Video

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https://www.dropbox.com/s/zz4kf2w0p2j8mzh/Enriquez_MAT2680_Project_3.MOV?dl=0

Pictures

Written work for Problem 6.3.6

Steps

Use “Undetermined coefficients” method

  1. Solve for the roots of the characteristic equation.
    • use quadratic formula
    • use the the roots of the characteristic equation to find the solution of the complimentary equation
      • refer back to solutions of “constant coefficient second order linear equations”
  2. Find a suitable general form for a “particular solution”, y_{p} , that matches the “right side of the given equation” (y_{p} is the same as Q_{p}, which is function representing the amount of charge of a sytem).
    • if the right side of the equation is cos(ct), sin(ct), or cos(ct)+sin(ct), then the particular solution is of the form Acos(ct)+Bsin(ct)
    • since no terms in y_{p} is a solution of the complimentary equation, no extra steps are necessary in deciding the form of the particular solution
  3. Differentiate the chosen particular solution twice to find y_{p}^' and y_{p}^''.
  4. Substitute the values of y_{p} , y_{p}^' , and y_{p}^'' into Q, Q', and Q''.
  5. Simplify the new equation.
  6. Rearrange the terms of the equation (factoring may be necessary) so that the “left side of the equation” matches the form of the “right side of the equation”.
  7. Create and solve a system of equations to solve for the undetermined coefficients.
  8. Plug in the values of the undetermined coefficients into the chosen particular solution to determine Q_{p}.
  9. Since the steady steady state current, I_{p} , is equal to Q_{p}^' , we must differentiate Q_{p} to solve for the steady state current of the system.