Solving Second Order Differential Equations

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  • čas přidán 9. 02. 2021
  • engineers.academy/level-5-hig...
    This video continues from previous videos on solving first-order differential equations and introduces the second order differential equation. This video addresses how we can solve homogenous second order differential equations with three worked examples.
    First Order Differential Equation Videos:
    Part 1: • Solving First Order Di...
    Part 2: • Solving First Order Di...
    More Second Order Differential Equation Videos:
    Example 1 - Linear Motion: • Second Order Different...
    Example 2 - Mechanical System: • Second Order Different...
    Example 3 - RLC Circuit: • Second Order Different...
    Non Homogenous 2ODE: • Solving Non-Homogenous...
    To learn more about our courses and programmes please visit: engineers.academy/

Komentáře • 31

  • @realdka13
    @realdka13 Před rokem +17

    I've spent like 2 hours on youtube trying to find a video to review this method of solving DE's. Its been 2 years since I've had DE. Thank you for pulling through

  • @jamaljaffer8412
    @jamaljaffer8412 Před rokem +3

    It's a brilliant method of explaining the DE especially in solving RLC Circuit Problems / Analysis. many thanks.

  • @aladuraagboola4599
    @aladuraagboola4599 Před rokem +4

    Wherever you are. May God bless you for this tutorial 🥺🙏🏼

  • @WisdomAfrica1990
    @WisdomAfrica1990 Před 2 lety +9

    This is really helpful honestly. Thank you

  • @keithlungile
    @keithlungile Před 2 lety +10

    This is excellent, well explained!

  • @erlingmadman6065
    @erlingmadman6065 Před 2 lety +7

    Awesome stuff. Thank you!

  • @clapyocheeks
    @clapyocheeks Před 4 měsíci

    I've been struggling to understand RLC circuits since I took differential equations 3-4 semesters ago. I didn't find any luck with my book, lecture, or other youtube videos. This video made me really understand what's going on. Thank you so much!

    • @EngineersAcademyLTD
      @EngineersAcademyLTD  Před 4 měsíci

      Glad it was helpful - there is a specific RLC example linked in the description also.

  • @lesserspottedmugwump.363
    @lesserspottedmugwump.363 Před 2 lety +5

    Excellently explained. Very clear.

  • @naizfayasmin7551
    @naizfayasmin7551 Před 3 lety +7

    Excellent teaching Sir

  • @sheilamawasa8552
    @sheilamawasa8552 Před 2 lety +7

    Great teaching

  • @jokerins7660
    @jokerins7660 Před rokem +2

    Very useful video and examples for exam preparation. Thank you!

  • @christinewanja8344
    @christinewanja8344 Před 2 lety +5

    excellent explanation
    Thankyou

  • @childofgod310
    @childofgod310 Před rokem +2

    wow the best teaching ever

  • @ziadnasser9197
    @ziadnasser9197 Před 2 lety +5

    Excellent
    🥰🥰

  • @DAHBLIFE
    @DAHBLIFE Před rokem +3

    Thank you! I can now use this in solving RLC Circuit Problems / Analysis

    • @EngineersAcademyLTD
      @EngineersAcademyLTD  Před rokem

      Glad it is helpful - there is a specific RLC example here also czcams.com/video/w9Z4Ly98Lcw/video.html

  • @judymueni7165
    @judymueni7165 Před 4 měsíci +2

    God bless you 🙏🙏

  • @muhammadabdulrahman8665
    @muhammadabdulrahman8665 Před rokem +2

    Thank you so much

  • @NoobXgaming01541
    @NoobXgaming01541 Před 6 měsíci +1

    Thnku Sir🙌

  • @enocknyaanga4887
    @enocknyaanga4887 Před rokem

    Wonderful presentation

  • @James-eo2wh
    @James-eo2wh Před 2 lety +3

    Very good video and very useful leading up to my exam. Ps I think it’s homogeneous not homogenous

  • @descendantsofbookworm2127

    very useful video

  • @masonalloway5964
    @masonalloway5964 Před 3 měsíci +1

    Hi. For example 3, when you substituted the imaginary component of m in for Beta, why did you disregard the negative imaginary component? Surely you could sub in - sqrt(11)/2 and achieve a completely different but valid answer? (great video btw)

    • @EngineersAcademyLTD
      @EngineersAcademyLTD  Před 3 měsíci +1

      Its the magnitude of the imaginary component that is used as beta here in this particular formulation, ignoring sign.

  • @alphaamorrivera1746
    @alphaamorrivera1746 Před 2 lety +4

    Hello sir, in the 2nd example, how did you get the 9/4 and -9/4? Thank you!

    • @EngineersAcademyLTD
      @EngineersAcademyLTD  Před 2 lety

      By solving the two simultaneous equations to find A and B. See here: czcams.com/video/fftnk5tVMTg/video.html

  • @dollyarthiya7661
    @dollyarthiya7661 Před 2 lety +2

    What if the 2nd term is not given?

    • @EngineersAcademyLTD
      @EngineersAcademyLTD  Před 2 lety +2

      If you mean that there is no dy/dx term in the equation then you can still use the auxiliary eqn but the value of b would be zero.

  • @sydneyoduli7937
    @sydneyoduli7937 Před 8 měsíci

    Thank you so much