Question

Bernoulli's Equation. $$y^{\prime}=y-x y^{2}(x+2)$$

Short answer

The Bernoulli differential equation must be transformed into a linear differential equation using the substitution method, then solved with the integrating factor method, and finally revert back to the original variable to find the solution for \(y\).

Step-by-step solution

Identify the Type of Differential Equation

The given equation is of the form that suggests it can be approached using Bernoulli's equation. Bernoulli's equations are nonlinear differential equations that can be solved by transforming them into a linear differential equation. The general form of Bernoulli's equation is \(y'+P(x)y=Q(x)y^n\), where \(n\) is any real number. In our case, \(n = 2\).

Rewrite the Differential Equation in Bernoulli's Form

To get the equation in the standard form of Bernoulli's equation, we separate y terms on one side. The given equation is already in the proper form where \(P(x) = -x(x+2)\) and \(Q(x) = -x(x+2)\), with \(n = 2\).

Perform a Variable Substitution

We introduce a substitution \(v = y^{1-n} = y^{-1}\) which implies that \(y = v^{-1}\) and then find the derivative \(y' = -v^{-2}v'\). Substitute \(y\) and \(y'\) into the original differential equation to get an equation in \(v\) and \(v'\).

Substitute and Obtain a Linear Differential Equation

Substituting \(y = v^{-1}\) and \(y' = -v^{-2}v'\) into the original equation yields \(-v^{-2}v' = v^{-1} - xv^{-2} + 2x\). Multiply through by \(-v^{2}\) to clear the fractions and get \(v' = v - x + 2xv^{2}\), which is a linear differential equation in \(v\).

Solve the Linear Differential Equation for \(v\)

The obtained linear differential equation is in the form \(v' + P(x)v = Q(x)\), where \(P(x) = -1\) and \(Q(x) = x(2v-1)\). Solve this equation using an integrating factor, \(\mu(x) = e^{\int P(x) \, dx} = e^{-x}\), and the formula for the integrating factor method, \(v = \frac{1}{\mu(x)}\left(\int \mu(x)Q(x)dx + c\right)\), where \(c\) is the constant of integration.

Revert Back to Original Variable

After finding the general solution for \(v\), revert back to \(y\) by using the original substitution \(y = v^{-1}\). The final solution will be expressed in terms of \(y\).

Find the Particular Solution

If initial conditions are provided, use them to find the particular solution for \(y\) by substituting the initial condition into the general solution and solving for the constant of integration, \(c\).

Key concepts

Differential Equations

A differential equation is a mathematical equation that involves functions and their derivatives. It's a tool that describes the relationship between a function and its rates of change, enabling us to model and analyze a variety of dynamic systems such as population growth, motion, and heat conduction.

Many physical phenomena can be formulated as differential equations. One type of these equations is the Bernoulli's differential equation, which has a standard form of \(y'+P(x)y=Q(x)y^n\), where \(P(x)\) and \(Q(x)\) are functions of \(x\) and \(n\) is any real number. When \(n=0\) or \(n=1\), the equation is linear; otherwise, it's non-linear.

When faced with a differential equation like the one in the original exercise, one of the key steps is to manipulate the equation into a recognizable form. In this case, we recognized the Bernoulli's form and strategized on transforming it into a linear equation by means of a substitution, which simplifies the equation and makes finding a solution possible.

Variable Substitution

In mathematics, variable substitution is a technique used to simplify problems. Within the context of differential equations, substitution is often employed to transform a complex, non-linear equation into one that's easier to solve.

In the given exercise, we utilize a substitution to convert the given Bernoulli's equation into a linear one by introducing a new variable \(v\). Specifically, the substitution \(v = y^{1-n} = y^{-1}\) translates the original variables into terms involving \(v\) and \(v'\), the derivative of \(v\) with respect to \(x\). This substitution results in a linear differential equation, which can often be solved by conventional methods like separation of variables or integrating factors.

The ability to recognize the appropriate substitution and the know-how to perform it correctly are essential skills when dealing with differential equations. Substitution simplifies the equation and paves the way for finding a general solution before reverting back to the original variable for the final answer.

Integrating Factor Method

The integrating factor method is a powerful technique used to solve linear differential equations of the form \(v' + P(x)v = Q(x)\). The process involves multiplying the differential equation by an integrating factor that will allow it to be written in a form where the left-hand side becomes the derivative of a product of two functions. This integrating factor, denoted as \(\mu(x)\), is typically \(e^{\int P(x) dx}\).

For the exercise at hand, once we obtained the equation in terms of the substitution variable \(v\), we calculated the integrating factor \(\mu(x) = e^{-x}\), and used it to transform the equation into a form where the solution \(v\) could be isolated. Integrating both sides then provided us with the general solution for \(v\). The final step involved reverting back to the original variable \(y\) using the inverse of our initial substitution.

Mastering the integrating factor method is crucial when tackling first-order linear differential equations, as it not only aids in solving the equation but also in understanding the behavior of the solutions across various contexts.