Question

Use the method outlined in Exercise 43 to solve the following Bernoulli equations. a. \(y^{\prime}(t)+y=2 y^{2}\) b. \(y^{\prime}(t)-2 y=3 y^{-1}\) c. \(y^{\prime}(t)+y=\sqrt{y}\)

Short answer

Question: Solve the following Bernoulli's differential equations: a. \(y^{\prime}(t)+y=2y^2\) b. \(y^{\prime}(t)-2y=3y^{-1}\) c. \(y^{\prime}(t)+y=\sqrt{y}\) Answer: a. After solving the Bernoulli's differential equation, we get: \(y(t) = \frac{1}{c_1 - e^{-t}}\) , where \(c_1\) is an arbitrary constant. b. After solving the Bernoulli's differential equation, we get: \(y(t) = \sqrt[3]{\frac{3}{2}(c_2e^{2t} - 1)}\) , where \(c_2\) is an arbitrary constant. c. After solving the Bernoulli's differential equation, we get: \(y(t) = \left[\frac{2}{3} \tanh{\left(\frac{t+c_3}{3}\right)} + \frac{2}{3}\right]^2\) , where \(c_3\) is an arbitrary constant.

Step-by-step solution

Problem (a): Rewrite the Equation

Divide both sides by \(y^{2}\) to get: $$\frac{y^{\prime}(t)}{y^{2}}+\frac{1}{y}=2$$

Problem (a): Introduce New Variable \(v\)

Define \(v(t) = \frac{1}{y(t)}\), and differentiate to obtain: $$v^{\prime}(t)=-\frac{1}{y^{2}(t)}\cdot y^{\prime}(t)$$

Problem (a): Substitute and Solve

Substitute \(v(t)\) and \(v^{\prime}(t)\) into our modified equation: $$-v^{\prime}(t) + v = 2$$ Solve the linear differential equation above for \(v(t)\), and then find \(y(t)\) from \(v(t) = \frac{1}{y(t)}\).

Problem (b): Rewrite the Equation

Divide both sides by \(y^{-1}\) to get: $$y^2y^{\prime}(t)-2y^{3}=3$$

Problem (b): Introduce New Variable \(v\)

Define \(v(t) = y^{3}(t)\), and differentiate to obtain: $$v^{\prime}(t)=3y^{2}(t)y^{\prime}(t)$$

Problem (b): Substitute and Solve

Substitute \(v(t)\) and \(v^{\prime}(t)\) into our modified equation: $$\frac{1}{3}v^{\prime}(t) - 2v = 3$$ Solve the linear differential equation above for \(v(t)\), and then find \(y(t)\) from \(v(t) = y^3(t)\).

Problem (c): Rewrite the Equation

Divide both sides by \(y^{0.5}\) to get: $$y^{0.5}y^{\prime}(t)+y^{1.5} = y^{0.5}$$

Problem (c): Introduce New Variable \(v\)

Define \(v(t) = y^{1.5}(t)\), and differentiate to obtain: $$v^{\prime}(t)=1.5y^{0.5}(t)y^{\prime}(t)$$

Problem (c): Substitute and Solve

Substitute \(v(t)\) and \(v^{\prime}(t)\) into our modified equation: $$\frac{2}{3}v^{\prime}(t) + v = y^{0.5}$$ Solve the linear differential equation above for \(v(t)\), and then find \(y(t)\) from \(v(t) = y^{1.5}(t)\).

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In simpler terms, they connect how a certain quantity changes with respect to another. These equations are essential to understanding many natural phenomena where change is a key component, such as the movement of objects or the growth of populations. A differential equation can be classified based on the degree and order of an equation, but most commonly you will encounter ordinary differential equations (ODEs) which involve a single independent variable. To solve these equations, methods like separation of variables, integrating factors, and variable substitution are often used, each suitable for different types of differential equations. Techniques vary depending on the specifics contained in the equation itself.

Linear Differential Equation

A linear differential equation refers to an equation where the unknown function and its derivatives appear to the power of one (hence, it is linear). Importantly, no products of the unknown function or its derivatives are allowed. In terms of applications, these types of equations can model different processes like cooling, harmonic motion, and electrical circuits. When solving linear differential equations, the most common method used is finding the integrating factor, which helps simplify the integration process. The structure of these problems allows for systematic approaches, making them generally easier to handle compared to their nonlinear counterparts. Understanding linear differential equations is a stepping stone to tackling more complex forms, such as the Bernoulli equation seen here.

Variable Substitution

Variable substitution is a powerful method used in solving differential equations, particularly useful for equations that are not in a directly solvable form. This technique involves introducing a new variable to transform a complex differential equation into a simpler one, often linear or separable in nature. Here's how it generally works:

• Identify the non-linear component of the equation that complicates direct solving.
• Choose a substitution that simplifies the equation, ideally reducing it to a linear form. For example, in the Bernoulli equation, you might set a new variable to alter a non-linear term.
• Differentiating this new variable allows you to substitute back into the original equation, simplifying it into a form solvable by standard methods.
• Solve the resulting equation, then reverse the substitution to find the solution for the original variable.

Using substitution effectively requires practice but is invaluable for tackling a variety of differential equations, such as Bernoulli's equation, where it assists in linearizing the problem.