Differential Equations - Bernoulli Differential Equation & its Reduction to a Linear DE

  • Bernoulli differential equation is of the form:
    • dy/dx + P(x)y = Q(x)y^n
  • To solve a Bernoulli differential equation, we perform a substitution which reduces it to a linear differential equation.
  • The substitution used for reduction is:
    • u = y^(1-n)
  • By performing this substitution, the Bernoulli differential equation can be transformed into a linear differential equation.
  • The steps for reduction of a Bernoulli differential equation to a linear differential equation are as follows:
    1. Write the given Bernoulli differential equation in standard form: dy/dx + P(x)y = Q(x)y^n

    2. Perform the substitution: u = y^(1-n)

    3. Differentiate both sides of the substitution equation with respect to x:

      • du/dx = (1-n)y^(-n)(dy/dx)

Differential Equations - Bernoulli Differential Equation & its Reduction to a Linear DE (cont.)

  • Substituting this result into the original Bernoulli differential equation, we get: (1-n)y^(-n)(dy/dx) + P(x)y = Q(x)y^n

    Simplifying this equation, we get:

    (1-n)du/dx + P(x)y^(1-n) = Q(x)y^n

  • Multiply the entire equation by (1-n): du/dx + (1-n)P(x)y^(1-n) = (1-n)Q(x)y^n

  • The resulting equation is a linear differential equation.

  • Solve this linear differential equation using the appropriate methods for linear differential equations.

  • Once the solution is found, we substitute back the value of y^(1-n) to get the solution for the original Bernoulli differential equation.

  • Let’s look at an example to better understand the process. Example: Solve the Bernoulli differential equation: dy/dx + 2xy = x^2y^3

Example: Bernoulli Differential Equation

  • Given equation: dy/dx + 2xy = x^2y^3
  • Perform the substitution: u = y^(1-3) = y^(-2)
  • Differentiate both sides of the substitution equation with respect to x:
    • du/dx = -2y^(-3)(dy/dx)
  • Substitute the values into the original equation:
    • (-2y^(-3)(dy/dx)) + 2xy = x^2y^3
  • Multiply the equation by (-2):
    • 2y^(-3)(dy/dx) - 4xy = -2x^2y^3

Example: Bernoulli Differential Equation (cont.)

  • Simplify the equation:
    • 2(dy/dx)/y^3 - 4xy = -2x^2y^3
  • Rearrange the terms:
    • 2(dy/dx)/y^3 = 4xy - 2x^2y^3
  • Divide both sides by 2:
    • (dy/dx)/y^3 = 2xy - x^2y^3
  • This is now a linear differential equation.
  • Solve this linear differential equation to find the solution for u.
  • Once u is found, substitute back the value of u to get the solution for the original Bernoulli differential equation.
  • Let’s solve this step by step.

Steps: Solving the Linear Differential Equation

  1. Write the given linear differential equation in standard form: (dy/dx)/y^3 = 2xy - x^2y^3
  1. Multiply both sides by y^3:
    • dy/dx = 2xy^4 - x^2y^6
  1. Separate the variables and integrate:
    • ∫(1/y^4)dy = ∫(2xy - x^2y^3)dx
  1. Evaluate the integrals on both sides:
    • (-1/3)y^(-3) = x^2y^2 - (1/3)x^3y^4 + C
  1. Simplify the equation by multiplying through by 3:
    • -y^(-3) = 3x^2y^2 - x^3y^4 + 3C

Steps: Solving the Linear Differential Equation (cont.)

  1. Rearrange the terms to isolate y:
    • y^(-3) - 3x^2y^2 + x^3y^4 = -3C
  1. Multiply through by y^3 to get rid of the negative exponent:
    • 1 - 3x^2y^5 + x^3y^7 = -3Cy^3
  1. Simplify the equation:
    • x^3y^7 - 3x^2y^5 - 3Cy^3 + y^3 - 1 = 0
  1. This equation is the solution for u.
  1. Substitute back the value of u = y^(-2) to get the solution for the original Bernoulli differential equation.
  • The above steps demonstrate the process of solving a Bernoulli differential equation using reduction to a linear differential equation.
  • Practice various examples to gain proficiency in solving Bernoulli differential equations.

Differential Equations - Bernoulli Differential Equation & its Reduction to a Linear DE

  • Bernoulli differential equation is of the form: dy/dx + P(x)y = Q(x)y^n
  • To solve a Bernoulli differential equation, we perform a substitution which reduces it to a linear differential equation.
  • The substitution used for reduction is: u = y^(1-n)
  • By performing this substitution, the Bernoulli differential equation can be transformed into a linear differential equation.
  • The steps for reduction of a Bernoulli differential equation to a linear differential equation are as follows:
    1. Write the given Bernoulli differential equation in standard form: dy/dx + P(x)y = Q(x)y^n
    2. Perform the substitution: u = y^(1-n)
    3. Differentiate both sides of the substitution equation with respect to x:
      • du/dx = (1-n)y^(-n)(dy/dx)

Differential Equations - Bernoulli Differential Equation Reduction (cont.)

  • Substituting this result into the original Bernoulli differential equation, we get: (1-n)y^(-n)(dy/dx) + P(x)y = Q(x)y^n

  • Simplifying this equation, we get:

    (1-n)du/dx + P(x)y^(1-n) = Q(x)y^n

  • Multiply the entire equation by (1-n):

    du/dx + (1-n)P(x)y^(1-n) = (1-n)Q(x)y^n

  • The resulting equation is a linear differential equation.

  • Solve this linear differential equation using the appropriate methods for linear differential equations.

  • Once the solution is found, we substitute back the value of y^(1-n) to get the solution for the original Bernoulli differential equation.

  • Let’s look at an example to better understand the process.

Example: Bernoulli Differential Equation

  • Given equation: dy/dx + 2xy = x^2y^3
  • Perform the substitution: u = y^(1-3) = y^(-2)
  • Differentiate both sides of the substitution equation with respect to x:
    • du/dx = -2y^(-3)(dy/dx)
  • Substitute the values into the original equation:
    • (-2y^(-3)(dy/dx)) + 2xy = x^2y^3
  • Multiply the equation by (-2):
    • 2y^(-3)(dy/dx) - 4xy = -2x^2y^3
  • Simplify the equation:
    • 2(dy/dx)/y^3 - 4xy = -2x^2y^3
  • Rearrange the terms:
    • 2(dy/dx)/y^3 = 4xy - 2x^2y^3

Steps: Solving the Linear Differential Equation

  1. Write the given linear differential equation in standard form: (dy/dx)/y^3 = 2xy - x^2y^3
  1. Multiply both sides by y^3:
    • dy/dx = 2xy^4 - x^2y^6
  1. Separate the variables and integrate:
    • ∫(1/y^4)dy = ∫(2xy - x^2y^3)dx
  1. Evaluate the integrals on both sides:
    • (-1/3)y^(-3) = x^2y^2 - (1/3)x^3y^4 + C
  1. Simplify the equation by multiplying through by 3:
    • -y^(-3) = 3x^2y^2 - x^3y^4 + 3C

Steps: Solving the Linear Differential Equation (cont.)

  1. Rearrange the terms to isolate y:
    • y^(-3) - 3x^2y^2 + x^3y^4 = -3C
  1. Multiply through by y^3 to get rid of the negative exponent:
    • 1 - 3x^2y^5 + x^3y^7 = -3Cy^3
  1. Simplify the equation:
    • x^3y^7 - 3x^2y^5 - 3Cy^3 + y^3 - 1 = 0
  1. This equation is the solution for u.
  1. Substitute back the value of u = y^(-2) to get the solution for the original Bernoulli differential equation.
  • The above steps demonstrate the process of solving a Bernoulli differential equation using reduction to a linear differential equation.
  • Practice various examples to gain proficiency in solving Bernoulli differential equations.

Slide 21: Solving Bernoulli Differential Equations (Examples)

  • Example 1:

    • Given equation: dy/dx - 5xy^2 = 2x^5y^8
    • Reduce the equation by substituting u = y^(1-2) = y^(-1)
    • Differentiate both sides with respect to x: du/dx = -y^(-2) * dy/dx
    • Substitute the values back into the equation and simplify.
  • Example 2:

    • Given equation: dy/dx + x^2y^3 = 6xy^2
    • Reduce the equation by substituting u = y^(1-3) = y^(-2)
    • Differentiate both sides with respect to x: du/dx = -2y^(-3) * dy/dx
    • Substitute the values back into the equation and simplify.
  • Example 3:

    • Given equation: dy/dx + y/x = (e^x)y^2
    • Reduce the equation by substituting u = y^(1-2) = y^(-1)
    • Differentiate both sides with respect to x: du/dx = -y^(-2) * dy/dx
    • Substitute the values back into the equation and simplify.

Slide 22: Benefits of Using Bernoulli Differential Equation

  • Reduction of a Bernoulli differential equation to a linear differential equation makes it easier to solve.
  • By reducing to a linear differential equation, we can utilize various methods like separation of variables, integrating factors, or variation of parameters.
  • Bernoulli differential equation reduction allows us to solve a wider range of differential equations, especially those that are difficult to solve directly.
  • It provides a systematic approach to solving differential equations that involve non-linear terms and power functions.
  • The process of reduction helps in gaining a deeper understanding of the behavior and solutions of nonlinear differential equations.

Slide 23: Applications of Bernoulli Differential Equation

  • Bernoulli differential equations have various applications in different fields of science and engineering.
  • In population dynamics, the Bernoulli differential equation can model exponential growth or decay.
  • In physics, Bernoulli equations are used to analyze fluid flow and aerodynamics.
  • In economics, the Bernoulli differential equation can describe growth and decay in monetary systems.
  • In pharmacokinetics, the Bernoulli equation can be used to model drug absorption and elimination processes.
  • Bernoulli differential equations also have applications in control systems, chemistry, and biology.

Slide 24: Limitations of Bernoulli Differential Equation Reduction

  • Bernoulli differential equation reduction may not be possible if the value of n is not 1 or 2.
  • The reduction process can be complicated for higher values of n or when the equation includes higher order derivatives.
  • There may be cases where the original Bernoulli equation cannot be reduced to a linear differential equation.
  • Some Bernoulli differential equations may require additional techniques or approximations to solve, beyond the reduction process.
  • It is important to carefully analyze the given equation before attempting to reduce it to a linear differential equation.

Slide 25: Summary

  • Bernoulli differential equations involve nonlinear terms and can be reduced to linear differential equations using a substitution method.
  • The substitution used is u = y^(1-n).
  • By substituting this value into the original equation and simplifying, the Bernoulli equation becomes a linear differential equation.
  • Solving the resulting linear differential equation provides the solution for u.
  • Substituting the value of u back into the original substitution equation gives the solution for the original Bernoulli differential equation.
  • Bernoulli differential equations have various applications and can be solved using reduction techniques for easier analysis.

Slide 26: Additional Resources

  • Differential Equations: An Introduction by H. T. Davis
  • Differential Equations with Applications and Historical Notes by George F. Simmons
  • Differential Equations for Dummies by Steven Holzner
  • Khan Academy: Differential Equations
  • MIT OpenCourseWare: Ordinary Differential Equations

Slide 27: Q&A

  • Are there any questions about the reduction of Bernoulli differential equations?
  • Can you provide an example of a Bernoulli differential equation that cannot be reduced to a linear differential equation?
  • What are some real-world applications of Bernoulli differential equations?
  • Are there any alternative methods to solve Bernoulli differential equations?
  • How can we determine if a given differential equation is a Bernoulli equation?

Slide 28: References

  • Boyce, W. E., & DiPrima, R. C. (2012). Elementary differential equations and boundary value problems. Wiley.
  • Nagle, R. K., Saff, E. B., & Snider, A. D. (2016). Fundamentals of differential equations. Pearson Education.
  • Tenenbaum, M., & Pollard, H. (2018). Ordinary differential equations. Courier Dover Publications.

Slide 29: Thank You

  • Thank you for attending this lecture on Bernoulli differential equations and their reduction to linear differential equations.
  • If you have any further questions or need clarification, please feel free to reach out.
  • Stay tuned for more lectures on differential equations and other topics in mathematics.
  • Good luck with your studies and upcoming exams!