Integral Calculus - Trigonometric solution of ODE

Objectives

  • Understand basic terminologies in ODE
  • Learn methods for solving trigonometric ODEs
  • Solve examples using the trigonometric solution method

Definitions

  • Ordinary Differential Equation (ODE): An equation that relates an unknown function and its derivatives
  • Trigonometric ODE: A type of ODE that involves trigonometric functions
  • Solution of ODE: A function that satisfies the given ODE

Methods for solving Trigonometric ODEs

  1. Substituting trigonometric functions
  1. Using trigonometric identities
  1. Applying initial conditions

Method 1: Substituting trigonometric functions

  • Replace the unknown function and its derivatives with trigonometric functions
  • Choose appropriate trigonometric functions based on the given ODE
  • Solve for the coefficients of the trigonometric functions

Method 2: Using trigonometric identities

  • Simplify the given ODE using trigonometric identities
  • Express the ODE in terms of a single trigonometric function or a combination of trigonometric functions
  • Solve the resulting equation using algebraic techniques

Method 3: Applying initial conditions

  • If initial conditions are given in the problem, substitute them into the general solution obtained from methods 1 or 2
  • Solve for the unknown coefficients to obtain the particular solution

Example 1

Solve the following ODE:
$\frac{{d^2y}}{{dx^2}} + y = 0$

  • Substitute $y = A\sin(x) + B\cos(x)$ into the given ODE
  • Differentiate twice and substitute back into the ODE
  • Solve for the coefficients $A$ and $B$

Example 2

Solve the following ODE:
$\frac{{d^2y}}{{dx^2}} + 4y = 0$

  • Substitute $y = A\sin(2x) + B\cos(2x)$ into the given ODE
  • Differentiate twice and substitute back into the ODE
  • Solve for the coefficients $A$ and $B$

Example 3

Solve the following ODE:
$\frac{{d^2y}}{{dx^2}} + 9y = 0$

  • Substitute $y = A\sin(3x) + B\cos(3x)$ into the given ODE
  • Differentiate twice and substitute back into the ODE
  • Solve for the coefficients $A$ and $B$

Summary

  • Trigonometric solution of ODE involves substituting trigonometric functions or using trigonometric identities to solve the ODE
  • Applying initial conditions helps in obtaining the particular solution
  • Practice solving various examples to gain proficiency in this method
  1. Example 1 (contd.)
  • Substitute $y = A\sin(x) + B\cos(x)$ into the given ODE: $\frac{{d^2}}{{dx^2}}(A\sin(x) + B\cos(x)) + (A\sin(x) + B\cos(x)) = 0$
  • Differentiate twice: $(A\cos(x) - B\sin(x)) + (A\sin(x) + B\cos(x)) = 0$
  • Substitute back into the ODE: $A\cos(x) - B\sin(x) + A\sin(x) + B\cos(x) = 0$
  • Combine like terms: $(A+B)\cos(x) + (A-B)\sin(x) = 0$
  • Since this equation must hold for all values of $x$, we must have: $A+B=0$ and $A-B=0$
  • Solve for $A$ and $B$: $A = 0$ and $B = 0$
  • Therefore, the general solution is $y = 0$
  1. Example 2 (contd.)
  • Substitute $y = A\sin(2x) + B\cos(2x)$ into the given ODE: $\frac{{d^2}}{{dx^2}}(A\sin(2x) + B\cos(2x)) + 4(A\sin(2x) + B\cos(2x)) = 0$
  • Differentiate twice: $-4(A\sin(2x) + B\cos(2x)) + 4(A\sin(2x) + B\cos(2x)) = 0$
  • Substitute back into the ODE: $-4(A\sin(2x) + B\cos(2x)) + 4(A\sin(2x) + B\cos(2x)) = 0$
  • Simplify: $0 = 0$
  • The equation holds for all values of $x$
  • Therefore, the general solution is $y = A\sin(2x) + B\cos(2x)$
  1. Example 3 (contd.)
  • Substitute $y = A\sin(3x) + B\cos(3x)$ into the given ODE: $\frac{{d^2}}{{dx^2}}(A\sin(3x) + B\cos(3x)) + 9(A\sin(3x) + B\cos(3x)) = 0$
  • Differentiate twice: $-9(A\sin(3x) + B\cos(3x)) + 9(A\sin(3x) + B\cos(3x)) = 0$
  • Substitute back into the ODE: $-9(A\sin(3x) + B\cos(3x)) + 9(A\sin(3x) + B\cos(3x)) = 0$
  • Simplify: $0 = 0$
  • The equation holds for all values of $x$
  • Therefore, the general solution is $y = A\sin(3x) + B\cos(3x)$
  1. Summary
  • Trigonometric solution of ODE involves substituting trigonometric functions or using trigonometric identities to solve the ODE
  • Applying initial conditions helps in obtaining the particular solution
  • Practice solving various examples to gain proficiency in this method
  1. Additional Example
  • Solve the following ODE: $\frac{{d^2y}}{{dx^2}} + 2 \frac{{dy}}{{dx}} + 2y = 0$
  • Assume a solution of the form $y = e^{rx}$
  • Substitute into the ODE and solve for $r$: $r^2e^{rx} + 2re^{rx} + 2e^{rx} = 0$
  • Simplify: $e^{rx}(r^2 + 2r + 2) = 0$
  • Since $e^{rx} \neq 0$ for any $x$, we must have $r^2 + 2r + 2 = 0$
  • Solve the quadratic equation: $r = -1 \pm i$
  • The general solution is: $y = Ae^{-x}\cos(x) + Be^{-x}\sin(x)$
  1. Method 3: Applying initial conditions
  • If initial conditions are given in the problem, substitute them into the general solution obtained from methods 1 or 2
  • Solve for the unknown coefficients to obtain the particular solution
  • Example: Solve the following ODE with initial conditions: $\frac{{d^2y}}{{dx^2}} - 9y = 0$ with $y(0) = 2$ and $y’(0) = 1$
  • Using method 1, we obtain the general solution: $y = A\sin(3x) + B\cos(3x)$
  • Substitute the initial conditions: $A\sin(0) + B\cos(0) = 2$ and $3A\cos(0) - 3B\sin(0) = 1$
  • Simplify: $B = 2$ and $3A = 1$
  • Therefore, the particular solution is: $y = \frac{1}{3}\sin(3x) + 2\cos(3x)$
  1. Example 4
  • Solve the following ODE with initial conditions: $\frac{{d^2y}}{{dx^2}} + 4\frac{{dy}}{{dx}} + 4y = 0$ with $y(0) = 1$ and $y’(0) = 0$
  • Using method 2, we obtain the general solution: $y = (A+Bx)e^{-2x}$
  • Substitute the initial conditions: $(A+B(0))e^{-(2)(0)} = 1$ and $(B - 2(A+B(0)))e^{-(2)(0)} = 0$
  • Simplify: $A = 1$ and $B = 2$
  • Therefore, the particular solution is: $y = (1 + 2x)e^{-2x}$
  1. Example 5
  • Solve the following ODE: $\frac{{d^2y}}{{dx^2}} - 2\frac{{dy}}{{dx}} + 10y = 0$
  • Assume a solution of the form $y = e^{rx}$
  • Substitute into the ODE and solve for $r$: $r^2e^{rx} - 2re^{rx} + 10e^{rx} = 0$
  • Simplify: $e^{rx}(r^2 - 2r + 10) = 0$
  • Since $e^{rx} \neq 0$ for any $x$, we must have $r^2 - 2r + 10 = 0$
  • Solve the quadratic equation: $r = \frac{{2 \pm \sqrt{(-2)^2 - 4(1)(10)}}}{2}$
  • Simplify: $r = 1 \pm 3i$
  • The general solution is: $y = e^x(C_1\cos(3x) + C_2\sin(3x))$
  1. Relationship between ODEs and Trigonometric Functions
  • Trigonometric functions play an important role in solving ODEs
  • The trigonometric solutions represent oscillations, vibrations, and periodic phenomena
  • Trigonometric identities can be used to simplify and solve ODEs
  1. Conclusion
  • Trigonometric solution of ODEs provides elegant and efficient methods to solve certain types of ODEs
  • The solution involves substituting trigonometric functions or using trigonometric identities
  • Applying initial conditions helps in obtaining the particular solution
  • Practice and understanding of trigonometric identities are important for solving ODEs
  1. Example 6
  • Solve the following ODE: $\frac{{d^2y}}{{dx^2}} + 2\frac{{dy}}{{dx}} + 5y = 0$
  • Assume a solution of the form $y = e^{rx}$
  • Substitute into the ODE and solve for $r$: $r^2e^{rx} + 2re^{rx} + 5e^{rx} = 0$
  • Simplify: $e^{rx}(r^2 + 2r + 5) = 0$
  • Since $e^{rx} \neq 0$ for any $x$, we must have $r^2 + 2r + 5 = 0$
  • Solve the quadratic equation: $r = \frac{{-2 \pm \sqrt{(2)^2 - 4(1)(5)}}}{2}$
  • Simplify: $r = -1 \pm 2i$
  • The general solution is: $y = e^{-x}(C_1\cos(2x) + C_2\sin(2x))$
  1. Example 7
  • Solve the following ODE: $\frac{{d^2y}}{{dx^2}} + 6\frac{{dy}}{{dx}} + 13y = 0$
  • Assume a solution of the form $y = e^{rx}$
  • Substitute into the ODE and solve for $r$: $r^2e^{rx} + 6re^{rx} + 13e^{rx} = 0$
  • Simplify: $e^{rx}(r^2 + 6r + 13) = 0$
  • Since $e^{rx} \neq 0$ for any $x$, we must have $r^2 + 6r + 13 = 0$
  • Solve the quadratic equation: $r = \frac{{-6 \pm \sqrt{(6)^2 - 4(1)(13)}}}{2}$
  • Simplify: $r = -3 \pm 2i$
  • The general solution is: $y = e^{-3x}(C_1\cos(2x) + C_2\sin(2x))$
  1. Example 8
  • Solve the following ODE: $\frac{{d^2y}}{{dx^2}} + 8\frac{{dy}}{{dx}} + 20y = 0$
  • Assume a solution of the form $y = e^{rx}$
  • Substitute into the ODE and solve for $r$: $r^2e^{rx} + 8re^{rx} + 20e^{rx} = 0$
  • Simplify: $e^{rx}(r^2 + 8r + 20) = 0$
  • Since $e^{rx} \neq 0$ for any $x$, we must have $r^2 + 8r + 20 = 0$
  • Solve the quadratic equation: $r = \frac{{-8 \pm \sqrt{(8)^2 - 4(1)(20)}}}{2}$
  • Simplify: $r = -4 \pm 2i$
  • The general solution is: $y = e^{-4x}(C_1\cos(2x) + C_2\sin(2x))$
  1. Example 9
  • Solve the following ODE: $\frac{{d^2y}}{{dx^2}} + 12\frac{{dy}}{{dx}} + 37y = 0$
  • Assume a solution of the form $y = e^{rx}$
  • Substitute into the ODE and solve for $r$: $r^2e^{rx} + 12re^{rx} + 37e^{rx} = 0$
  • Simplify: $e^{rx}(r^2 + 12r + 37) = 0$
  • Since $e^{rx} \neq 0$ for any $x$, we must have $r^2 + 12r + 37 = 0$
  • Solve the quadratic equation: $r = \frac{{-12 \pm \sqrt{(12)^2 - 4(1)(37)}}}{2}$
  • Simplify: $r = -6 \pm \sqrt(37)i$
  • The general solution is: $y = e^{-6x}(C_1\cos(\sqrt(37)x) + C_2\sin(\sqrt(37)x))$
  1. Example 10
  • Solve the following ODE: $\frac{{d^2y}}{{dx^2}} + 3\frac{{dy}}{{dx}} + 2y = 0$
  • Assume a solution of the form $y = e^{rx}$
  • Substitute into the ODE and solve for $r$: $r^2e^{rx} + 3re^{rx} + 2e^{rx} = 0$
  • Simplify: $e^{rx}(r^2 + 3r + 2) = 0$
  • Since $e^{rx} \neq 0$ for any $x$, we must have $r^2 + 3r + 2 = 0$
  • Solve the quadratic equation: $r = \frac{{-3 \pm \sqrt{(3)^2 - 4(1)(2)}}}{2}$
  • Simplify: $r = -1$ and $r = -2$
  • The general solution is: $y = C_1e^{-x} + C_2e^{-2x}$
  1. Conclusion
  • Trigonometric solution of ODE provides elegant and efficient methods to solve certain types of ODEs
  • The solutions involve exponential functions and trigonometric functions
  • Applying initial conditions helps in obtaining the particular solution
  • Understanding and practice of ODEs and trigonometric functions are essential for solving ODEs
  1. Recap
  • Ordinary Differential Equation (ODE): An equation that relates an unknown function and its derivatives
  • Trigonometric ODE: A type of ODE that involves trigonometric functions
  • Methods for solving trigonometric ODEs: substituting trigonometric functions, using trigonometric identities, and applying initial conditions
  • Examples: solved examples for different types of trigonometric ODEs
  • Relationship between ODEs and Trigonometric Functions: Trigonometric functions represent oscillations and periodic phenomena
  • Importance of Trigonometric ODEs: Trigonometric solutions provide insights into various physical phenomena
  1. Next Steps
  • Practice solving more examples of trigonometric ODEs
  • Explore other methods of solving ODEs, such as power series solutions and Laplace transforms
  • Apply the knowledge gained to solve real-life problems that involve ODEs
  1. Resources
  • Textbooks: Recommended textbooks for 12th Boards Maths exam
  • Online Resources: Websites, videos, and tutorials for further study
  • Practice Questions: Additional practice questions for better understanding
  1. Questions
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