🔵14 - Non Exact Differential Equations and Integrating Factors 2

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  • Опубликовано: 13 янв 2025

Комментарии • 51

  • @sunil68194
    @sunil68194 4 месяца назад +1

    Extremely well explained sir, genuinely cleared my doubts about non exact DEs. Subscribed 🤝

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  4 месяца назад

      You are most welcome.
      Thanks so much for watching. Keep watching for more

  • @allfootball.highlights
    @allfootball.highlights Год назад +2

    That's real stuff right there ur too gud nd u left me with no choice but to follow nd share this stuff 🤞

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  Год назад +1

      Awwww thanks so much

    • @reamabdulsalam524
      @reamabdulsalam524 11 месяцев назад

      Hi can you please give an example of if the integration factor is not subject to x , you have mentioned the second case but you did not show it please do because I could not understand anything from my tutor you are marvellous

  • @Karen-q2u8e
    @Karen-q2u8e 2 месяца назад +1

    how do you know which rule/case to follow before computing the integrating factor?? Or can you use any case on any question

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  2 месяца назад +2

      It really focuses on the question. And solving so on and forth, you will get the experience.

  • @syprinepamba6938
    @syprinepamba6938 9 месяцев назад +1

    Thanks 🙏🏾

  • @pleaseinsertname1893
    @pleaseinsertname1893 Год назад +2

    Very good video. Thank you.

  • @williamstechtips9726
    @williamstechtips9726 10 месяцев назад +1

    really helpful

  • @CeylonTleaves
    @CeylonTleaves 11 месяцев назад

    This was super helpful, thank you very much !!

  • @eucliddankwah1557
    @eucliddankwah1557 Год назад +5

    Why didn't you use integration by parts when integrating 2ye^y²
    We are integrating two products and one of the products is not a constant

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  Год назад +4

      Well it's true, but you can as well use u integration, and that's what I did, which is very fast and simple.
      Let me explain how
      Integral of 2ye^(y^2)
      Let u = y^2
      du/dy = 2y, dy = du/2y.
      Therefore
      Integral of 2ye^(y^2) becomes
      Integral of 2ye^(u).du/2y
      Integral of e^u dy
      Which is equal to: e^u + c, you sub u = y^2 back.
      e^y^2.
      Hope you get it. Thanks

    • @eucliddankwah1557
      @eucliddankwah1557 Год назад +1

      @@SkanCityAcademy_SirJohn Yh but you ve taken the 2y as a constant
      I used integration by parts and didn't get the same answer as yours

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  Год назад +2

      @eucliddankwah1557 I think I've just explained to you how I used u integration to do it, when you have a question like this ay times and exponential function, my method usually works. But not all cases. So refine your solution process on integration by part to see if you will get what I had. I think the problem is your solution process.

    • @eucliddankwah1557
      @eucliddankwah1557 Год назад +1

      @@SkanCityAcademy_SirJohn okay
      Thanks

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  Год назад +2

      @@eucliddankwah1557 you are welcome

  • @AbdulRahamanAbubakari-h2w
    @AbdulRahamanAbubakari-h2w 8 месяцев назад +1

    The method of substitution is used when a function and its derivative can be found in the function we are to integrate. Here in this case, the derivative of 2y is 2 w.r.t y and that of e^y^2 is 2ye^y^2. So, it is wrong to use the method of substitution here because the rule that should be hold when applying it is not found in the function.

  • @priyanshusharma7686
    @priyanshusharma7686 7 месяцев назад

    that was very helpful and great explanation🤩

  • @sekeychristian3458
    @sekeychristian3458 9 месяцев назад +1

    I'm so so grateful, but i also think the one on the right is supposed to be integration by parts. Thanks

  • @MbahKelly-td2ip
    @MbahKelly-td2ip Месяц назад +1

    Thank you sir

  • @graphingwithgeogebra
    @graphingwithgeogebra Год назад +2

    could you please talk about case 3 also? i really need it please

  • @calculusjoe53
    @calculusjoe53 6 месяцев назад +1

    In the 2nd eg, when integrating Ndy; the first term was a product of functions of y thus 2ye^y^2 so I was expecting the integration of the first term of N to be done by parts...? My humble concern..

  • @spiritedsaiyan3247
    @spiritedsaiyan3247 11 дней назад

    hello i have this question. y(1+xy)dx (- x)dy = 0. I am unable to solve with these two cases that u made the video. can u please solve this?

  • @williamstechtips9726
    @williamstechtips9726 10 месяцев назад +1

    I am trying to solve this
    y dx + (x2y3 + x)dy = 0
    I know this equation can be solved by inspection.(the integrating factor is
    x^-2 y^-2)
    But I try to use your method, I cannot find a function that's form by x or y only.(case1 and case2 are both cannot be fullfilled)
    Do I make anything wrong?Or just there's an exemption of this method?
    Thanks

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  10 месяцев назад +1

      Yes, I think there is something wrong with your solution. Try it again and let's see

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  10 месяцев назад

      Integrating factor is u(y) = 1/y

    • @williamstechtips9726
      @williamstechtips9726 10 месяцев назад +1

      @@SkanCityAcademy_SirJohn sorry,maybe I type the question wrong and made some misunderstanding.
      the question should be
      ydx + ((x^2)(y^3)+ x)dy=0
      and the textbook just told me this can be solved by inspection and give the answer of the integrating factor = (x^-2)(y^-2)

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  10 месяцев назад

      @williamstechtips9726 oh okay

  • @MakoMavuwa
    @MakoMavuwa 2 месяца назад +1

    the last integration , confused me 😵‍💫

  • @reamabdulsalam524
    @reamabdulsalam524 11 месяцев назад +1

    Many thanks you are a star , please do mention why the integration factor in case of x you divide by M , and in case of y you have divided by -M only this point is not clear to me I need to know the reason thanks

    • @SkanCityAcademy_SirJohn
      @SkanCityAcademy_SirJohn  11 месяцев назад

      I understand you, but that has been a formula proven and defined for us, so we just use that, not to border ourselves much