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Chapter 6: Problem 27

Solve the Bernoulli differential equation. $$ y^{\prime}+\left(\frac{1}{x}\right) y=x y^{2} $$

Short Answer

Expert verified
The solution to the Bernoulli differential equation \(y' + \frac{1}{x} y = x y^2\) is given by \(y = 2/(x (x + 2C))\) for a constant C.

Step by step solution

01

Transforming the Bernoulli differential equation

We can transform the Bernoulli differential equation \(y' + \frac{1}{x} y = x y^2\) into a linear differential equation by a substitution. We will do a change of variables from y to v, where v = 1/y. This gives us, \(v' = - y^{-2} y'\). So the new equation becomes \( -v' - \frac{1}{x}v = x \).
02

Solving the linear differential equation

We have transformed the Bernoulli differential equation into a linear differential equation that can be solved using integrating factors. The integrating factor is given by \(e^{-\int \frac{1}{x} dx}\), which simplifies to \(1/x\). We multiply this through the equation giving us \(-\frac{1}{x} v' + \frac{1}{x^2} v = 1\). Then the left hand side is the derivative of \(v/x\). We can integrate this equation to find \(v = x \int dx = x^2 / 2 + Cx\) for some constant C.
03

Substitute back to solve for y

Substituting back in gives 1/y = v = x^2 / 2 + Cx. Hence, y = 1/(x^2 / 2 + Cx) or y = 2/(x (x + 2C)).

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Key Concepts

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

Linear Differential Equation
A linear differential equation is one of the easiest types of differential equations to work with. It means that the function and its derivatives only appear to the power of one, without any powers or tricks like roots or products of the derivatives. That's why they are called "linear," much like a straight line. In this problem, after a clever change of the variables, the more complex equation was simplified to a linear form:
  • The term \(v\) replaces \(y\) in the substitution.
  • The presence of \(v\) instead of \(y\) transforms the structure of the problem entirely.
  • The resulting differential equation becomes easy to solve, like straightening out a puzzle.
Understanding how transforming a complex equation into a linear one provides an opportunity to leverage simpler methods like the method of integrating factors, which simplifies future steps.
Integrating Factor
The integrating factor is a neat trick used to solve linear differential equations. It serves like a magical multiplier simplifying the equation, making it readily solvable. In our transformed equation, after converting to a linear form, we use an integrating factor:
  • To find the integrating factor here, we look at the coefficient of \(v\).
  • Calculate the exponential \(e^{- rac{1}{x}}\), which simplifies to \(1/x\) in this case.
  • Multiplying every term in the equation by the integrating factor gives a neat expression, which can be directly integrated.
This clever use of integrating factors assists in solving the step seamlessly, giving us the solution to the linear differential equation, which eventually helps us solve for \(y\) after back substitution.
Change of Variables
One powerful technique when facing seemingly complex equations is the change of variables. It's the concept of substituting a new variable for an existing one to simplify the equation's form. In our exercise, the change from \(y\) to \(v = 1/y\) transformed the Bernoulli differential equation into a manageable linear form:
  • Replacing \(y\) with \(v = 1/y\) switched the nonlinear aspect of the equation into a linear one.
  • This substitution allowed us to use simple methods like integrating factors to solve the equation.
  • Finally, after solving in terms of \(v\), we could substitute back to get the actual solution for \(y\).
The change of variables is a crucial technique that can simplify complex problems into more digestible formats. It's a fundamental concept in solving Bernoulli equations and is essential for breaking down overwhelming equations into parts you can easily solve step by step.

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Most popular questions from this chapter

The logistic equation models the growth of a population. Use the equation to (a) find the value of \(k\), (b) find the carrying capacity, (c) find the initial population, (d) determine when the population will reach \(\mathbf{5 0 \%}\) of its carrying capacity, and (e) write a logistic differential equation that has the solution \(P(t)\). \(P(t)=\frac{5000}{1+39 e^{-0.2 t}}\)

In your own words, describe the difference between a general solution of a differential equation and a particular solution.

Solve the Bernoulli differential equation. $$ y^{\prime}+3 x^{2} y=x^{2} y^{3} $$

At time \(t=0\), a bacterial culture weighs 1 gram. Two hours later, the culture weighs 2 grams. The maximum weight of the culture is 10 grams. (a) Write a logistic equation that models the weight of the bacterial culture. (b) Find the culture's weight after 5 hours. (c) When will the culture's weight reach 8 grams? (d) Write a logistic differential equation that models the growth rate of the culture's weight. Then repeat part (b) using Euler's Method with a step size of \(h=1\). Compare the approximation with the exact answers. (e) At what time is the culture's weight increasing most rapidly? Explain.

Give the standard form of the Bernoulli equation. Describe how one reduces it to a linear equation.
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