Results 1 to 7 of 7

Math Help - Is there a method for finding the general solution in this case?

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    17

    Is there a method for finding the general solution in this case?

    I am only given y= x/3 and am asked to find the general solution. So what I did was get the derivative up until the second order and I multiplied each so that if I were to add them up they would equal zero. The question I have is...how do you choose what to multiply the derivatives by? Because I took a quiz and the first time I did they had x^2y''+7xy'+9y=0 and the second time they had switched it to x^2y''+3xy-3y=0.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jukodan View Post
    I am only given y= x/3 and am asked to find the general solution. So what I did was get the derivative up until the second order and I multiplied each so that if I were to add them up they would equal zero. The question I have is...how do you choose what to multiply the derivatives by? Because I took a quiz and the first time I did they had x^2y''+7xy'+9y=0 and the second time they had switched it to x^2y''+3xy-3y=0.
    Perhaps some clarification is needed for this post. I for one find your question incomprehensible.

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2009
    Posts
    397
    As for myself, i think y = x/3 is the general solution. To be more precise, particular solution...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2009
    Posts
    17
    Please excuse my lack of clarity. I was up all night doing DE. Now here is a link to a very similar problem. http://i9.photobucket.com/albums/a92...dadsadsaad.jpg
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jukodan View Post
    I am only given y= x/3 and am asked to find the general solution. So what I did was get the derivative up until the second order and I multiplied each so that if I were to add them up they would equal zero. The question I have is...how do you choose what to multiply the derivatives by? Because I took a quiz and the first time I did they had x^2y''+7xy'+9y=0 and the second time they had switched it to x^2y''+3xy-3y=0.
    Find the differential equation whos solution is y=x/3:

    y'=1/3, y(0)=0

    or:

    y''=0, y'(0)=1/3, y(0)=0

    Will do, I'm sure there are others:

    CB
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Aug 2009
    Posts
    143
    General solution to x^2y''+3xy'-3y=0 is:

    y(x)=c_1x+\frac{c_2}{x^3}

    I wonder why you are given y=\frac{x}{3} instead of y=x
    Last edited by mr fantastic; September 19th 2009 at 12:23 AM. Reason: Restored original reply
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Let the ED equation be...

    y^{''} + 3\cdot \frac{y^{'}}{x} - 3\cdot \frac{y}{x^{2}}=0 (1)
    ... and we suppose to know a particular solution of it... y=x in this case. A general procedure to find a second solution independent from the first is illustrated now...

    If u and v are both solutions of (1) is...

    u^{''} + 3\cdot \frac{u^{'}}{x} - 3\cdot \frac{u}{x^{2}}=0

    v^{''} + 3\cdot \frac{v^{'}}{x} - 3\cdot \frac{v}{x^{2}}=0 (2)

    Multiplying the first of (2) by v, the second by u and taking the difference we obtain...

    u^{''}\cdot v - v^{''}\cdot u + \frac{3}{x} \cdot (u'\cdot v - v'\cdot u)=0 (3)

    Setting \varphi = u'\cdot v - v'\cdot u the (3) is written as...

     \varphi ' + \frac{3}{x}\cdot \varphi=0 (4)

    ... and its silution is...

    \varphi = \frac{c_{1}}{x^{3}} (5)

    Deviding both termes of (4) by v^{2} and taking into account (5) we have...

    \frac{u'\cdot v - v'\cdot u}{v^{2}} = \frac {d}{dx} \frac{u}{v} = \frac{c_{1}}{x^{3}\cdot v^{2}} (6)

    ... the solution of which is...

    \frac{u}{v}= c_{2} + c_{1} \int \frac{dx}{x^{3}\cdot v^{2}} (7)

    If we observe (7) is easy to see that if v is solution of (1), then...

    u=v \int \frac{dx}{x^{3}\cdot v^{2}} (8)

    ... is also solution of (1). In particular...

     v=x \rightarrow u= \frac{1}{x^{3}} (9)

    ... so that the general solution of (1) is...

    y= c_{1} x + \frac{c_{2}}{x^{3}} (10)

    Kind regards

    \chi  \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the general solution of the D.E.
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: June 3rd 2011, 12:58 PM
  2. Finding the General Solution
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 27th 2009, 06:59 PM
  3. General Solution of PDE using Fourier's method
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 17th 2009, 08:40 PM
  4. Finding the general solution from a given particular solution.
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: October 7th 2009, 01:44 AM
  5. Finding general solution
    Posted in the Calculus Forum
    Replies: 7
    Last Post: May 23rd 2008, 02:10 PM

Search Tags


/mathhelpforum @mathhelpforum