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

Short Answer

Expert verified
The solution to the differential equation is \( y = \frac{C_1}{1-C_1} \), where \( C_1 \) is a constant.

Step by step solution

01

Separate Variables

The given differential equation is \( \frac{d y}{d x}=y(1+y) \). To solve this, we start by separating variables. Divide both sides by \( y(1+y) \) and multiply both sides by \( dx \). This gives us:\[\frac{1}{y(1+y)} \, dy = dx\]
02

Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \). This means:\[\int \frac{1}{y(1+y)} \, dy = \int \, dx\]
03

Partial Fraction Decomposition

Before integrating the left side, perform partial fraction decomposition:\[\frac{1}{y(1+y)} = \frac{1}{y} - \frac{1}{1+y}\]
04

Integrate Using Partial Fractions

Integrate the decomposed fractions separately:\[\int \left( \frac{1}{y} - \frac{1}{1+y} \right) \, dy = \ln |y| - \ln |1+y|\]For the right side:\[\int \, dx = x + C\]
05

Combine and Simplify

After integration, equate and simplify:\[\ln |y| - \ln |1+y| = x + C\]Using logarithm properties, combine:\[\ln \left( \frac{|y|}{|1+y|} \right) = x + C\]
06

Exponentiate to Solve for y

Exponentiate both sides to solve for \( y \):\[\frac{|y|}{|1+y|} = e^{x+C} = Ce^x\]Here \( C \) is an arbitrary constant that represents \( e^C \). Let \( C_1 \) be \( Ce^x \), simplify further:\[|y| = C_1 |1+y|\]
07

Simplified General Solution

Depending on the values/signs of \( y \) and \( 1+y \), resolve the absolute values:\[y = C_1 (1+y)\]Solve for \( y \):\[y(1-C_1) = C_1\]\[y = \frac{C_1}{1-C_1}\]Thus, the solution describes a family of solutions parameterized by \( C_1 \).

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Key Concepts

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

Separation of Variables
The technique of separation of variables is a powerful method used to solve differential equations. When you have an equation like \( \frac{d y}{d x}=y(1+y) \), separation of variables allows you to rearrange it so each variable appears on opposite sides of the equation. This way, you can independently integrate both sides.

In this case, the original differential equation is separated by dividing both sides by \( y(1+y) \) and multiplying by \( dx \). The result is \( \frac{1}{y(1+y)} \, dy = dx \). This creates an equation that can be integrated with respect to its own variable, making it possible to solve the differential equation step by step.

Separation of variables is particularly useful because it simplifies the integration process by reducing the problem to two independent integrals. This method can tackle a variety of differential equations, provided they can be rewritten with the separated forms.
Partial Fraction Decomposition
When faced with complex rational expressions during integration, partial fraction decomposition can be a lifesaver. This technique involves breaking down a complicated fraction into simpler fractions that are easier to integrate. In the context of this problem, the expression \( \frac{1}{y(1+y)} \) is decomposed into \( \frac{1}{y} - \frac{1}{1+y} \).

To perform partial fraction decomposition, you consider the denominator's factors. Here, \( y(1+y) \) is expressed as two separate terms, \( y \) and \( 1+y \). By writing the initial fraction as a sum or difference of individual fractions corresponding to these terms, the resulting expressions become more straightforward to handle during integration.

This method is particularly useful for rational expressions where the degree of the numerator is less than the degree of the denominator. It’s a staple in calculus because it simplifies integrals, making them more approachable for students.
Integration Techniques
Integration is a vital part of solving differential equations, turning derivative expressions into function statements. Here, once variables in the differential equation are separated and partial fraction decomposition is applied, integration can proceed smoothly.

For the decomposed expressions \( \frac{1}{y} \) and \( \frac{1}{1+y} \), the integration results in natural logarithms: \( \int \frac{1}{y} \, dy = \ln |y| \) and \( \int \frac{1}{1+y} \, dy = \ln |1+y| \).

These results are obtained using a standard rule of integration for \( \frac{1}{u} \), which equals \( \ln |u| + C \). This technique is common when dealing with rational expressions.

Additionally, integrating \( dx \) results in \( x + C \), where \( C \) is the constant of integration. This constant captures the family of solutions related to initial conditions or boundary values.

Mastering integration techniques is crucial in calculus, as they enable the connection between derivatives and antiderivatives, opening doors to solving a wide array of mathematical problems.

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Most popular questions from this chapter

Suppose that a tank holds 1000 liters of water, and \(2 \mathrm{~kg}\) of salt is poured into the tank. (a) Compute the concentration of salt in g liter \(^{-1}\). (b) Assume now that you want to reduce the salt concentration. One method would be to remove a certain amount of the salt water from the tank and then replace it by pure water. How much salt water do you have to replace by pure water to obtain a salt concentration of \(1 \mathrm{~g}\) liter \(^{-1}\) ? (c) Another method for reducing the salt concentration would be to hook up an overflow pipe and pump pure water into the tank. That way, the salt concentration would be gradually reduced. Assume that you have two pumps, one that pumps water at a rate of 1 liter \(\mathrm{s}^{-1}\), the other at a rate of 2 liter \(\mathrm{s}^{-1} .\) For each pump, find out how long it would take to reduce the salt concentration from the original concentration to \(1 \mathrm{~g}\) liter \(^{-1}\) and how much pure water is needed in each case. (Note that the rate at which water enters the tank is equal to the rate at which water leaves the tank.) Compare the amount of water needed using the pumps with the amount of water needed in part (b).

Solve the given autonomous differential equations. \(\frac{d N}{d t}=5-N\), where \(N(2)=3\)

Suppose that \(N(t)\) denotes the size of a population at time \(t .\) The population evolves according to the logistic equation, but, in addition, predation reduces the size of the population so that the rate of change is given by $$\frac{d N}{d t}=N\left(1-\frac{N}{50}\right)-\frac{9 N}{5+N}$$ The first term on the right-hand side describes the logistic growth; the second term describes the effect of predation. (a) Set $$g(N)=N\left(1-\frac{N}{50}\right)-\frac{9 N}{5+N}$$ and graph \(g(N)\). (b) Find all equilibria of \((8.65)\). (c) Use your graph in (a) to determine the stability of the equilibria you found in (b). (d) Use the method of eigenvalues to determine the stability of the equilibria you found in (b).

Solve the given autonomous differential equations. \(\frac{d y}{d x}=3 y\), where \(y_{0}=2\) for \(x_{0}=0\)

Solve each differential equation with the given initial condition. \(\frac{d y}{d x}=(y+1) e^{-x}\), with \(y_{0}=2\) if \(x_{0}=0\)

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