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Chapter 1: Problem 22

Find general solutions of the differential equations. Primes denote derivatives with respect to \(x\) throughout. $$ x^{2} y^{\prime}+2 x y=5 y^{4} $$

Short Answer

Expert verified
The general solution is given by \( y = \pm \left(2(5x^{-1} - 2\ln|x| - C)\right)^{-1/2}. \)

Step by step solution

01

Rearrange the Equation

Rearrange the given differential equation \[ x^2 y' + 2xy = 5y^4 \] into a form that makes it easier to separate the variables. Divide every term by \( x^2 y^4 \) to prepare for separation of variables: \[ y' = \frac{5y^3}{x^2} - \frac{2y}{x}. \]
02

Separate Variables

Separate the variables, \( y \) and \( x \). The equation becomes \[ \frac{y'}{y^3} - \frac{2}{xy^3} = \frac{5}{x^2}. \] Or, \[ \left( \frac{1}{y^3} \right) y' = \frac{5}{x^2} + \frac{2}{xy^3}. \] Multiply the entire equation by \( y^3 \) to simplify: \[ y^{-3} y' = \frac{5}{x^2} + \frac{2}{x} \left( \frac{1}{y^3} \right). \]
03

Integrate Both Sides

Integrate both sides with respect to \( x \). The left side becomes \[\int y^{-3} dy = - \frac{1}{2} y^{-2} + C_1. \] The right side involves the integration of \[\int \left( \frac{5}{x^2} + \frac{2}{x} \right) dx. \] This results in \[-\frac{5}{x} + 2\ln|x| + C_2.\]
04

Combine Integration Results

Equate the results of the integration from both sides. Set the constants \( C \) to combine and simplify:\[-\frac{1}{2} y^{-2} = -\frac{5}{x} + 2\ln|x| + C.\]
05

Solve for y

Solve for \( y \) by rearranging the expression obtained. First, express \( y^{-2} \):\[ y^{-2} = 2 \left(\frac{5}{x} - 2\ln|x| - C\right).\] Finally, solve for \( y \):\[ y = \left(\frac{1}{2 (5x^{-1} - 2 \ln|x| - C)}\right)^{-1/2}. \] This expresses \( y \) in terms of \( x \).

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Key Concepts

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

General Solutions
When tackling differential equations, one often seeks the general solution. A general solution is a form of the equation that incorporates an arbitrary constant, denoted usually as \(C\). This constant accounts for the various possible solutions that the differential equation might have.
  • Represents a family of solutions.
  • Typically involves an arbitrary constant \(C\).
  • Formed after integration or other solution methods.
Finding the general solution is akin to finding the overall pattern or formula which all specific solutions of the differential equation follow. In the example given, after performing a series of steps involving rearrangement, separation, and integration, the solution incorporates a \(C\), allowing for flexibility and application to boundary conditions.
Separation of Variables
Separation of Variables is a strategic method used to simplify solving a differential equation. It involves rearranging the terms so that all occurrences of a particular variable, such as \(y\), are on one side of the equation, and all occurrences of the other variable, such as \(x\), are on the other side.
  • This simplification allows each side to be integrated separately.
  • Useful for first-order differential equations.
  • Facilitates integration by isolating variables.
In our original problem, we aimed to isolate \(y\) and its derivatives on one side of the equation, and \(x\) and its derivatives on the other. This separation allows us to handle each variable's behavior separately and solve for the general solution more straightforwardly.
Integration
Integration is a mathematical process used to find the integral of a function, which can be thought of as the inverse operation of differentiation. During the solution of the differential equation, both sides are integrated with respect to their respective variables.
  • Left side integration involves terms such as \(\int y^{-3} dy\).
  • Right side involves \(\int \left( \frac{5}{x^2} + \frac{2}{x} \right) dx\).
  • Results include arbitrary constants \(C_1\) and \(C_2\) on each side.
The results of these integrations yield expressions that form part of our general solution. Solving these integrated expressions often involves simplifying or rearranging terms to solve for the dependent variable, which in this case is \(y\).
Boundary Value Problems
Boundary value problems (BVP) involve finding solutions to differential equations which also satisfy certain pre-set conditions, known as boundary conditions. These conditions typically specify the values of the solution at specific points in the domain.
  • Boundary conditions ensure solutions meet specific criteria.
  • Helps pinpoint particular solutions within the general solution family.
  • Often involves applying physical, initial, or boundary constraints.
While the original example focused on the general solution, the principles learned there apply directly to boundary value problems. Once a general solution is known, boundary conditions can be applied to solve for the constants, narrowing the solution to meet specific problem requirements.

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Most popular questions from this chapter

Find explicit particular solutions of the initial value problems $$ \frac{d y}{d x}=2 x y^{2}+3 x^{2} y^{2}, \quad y(1)=-1 $$

The amount \(A(t)\) of atmospheric pollutants in a certain mountain valley grows naturally and is tripling every \(7.5\) years. (a) If the initial amount is 10 pu (pollutant units), write a formula for \(A(t)\) giving the amount (in pu) present after \(t\) years. (b) What will be the amount (in pu) of pollutants present in the valley atmosphere after 5 years? (c) If it will be dangerous to stay in the valley when the amount of pollutants reaches 100 pu, how long will this take?

The intensity \(I\) of light at a depth of \(x\) meters below the surface of a lake satisfies the differential equation \(d I / d x=(-1.4) I .\) (a) At what depth is the intensity half the intensity \(I_{0}\) at the surface (where \(\left.x=0\right) ?\) (b) What is the intensity at a depth of \(10 \mathrm{~m}\) (as a fraction of \(I_{0}\) )? (c) At what depth will the intensity be \(1 \%\) of that at the surface?

A pitcher of buttermilk initially at \(25^{\circ} \mathrm{C}\) is to be cooled by setting it on the front porch, where the temperature is \(0^{\circ} \mathrm{C}\). Suppose that the temperature of the buttermilk has dropped to \(15^{\circ} \mathrm{C}\) after \(20 \mathrm{~min}\). When will it be at \(5^{\circ} \mathrm{C}\) ?

Verify that the given differential equation is exact; then solve it. $$ \left(3 x^{2} y^{3}+y^{4}\right) d x+\left(3 x^{3} y^{2}+y^{4}+4 x y^{3}\right) d y=0 $$
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