Results 1 to 2 of 2

Math Help - Where am I going wrong in this exact equations problem?

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    79

    Where am I going wrong in this exact equations problem?

    Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Solve the equation.

    (x+2)sinydx+xcosydy)=0, (x,y)=xe^x

    So first, M=(x+2)siny and N=xcosy. Thus, M_y=(x+2)cosy and N_x=cosy
    So this is inexact.

    Then multiplying everything by we get M=e^x(x^2+2x)siny and N=x^2e^xcosy. Thus, M_y=e^x(x^2+2x)cosy and N_x=e^x(x^2+2x)cosy. So now they are exact

    I have chosen to integrate N first since it's a bit easier than M. So integrating N I get x^2e^xsiny+h(x)

    Then I differentiated this with respect to x getting 2xe^xsiny+h'(x)
    Then I compared this to M. 2xe^xsiny+h'(x)=M=e^x(x^2+2x)siny. So from this I get h'(x)=x^2e^x and then h(x)=1/3x^3e^x making my final answer x^2e^xsiny+(1/3)x^3e^x=c.

    But according to my book the final answer is only x^2e^xsiny=c. I'm not sure where the (1/3)x^3e^x is going. Help please
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,357
    Thanks
    36
    Quote Originally Posted by steph3824 View Post
    Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Solve the equation.

    (x+2)sinydx+xcosydy)=0, (x,y)=xe^x

    So first, M=(x+2)siny and N=xcosy. Thus, M_y=(x+2)cosy and N_x=cosy
    So this is inexact.

    Then multiplying everything by we get M=e^x(x^2+2x)siny and N=x^2e^xcosy. Thus, M_y=e^x(x^2+2x)cosy and N_x=e^x(x^2+2x)cosy. So now they are exact

    I have chosen to integrate N first since it's a bit easier than M. So integrating N I get x^2e^xsiny+h(x)

    Then I differentiated this with respect to x getting {\color{red}2xe^x}siny+h'(x)
    Then I compared this to M. 2xe^xsiny+h'(x)=M=e^x(x^2+2x)siny. So from this I get h'(x)=x^2e^x and then h(x)=1/3x^3e^x making my final answer x^2e^xsiny+(1/3)x^3e^x=c.

    But according to my book the final answer is only x^2e^xsiny=c. I'm not sure where the (1/3)x^3e^x is going. Help please
    The part in red above. You need to use the product rule so there's another piece.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Exact Equations
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 10th 2011, 08:24 PM
  2. Replies: 2
    Last Post: November 2nd 2010, 06:04 AM
  3. ODE exact equations
    Posted in the Calculus Forum
    Replies: 0
    Last Post: August 3rd 2008, 06:30 AM
  4. Are these differential equations exact?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 11th 2008, 08:25 AM
  5. Exact Equations
    Posted in the Calculus Forum
    Replies: 0
    Last Post: April 10th 2007, 04:13 PM

Search Tags


/mathhelpforum @mathhelpforum