Results 1 to 7 of 7

Math Help - Differential Equation

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    15

    Differential Equation

    Hi, I'm stuck on the following question:

    Find the general solution to the following differential equation:

    (dy/dx)+y.secx=secx


    Any help available would be greatly appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2010
    Posts
    44
    Hi AlanC877,

    \frac{dy}{dx}+ysecx\;=\;secx

    \frac{dy}{dx}\;=\;\left(1-y\right)secx

    \frac{dy}{\left(1-y\right)}\;=\;secxdx

    \frac{dy}{\left(y-1\right)}+secxdx\;=\;0

    From now on, it can be solved as seperable differential equation. If you still have problem, write again.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    is it correct
    dy/dx=secx(1-y) ?
    If so - seperate variables.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2010
    Posts
    15
    That does make sense but I have been asked to find it by finding the integrating factor first.

    i.e. P(x) would be set to equal secx, then the integral of this (secx) would be found. The integrating factor would be e^(the integral).
    You then multiply both sides of the equation by the integrating factor, integrate both sides and rearrange to get an equation for y. Do you know this method?!

    Really what I'm stuck on is integrating secx is suppose!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2010
    Posts
    44
    I know that type of solution but -i think- this is much easier.

    And \int{secx} = \ln{|secx+tanx| + C}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2010
    Posts
    15
    Yep, I would agree but its the stupid way that we have to for the syllabus.

    How did you get that integral of secx? I think I'm just bein stupid but I can't remember the method.

    Thanks for your help so far.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2010
    Posts
    44
    \int sec x dx = \int sec x \frac{sec x + tan x}{sec x + tan x}dx

    set,  u=sec x + tan x

    then we find, du=(sec x\; tan x + sec^2x) dx

    substitute du = (sec x\; tan x + sec^2x) dx, u = sec x + tan x

    After substitution we get;
    <br />
\int secx\frac{sec x + tan x}{sec x + tan x}dx\;=\;\int\frac{\left(sec^2x+secx\;tanx\right)d  x}{secx+tanx}\;=\;\int\frac{du}{u}

    Solve integral, we get; \ln|u|+C

    Then substitute back u=tanx+secx, and we get;

    \int secx\;=\;\ln|secx+tanx| + C
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Partial Differential Equation satisfy corresponding equation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 16th 2011, 08:15 PM
  2. Replies: 4
    Last Post: May 8th 2011, 01:27 PM
  3. Replies: 1
    Last Post: April 11th 2011, 02:17 AM
  4. Partial differential equation-wave equation(2)
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 6th 2009, 09:54 AM
  5. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 12:39 PM

Search Tags


/mathhelpforum @mathhelpforum