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Solve the differential equations in Problems by regarding \(y\) as the independent variable rather than \(x\). \((1+2 x y) \frac{d y}{d x}=1+y^{2}\)

Short Answer

Expert verified
Rearrange: \(\frac{dx}{dy} = \frac{1+y^2}{1+2xy}\); integrate with respect to \(y\), solve the integral, and find the general solution "x = f(y) + C".

Step by step solution

01

Rearrange the Differential Equation

Given the differential equation \((1+2xy)\frac{dy}{dx} = 1 + y^2\), we want to treat \(y\) as the independent variable. To do this, rearrange the equation by dividing both sides by \(\frac{dy}{dx}\) to express \(\frac{dx}{dy}\): \[ \frac{dx}{dy} = \frac{1+y^2}{1+2xy}. \]
02

Integrate with Respect to y

We will integrate both sides with respect to \(y\). The equation \(\frac{dx}{dy} = \frac{1+y^2}{1+2xy}\) can be rewritten as \(\int dx = \int \frac{1+y^2}{1+2xy} dy\). This requires a separation of variables or a suitable substitution method to simplify.
03

Solve the Integral

To solve \(\int \frac{1+y^2}{1+2xy} dy\), observe that a potential method involves expressing the right side in a form that allows integration, potentially using partial fractions or substitution. Recognize \(1+2xy\) might suggest \(x\) as a constant or substitute \(t=1+2xy\) when integrating.
04

Determine the General Solution

Once the integral is calculated, express \(x\) in terms of \(y\) plus a constant of integration. This produces a function \(x=f(y)+C\) where \(C\) is an arbitrary constant reflecting the family of solutions.

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Key Concepts

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

Independent Variable Substitution
In differential equations, sometimes it is beneficial to change our perspective by treating one variable as independent instead of the one initially given. When we change the independent variable from \(x\) to \(y\), it means we look at how changes in \(y\) affect \(x\) instead of the other way around. This is called independent variable substitution.
This approach was used in the given differential equation \((1+2xy)\frac{dy}{dx} = 1+y^2\). By rearranging, we aim to express \(\frac{dx}{dy}\) instead of \(\frac{dy}{dx}\), transforming our perspective and giving new ways to tackle the problem. This rearrangement leads to:

\[ \frac{dx}{dy} = \frac{1+y^2}{1+2xy}. \]

Independent variable substitution can simplify and bring clarity, especially when integration becomes more intuitive when flipping the variables. This new view can transform challenging problems into solvable ones by tapping into more familiar integration techniques. Remember, this technique and perspective can be your gateway to new problem-solving paths.
Separation of Variables
Separation of variables is a method widely used for solving differential equations. The key idea is to rearrange the equation so that all components involving one variable and its differential are on one side and components involving the other variable are on the other side.
In the problem, \( \frac{dx}{dy} = \frac{1+y^2}{1+2xy} \), separating terms to facilitate integration implies arranging terms such that the left-hand side only involves \(dx\) and the right-hand side only employs \(dy\).

This separation allows each side to be integrated independently:
  • Integrate \(1\) with respect to \(x\), and
  • Integrate \(\frac{1+y^2}{1+2xy}\) with respect to \(y\).

This preparation creates paths for solving the integral, making the problem more approachable by breaking it into smaller, more manageable parts.
Integration Techniques
To solve differential equations using integration, understanding various techniques is helpful. Our problem encourages exploring integration with respect to \(y\) for the expression \( \frac{1+y^2}{1+2xy} \), which might not integrate neatly with simple methods. Here’s a peek into some techniques you might need:

  • Substitution Method: This is where we change variables to simplify the integrand. For our problem, substituting \(t = 1+2xy\) can simplify the integration process.
  • Partial Fraction Decomposition: If applicable, expressing the integrand as a sum or difference can make integration simpler.

The essence is tailoring an approach that considers the integral's structure. Sometimes you need to revisit the problem and decide the best method to apply based on the relationships you observe within the equation. Integration often reveals more when you choose the right technique.
Constant of Integration
When you integrate, especially with indefinite integrals, a constant of integration, often denoted by \(C\), becomes part of your solution. It represents a family of solutions rather than a single result. In our equation, once integration provided us with an expression like \(x=f(y)+C\), \(C\) embodies all specific solutions that fit the original equation.

Understanding \(C\) is crucial. It allows the solution to adjust to boundary conditions or specific initial conditions present in practical problems. It acts as a flexible component that ensures the comprehensive set of solutions covers all possibilities a differential equation can yield.
When solving real-world problems, knowing the constant lets you adapt your general solution to fit particular scenarios, proving the broad applicability and adaptability of the results you derive through integration.

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Most popular questions from this chapter

Show that the solution curves of the differential equation $$ \frac{d y}{d x}=-\frac{y\left(2 x^{3}-y^{3}\right)}{x\left(2 y^{3}-x^{3}\right)} $$ are of the form \(x^{3}+y^{3}=C x y\)

Find a general solution of each reducible second-order differential equation. Assume \(x, y\) and/or \(y^{\prime}\) positive where helpful (as in Example I1). $$ y y^{\prime \prime}+\left(y^{\prime}\right)^{2}=y y^{\prime} $$

Find a general solution of each reducible second-order differential equation. Assume \(x, y\) and/or \(y^{\prime}\) positive where helpful (as in Example I1). $$ x y^{\prime \prime}=y^{\prime} $$

(a) Find constants \(A\) and \(B\) such that \(y_{p}(x)=A \sin x+\) \(B \cos x\) is a solution of \(d y / d x+y=2 \sin x . \quad\) (b) Use the result of part (a) and the method of Problem 31 to find the general solution of \(d y / d x+y=2 \sin x .\) (c) Solve the initial value problem \(d y / d x+y=2 \sin x, y(0)=1\).

Problems deal with a shallow reservoir that has a one-square-kilometer water surface and an average water depth of 2 meters. Initially it is filled with fresh water, but at time \(t=0\) water contaminated with a liquid pollutant begins flowing into the reservoir at the rate of 200 thousand cubic meters per month. The well-mixed water in the reservoir flows out at the same rate. Your first task is to find the amount \(x(t)\) of pollutant (in millions of liters) in the reservoir after 1 months. The incoming water has a pollutant concentration of \(c(t)=10\) liters per cubic meter \(\left(\mathrm{L} / \mathrm{m}^{3}\right)\). Verify that the graph of \(x(t)\) resembles the steadily rising curve in Fig. 1.5.9, which approaches asymptotically the graph of the equilibrium solution \(x(t)=20\) that corresponds to the reservoir's long-term pollutant content. How long does it take the pollutant concentration in the reservoir to reach \(10 \mathrm{~L} / \mathrm{m}^{3} ?\)

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