Results 1 to 12 of 12

Thread: 3rd Order Non Homogenous diff equation check!

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    11

    3rd Order Non Homogenous diff equation check!

    Hi again all,

    After solving my last problem i have a new one for you all, i`ve done it but not sure if it is correct!

    $\displaystyle y1(x) = 1, y2(x) = cos 2x, y3(x) = sin 2x.$

    Wronskian is 8

    Find a (nonhomogeneous) third-order linear di erential equation with general so-
    lution using the above

    $\displaystyle y(x) = c1y1(x) + c2y2(x) + c3y3(x) + cos x:$

    This is what i have,

    $\displaystyle (m^2 + 4)(m-1)$

    there for the equation is $\displaystyle m^3 - m^2 + 4m -4 =0$

    hence Diff Equation is $\displaystyle y``` - y`` + 4y -4 = Cos(x)$

    thanks

    Funky
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    I think $\displaystyle e^{0}$ is another way to write one of your solutions. That will change your characteristic equation, won't it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2010
    Posts
    11
    no mate, i`m working backwards from the solutions i`ve been given to work of the 3rd order differential equations.

    thanks

    Funkybudduh
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    I`m working backwards from the solutions i`ve been given to work of the 3rd order differential equations.
    I realize that. What I'm saying is that if you have a characteristic equation $\displaystyle (m^2 + 4)(m-1)=0,$ then one solution is $\displaystyle m=1$ corresponding to $\displaystyle y=e^{x},$ not $\displaystyle y=1.$ Therefore, I think you need to change your characteristic equation you put forward en route to your DE.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2010
    Posts
    11
    i`m sorry i still dont understand....
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Ok. Let me start from a third-order DE, and work the problem in the usual way. I know you're solving the inverse problem of producing a DE, but bear with me. Let's say I have the DE $\displaystyle y'''-8y''+19y'-12y=0.$ The usual procedure with constant coefficient problems like this is to assume a solution of the form $\displaystyle y=e^{mx},$ plug it in, and solve for $\displaystyle m.$ We have $\displaystyle y'=me^{mx}, y''=m^{2}e^{mx},$ and finally $\displaystyle y'''=m^{3}e^{mx}.$ Plugging into the DE yields

    $\displaystyle m^{3}e^{mx}-8m^{2}e^{mx}+19me^{mx}-12e^{mx}=0.$ Factoring yields

    $\displaystyle e^{mx}(m^{3}-8m^{2}+19m-12)=0.$ Now $\displaystyle e^{mx}\not=0,$ hence we may divide out to obtain the characteristic equation

    $\displaystyle m^{3}-8m^{2}+19m-12=0.$ Now you can show that the roots of this equation are $\displaystyle m=4,3,1.$ That is, it factors as $\displaystyle (m-4)(m-3)(m-1)=0.$ Hence, the solution to this DE is as follows:

    $\displaystyle y=c_{1}e^{4x}+c_{2}e^{3x}+c_{1}e^{x}.$

    Do you see (and this is really where I'm going with this), that the root $\displaystyle m=1$ corresponds, because of our initial guess of $\displaystyle y=e^{mx}$, to the solution $\displaystyle y=e^{x}?$ It does NOT correspond to the solution $\displaystyle y=1.$

    If you don't believe me, think of it this way: according to the OP, you are claiming that $\displaystyle y=1$ solves the homogeneous ODE

    $\displaystyle y''' - y'' + 4y' -4y =0.$

    I altered your DE with some obvious corrections. Plug $\displaystyle y=1$ into this DE, and you will see that it does not satisfy it.

    Do you follow me now?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2010
    Posts
    11
    Ah now i get you, sorry about that. Well back to the drawing board, i`ll let you know if i have anything. Will probably start with a trig function and do it the way you`ve just showed me.

    thanks

    Funky
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    I gave you a decent hint in Post # 2. What could post # 2 tell you?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Oct 2010
    Posts
    11
    Then my new characterisctic equation is

    $\displaystyle m(m^2 + 4)=0$

    hence my 3rd order diff equation is

    $\displaystyle d^3y/dx^3 + 4dy/dx = g(x)$

    but how would i use this to find g(x) function?

    if the general solution of the 4rd order linear de (non homog) is
    $\displaystyle y(x)=c1y1(x) + c2y2(x) + c3y3(x) +Cos(x)$

    where y1, y2 and y3 are 1, cos(2x), sin(2x)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    I agree with your DE now.

    Well, you know what the LHS operator $\displaystyle \frac{d^{3}}{dx^{3}}+4\frac{d}{dx}$ does to the homogeneous part of your starting solution $\displaystyle c_{1}y_{1}+c_{2}y_{2}+c_{3}y_{3}.$ Find out what it does to the $\displaystyle \cos(x)$ part. Then just use that to determine the RHS of the DE. Be careful with the signs!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Oct 2010
    Posts
    11
    ok, my answer for g(x)= -3sin(x)

    i`ve taken cos(x) and found dy/dx and d^3/dx^3 then substituted it in dy^3/dx^3 + 4dy/dx = g(x). which yields g(x) = -3Sin(x)

    thanks again ackbeet
    Follow Math Help Forum on Facebook and Google+

  12. #12
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Yep, that looks right. You're welcome! Have a good one.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2nd order diff equation!
    Posted in the Differential Equations Forum
    Replies: 8
    Last Post: Oct 30th 2010, 04:31 PM
  2. [SOLVED] 2nd order diff. equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Sep 20th 2010, 01:11 PM
  3. Replies: 2
    Last Post: Mar 31st 2010, 12:24 AM
  4. First Order Homogenous Differential Equation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Feb 10th 2010, 06:05 AM
  5. Replies: 1
    Last Post: Feb 14th 2008, 10:11 AM

Search Tags


/mathhelpforum @mathhelpforum