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Chapter 2: Problem 16

Solve the given differential equations. $$ x \frac{d y}{d x}+y=-2 x^{6} y^{4} $$

Short Answer

Expert verified
The solution to the given differential equation \(x \frac{d y}{d x}+y=-2 x^{6} y^{4}\) is \(y(x) = \left(\frac{1}{\frac{1}{2}x^6 - 3C + 3\ln|x|}\right)^{\frac{1}{3}}\).

Step by step solution

01

Rearrange and Separate Variables

We are given the differential equation: \( x \frac{dy}{dx} + y = -2x^6y^4 \). To separate variables, divide both sides by \(x y^4\) and then multiply both sides by \(dx\): \( \frac{1}{y^4} \frac{dy}{dx} +\frac{1}{x} = -2x^5 dx\).
02

Integrate both sides

Now, we will integrate both sides of the equation with respect to x: \( \int \frac{1}{y^4} \frac{dy}{dx} dx + \int \frac{1}{x} dx = \int -2x^5 dx \). Let's integrate each term: \( \int y^{-4} dy + \int \frac{1}{x} dx = \int -2x^5 dx \).
03

Compute the integrals

Compute the integrals: \( -\frac{1}{3}y^{-3} + \ln|x| = -\frac{1}{6}x^6 + C \), where C is the constant of integration.
04

Solve for y(x)

Now we want to solve for \( y(x) \). We can do this by isolating the \( y \) term: \( -\frac{1}{3}y^{-3} = -\frac{1}{6}x^6 + C - \ln|x| \), \( y^{-3} = 3\left(\frac{1}{6}x^6 - C + \ln|x|\right) \), \( y(x) = \left(\frac{1}{\frac{1}{2}x^6 - 3C + 3\ln|x|}\right)^{\frac{1}{3}} \). The solution to the given differential equation is: \( y(x) = \left(\frac{1}{\frac{1}{2}x^6 - 3C + 3\ln|x|}\right)^{\frac{1}{3}} \).

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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 fundamental technique used in solving differential equations. It involves rearranging an equation so that each variable appears on only one side. This makes it easier to integrate. In this exercise, we started with the differential equation \( x \frac{dy}{dx} + y = -2x^6y^4 \).
To separate the variables, divide by \( xy^4 \) and multiply through by \( dx \), leading to the equation \( \frac{1}{y^4} dy + \frac{1}{x} dx = -2x^5 dx \).

This allows us to clearly distinguish the parts of the equation containing \( y \) and \( x \), preparing for the next step of integration.
Integration
Integration is the process of finding the integral of a function, essentially reversing differentiation. Once the variables are separated, you integrate each side independently. For the equation from our exercise,
we integrate:
  • \( \int y^{-4} dy \)
  • \( \int \frac{1}{x} dx \)
  • \( \int -2x^5 dx \)
Integrating each term results in:
  • \( -\frac{1}{3}y^{-3} \) for \( \int y^{-4} dy \)
  • \( \ln|x| \) for \( \int \frac{1}{x} dx \)
  • \( -\frac{1}{6}x^6 \) for \( \int -2x^5 dx \)
These expressions are then combined to form one equation, with a constant of integration added to account for any potential vertical shift.
Solving Differential Equations
Once you've integrated and combined all terms, you solve the resulting equation for one of the variables. Here is where it's crucial to manipulate the equation correctly. From our integrated expressions, we have the equation:
\( -\frac{1}{3}y^{-3} + \ln|x| = -\frac{1}{6}x^6 + C \).
We solve for \( y(x) \) by first isolating \( y^{-3} \), then finding \( y \) itself. Step by step manipulation leads us to:
  • \( y^{-3} = 3\left(\frac{1}{6}x^6 - C + \ln|x|\right) \)
  • Finally yielding \( y(x) = \left(\frac{1}{\frac{1}{2}x^6 - 3C + 3\ln|x|}\right)^{\frac{1}{3}} \)
This method results in a function \( y(x) \), the solution to our original differential equation.
Constant of Integration
The constant of integration, often represented as \( C \), plays a crucial role when solving indefinite integrals. It represents an unknown constant that could have been present before differentiation, capturing the family of all potential solutions.
In our case, after integration, we expressed \( C \) within our combined equation:
\( -\frac{1}{3}y^{-3} + \ln|x| = -\frac{1}{6}x^6 + C \).
This constant can be adjusted based on any initial conditions provided in related problems, ensuring our solution fits specific scenarios or constraints imposed initially. Without it, the equation loses generality, failing to provide all possible solutions.

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Most popular questions from this chapter

Solve the initial-value problems. $$ \frac{d y}{d x}+3 x^{2} y=x^{2}, \quad y(0)=2. $$

Solve the initial-value problems. Consider the Clairaut equation $$ y=p x+p^{2}, \quad \text { where } \quad p=\frac{d y}{d x} $$ (a) Find a one-parameter family of solutions of this equation. (b) Proceed as in the Remark of Exercise 20 and find an "extra" solution that is not a member of the one-parameter family found in part (a). (c) Graph the integral curves corresponding to several members of the one- parameter family of part (a); graph the integral curve corresponding to the "extra" solution of part (b); and describe the geometric relationship between the graphs of the members of the one-parameter family and the graph of the "extra" solution.

Solve the given differential equations. $$ x \frac{d y}{d x}+\frac{2 x+1}{x+1} y=x-1 $$

Solve the given differential equations. $$ x d y+(x y+y-1) d x=0 $$

Solve the initial-value problems. $$ \frac{d y}{d x}+y=f(x), \quad \text { where } \quad f(x)=\left\\{\begin{array}{ll} e^{-*}, & 0 \leq x<2, \\ e^{-2}, & x \geq 2, \end{array} \quad y(0)=1\right. $$
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