) is a function multiplying the derivative \(\frac{\mathrm{d}y}{\mathrm{d}x}\),
  • \(Q(x)\) is a function multiplying \(y\), and
  • \(R(x)\) is a function of \(x\) appearing on the right side.
These equations can be systematically solved using the integrating factor method. Understanding how to manipulate these equations is critical for students. It opens doors to various applications, ranging from physics to engineering problems.
Integrating Factor
The integrating factor is a vital tool used to simplify first-order linear differential equations. It transforms the equation into an easier form to integrate. The integrating factor \( \mu(x) \) is calculated using the formula:
  • \( \mu(x) = e^{\int \frac{Q(x)}{P(x)} \, dx} \).
This value is chosen to multiply through the entire differential equation. When applied, this factor converts the differential equation into one that can be integrated directly. In the problem, \( \mu(x) = e^{\int \frac{1}{x} \, dx} = |x| \), but for positive \(x\), \(\mu(x)\) simplifies to \(x\). This simplification helps convert the equation readily for the next steps in the solution process.
Product Rule
The product rule is an essential calculus rule used in differentiation. It allows us to differentiate products of two functions. The product rule states:
  • If \( u(x) \) and \( v(x) \) are functions of \( x \), then \( \frac{\mathrm{d}}{\mathrm{d} x}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x) \).
In our exercise, the expression \( x^2 \frac{\mathrm{d} y}{\mathrm{~d} x} + xy \) resembles the derivative of the product \( x^2y \). Recognizing such expressions is a significant step in applying the product rule effectively. This recognition simplifies the differential equation into a form \( \frac{\mathrm{d}}{\mathrm{d} x}(x^2 y) \), making it easier to integrate in subsequent steps.
Constant of Integration
In calculus, when we integrate a function, we always include a constant of integration, typically denoted as \(C\). This constant reflects the family of all possible solutions to an indefinite integral. Each particular value of \(C\) corresponds to a different curve on the graph, representing a particular solution. When integrating the equation:
  • \( \int x^5 \, dx = \frac{x^6}{6} + C \).
The \(C\) symbolizes all possible vertical shifts of the antiderivative. In solving differential equations, determining the constant of integration is crucial when applying initial conditions or boundary values, which allows us to pinpoint a unique solution within the infinite set of possibilities.

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Most popular questions from this chapter

Find the general solutions of the following equations: (a) \(\frac{\mathrm{d} y}{\mathrm{~d} x}=k x, \quad k\) constant (b) \(\frac{\mathrm{d} y}{\mathrm{~d} x}=-k y, \quad k\) constant (c) \(\frac{\mathrm{d} y}{\mathrm{~d} x}=y^{2}\) (d) \(y \frac{\mathrm{d} y}{\mathrm{~d} x}=\sin x\) (e) \(y \frac{\mathrm{d} y}{\mathrm{~d} x}=x+2\) (f) \(x^{2} \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y^{2}+y x\) (g) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{t^{4}}{x^{5}}\)

Find the general solution of the equation \(\frac{\mathrm{d} x}{\mathrm{~d} t}=t(x-2)\). Find the particular solution which satisfies \(x(0)=5\)

Find the general solution of $$ \frac{\mathrm{d}^{2} i}{\mathrm{~d} t^{2}}+8 \frac{\mathrm{d} i}{\mathrm{~d} t}+25 i=48 \cos 3 t-16 \sin 3 t $$

Obtain the power series solution of the equation \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+x y=0\), up to terms involving \(x^{7}\). This differential equation is known as an Airy equation.

The general solution of \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}+y=0\) is \(y=A x \mathrm{e}^{x}+B \mathrm{e}^{x} .\) Find the particular solution satisfying \(y(0)=0, \frac{\mathrm{dy}}{\mathrm{d} x}(0)=1\)
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