H2 Math Differential Equations

• knt

  22 Jan 12, 14:37

  Hello! I'm having some difficulty with solving a particular question and was wondering whether the answer is wrong or whether I missed out something in solving it. The question is as follows: Solve the differential equation sec x ( dy/dx) = e^sinx + y given that y=1 when x=0. The answer provided is y= e^sinx(sinx+1). Hope someone can help me with this. Thanks a lot! :)
• wee_ws

  22 Jan 12, 15:35

  Hi,

    For the answer you provided, I believe you meant

    y = (sin x + 1)e^(sin x).

    If you differentiate this solution, you will obtain the required DE.

    Thanks.

  Cheers,
  Wen Shih

• knt

  22 Jan 12, 15:47

  Hello For the answer, it is y= e^(sinx) X [sinx+1] For the DE, it's (secx)(dy/dx) = e^(sin x) + y Thanks a lot!

  Originally posted by wee_ws:

  Hi,

    For the answer you provided, do you mean

    y = e^[ (sin x)(sin x + 1) ] or something else?

    For the DE, do you mean

    (sec x)(dy / dx) = e^(sin x) + y or something else?

    Thanks.

  Cheers,
  Wen Shih

  
  
• knt

  22 Jan 12, 15:54

  Hello Thanks a lot! I would really appreciate it if you could show me the working because I'm not sure where I went wrong from the initial DE of (secx)(dy/dx) = e^(sin x) + y to get to y=e^(sinx)[sinx + 1] . Thanks again!
• knt

  22 Jan 12, 15:59

  Hello! Thanks a lot! I would really appreciate it if you could show me the working because I'm not sure where I went wrong. Thanks again!

  Originally posted by knt:

  Hello For the answer, it is y= e^(sinx) X [sinx+1] For the DE, it's (secx)(dy/dx) = e^(sin x) + y Thanks a lot!

  
  

  
  
• wee_ws

  22 Jan 12, 19:15

  Hi,

    I believe it is assessed under H3 Maths where the integrating factor method is to be used to solve the question.

    We rewrite the DE as

    (dy/dx) - (cos x) y = (cos x) e^(sin x).

    Integrating factor is e to the power of integral of -cos x dx, leading to

    e^(-sin x).

    Multiplying both sides of the DE, we obtain

    e^(-sin x) (dy/dx) - (cos x) e^(-sin x) y = cos x

    => d/dx [ y e^(-sin x) ] = cos x.

    Integrating both sides wrt x, we obtain

    y e^(-sin x) = integral of cos x dx

                       = sin x + c.

    When x = 0 and y = 1, c = 1.

    Now y e^(-sin x) = sin x + 1 is the particular solution, from which we arrive at

    y = (sin x + 1) e^(sin x).

    Thanks and have a great reunion dinner without worrying about this question :)

  Cheers,
  Wen Shih

• knt

  23 Jan 12, 12:46

  Dear Wen Shih Thank you so very much for the reply! I appreciate it. Are there any books/websites that you could refer so that can i learn more about the different integrating methods? Thanks~ Wishing you a Blessed Lunar New Year! Cheers!
• wee_ws

  23 Jan 12, 15:03

  Hi knt,

    Thanks for your well-wishes! ç¥�你龙年行大è¿�ï¼�

    In H2 Maths, students are expected to know these methods:

    1. direct integration once (first order DE) or twice (second order DE),

    2. variable separable,

    3. by a given substitution.

    Others are:

    1. integrating factor,

    2. by well-established substitutions,

    3. by Euler's method,

    and many more.

    If you wish to go in depth, be sure to visit:

    http://www.sosmath.com/diffeq/diffeq.html

    Thanks.

  Cheers,
  Wen Shih