Contents

Video

(click the link to view video)

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.