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Math Help - 3rd Order Non Homogenous diff equation check!

  1. #1
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    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!

    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

    y(x) = c1y1(x) + c2y2(x) + c3y3(x) + cos x:

    This is what i have,

    (m^2 + 4)(m-1)

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

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

    thanks

    Funky
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  2. #2
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    I think e^{0} is another way to write one of your solutions. That will change your characteristic equation, won't it?
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  3. #3
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    no mate, i`m working backwards from the solutions i`ve been given to work of the 3rd order differential equations.

    thanks

    Funkybudduh
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  4. #4
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    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 (m^2 + 4)(m-1)=0, then one solution is m=1 corresponding to y=e^{x}, not y=1. Therefore, I think you need to change your characteristic equation you put forward en route to your DE.
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  5. #5
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    i`m sorry i still dont understand....
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  6. #6
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    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 y'''-8y''+19y'-12y=0. The usual procedure with constant coefficient problems like this is to assume a solution of the form y=e^{mx}, plug it in, and solve for m. We have y'=me^{mx}, y''=m^{2}e^{mx}, and finally y'''=m^{3}e^{mx}. Plugging into the DE yields

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

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

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

    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 m=1 corresponds, because of our initial guess of y=e^{mx}, to the solution y=e^{x}? It does NOT correspond to the solution y=1.

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

    y''' - y'' + 4y' -4y =0.

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

    Do you follow me now?
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  7. #7
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    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
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  8. #8
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    I gave you a decent hint in Post # 2. What could post # 2 tell you?
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  9. #9
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    Then my new characterisctic equation is

    m(m^2 + 4)=0

    hence my 3rd order diff equation is

    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
    y(x)=c1y1(x) + c2y2(x) + c3y3(x) +Cos(x)

    where y1, y2 and y3 are 1, cos(2x), sin(2x)
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  10. #10
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    I agree with your DE now.

    Well, you know what the LHS operator \frac{d^{3}}{dx^{3}}+4\frac{d}{dx} does to the homogeneous part of your starting solution c_{1}y_{1}+c_{2}y_{2}+c_{3}y_{3}. Find out what it does to the \cos(x) part. Then just use that to determine the RHS of the DE. Be careful with the signs!
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  11. #11
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    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
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  12. #12
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    Yep, that looks right. You're welcome! Have a good one.
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