Variation of Parameters

Поделиться
HTML-код
  • Опубликовано: 28 сен 2024
  • Variation of Parameters
    Solving the second-order inhomogeneous differentiation equation y'' + y = sec(t) using variation of parameters, with the Wronskian and Cramer's rule
    Subscribe to my channel: / drpeyam
    Instagram: / peyamstagram
    Teespring merch: dr-peyam.creat...

Комментарии • 15

  • @alipourzand6499
    @alipourzand6499 4 дня назад +2

    Great lecture! Currently I'm dealing with linear diophantine equations and one way of solving them is to find the homogeneous and one particular solution. And then the general solution is the som of the particular and homogeneous, I guess this method is also used in other math topics.

  • @jaikumar848
    @jaikumar848 4 дня назад +1

    Hello sir ! Shouldn't we just take laplace then solve and just take inverse laplace of differential equation to find function y . is there any drawbacks of this approach ?

    • @drpeyam
      @drpeyam  4 дня назад +3

      But then what is Laplace of sec?

    • @carultch
      @carultch 2 дня назад

      @@drpeyam There is no Laplace of secant, because it isn't "of exponential order".
      It's very difficult to find examples using elementary functions, that fit the following criteria so you can have an interchangeable choice between using Laplace transform and Variation of Parameters to solve them.
      1. Has an RHS that is composed only of elementary functions of t.
      2. Has a solution that is limited to only using elementary functions
      3. Has a Laplace transform of the RHS that we can use, that is composed of elementary functions of s.
      4. Is beyond the scope of the families of functions where undetermined coefficients can work. I.e. constants, polynomials of t, exponentials of t, sinusoids of t, and linear/multiplicative combinations thereof.
      I searched very hard for an example with either natural log, or a non-integer power function of t, as the function of t on the RHS, that could be solved with both VOP and Laplace transforms. One such example that worked, was where I set it up so there'd be a repeated root for the homogeneous part, and to use that solution as a factor of the RHS function. This allows the Wronskian to cancel out the exponential terms, so that natural log can integrate with polynomial functions of t, instead of exponential or trig functions of t. This strategy also works with e^(-a*t)*sqrt(t) as well.
      The example I found, has the general form of the following, where a & b are constants:
      y" + 2*a*y' + a^2*y = 4*e^(-a*t)*ln(b*t)
      With the general solution of:
      y(t) = [(a*u + v)*t + u + 2*t^2*ln(b*t) - 3*t^2]*e^(-a*t)
      where y(0) = u and y'(0) = v

  • @phill3986
    @phill3986 2 дня назад +1

    Nice 👍. Minor thing - shouldn't that be a -sin(t) at 10:56 ?

    • @drpeyam
      @drpeyam  2 дня назад +2

      Oh yes, you’re right, thanks for noticing!!

  • @ElifnurKale-fj2ex
    @ElifnurKale-fj2ex 4 дня назад +1

    Nice ❤

  • @daewat6114
    @daewat6114 8 дней назад

    Thank you so much sir, I was really struggling with understanding how to use variation of parameters for a while now but I finally get it!

    • @drpeyam
      @drpeyam  8 дней назад +1

      Glad it helped!

  • @teoo8816
    @teoo8816 4 дня назад

    the GOAT 😭‼️‼️‼️

  • @viktor-kolyadenko
    @viktor-kolyadenko 7 часов назад

    In 6:19 we can use Cramer's rule.

  • @JarogniewBorkowski
    @JarogniewBorkowski 4 дня назад +1

    5:30 I really do not understand when you plug supposed function y_p Into main ode you get simple two equations....
    When I do it in my way I get u'' and v'' in further equstions because there is second derivative in ode.
    Could you explain?

    • @drpeyam
      @drpeyam  4 дня назад +1

      Yes there is a video actually on the playlist that explains how we can get rid of the u” and v” terms!

    • @JarogniewBorkowski
      @JarogniewBorkowski 4 дня назад

      @@drpeyam Ohh I see. There was an assumption. Thank You

  • @gdeivgui123
    @gdeivgui123 4 дня назад

    Great ❤