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X' = ax+b find general solution

  1. Jul 30, 2015 #1
    1. The problem statement, all variables and given/known data

    Find general solution of the given problem in the pic


    2. Relevant equations

    All I need to know is where they found the fundamental set of solutions.

    3. The attempt at a solution

    All I need to know is where they found the fundamental set of solutions. Please see pic. All I want to know is where did the x1 and x2 come from. If I have those then I can do the rest.

    See attachment image.
     

    Attached Files:

  2. jcsd
  3. Jul 30, 2015 #2

    HallsofIvy

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    There are a number of different ways to do a problem like that. Probably the simplest is:
    Letting [tex]x= \begin{pmatrix}x_1(t) \\ x_2(t)\end{pmatrix}[/tex]
    we can write the equation as
    [tex]\begin{pmatrix}x_1' \\ x_2'\end{pmatrix}= \begin{pmatrix}0 & 1 \\ -t^{-2} & t^{-1}\end{pmatrix}\begin{pmatrix}x_1\\ x_2\end{pmatrix}+ \begin{pmatrix}0 \\ 2t^{-1}\end{pmatrix}[/tex]
    [tex]\begin{pmatrix}x_1' \\ x_2'\end{pmatrix}= \begin{pmatrix}x_2 \\ -t^{-2}x_1+ t^{-1}x_2+ 2t^{-1}\end{pmatrix}[/tex]

    which gives the two equations [itex]x_1'= x_2[/itex] and [itex]x_2'= -t^{-2}x_1+t^{-1}x_2+ 2t^{-1}[/itex].

    To solve those equations, differentiate the first one a second time: [itex]x_1''= x_2'= -t^{-2}x_1+ t^{-1}x_2+ 2t^{-1}[/itex].
    From the first of the two equations, [itex]x_2= x_1'[/itex] so we can write this equation as [itex]x_1''= -t^{-2}x_1+ t^{-1}x_1'+ 2t^{-1}[/itex] which is the single second order equation [itex]x_1''- t^{-1}x_1'+ t^{-2}x_1= 2t^{-1}[/itex].

    Multiply by [itex]t^2[/itex] to get [itex]t^2x_1''- tx_1'+ x_1= 2t[/itex]. That is a "Cauchy type" or "equi-potential" equation (since each derivative is multiplied by t to the same power as the order of the derivative). The change of variable, u= ln(t), changes the equation to a corresponding "constant coefficients" equation, [itex]d^2x_1/du^2- 2 dx_1/du+ x_1= 2e^u[/itex].

    Can you solve that equation?
     
  4. Jul 31, 2015 #3
    Thank you so much! I solved the equation.

    x(t)=c1t+c2tln(t)+t[ln(t)]^2

    the matrix solution seems to be the wronstkian of the functions next to the c1 and c2, however the third term which doesn't have a constant I'm not sure what to do with it or if it has any significant importance.
     
  5. Jul 31, 2015 #4

    HallsofIvy

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    I assume you mean that is the solution for the function I called "x1". You still need to find x2 in order to find the vector function solution. Remember that you are solving a "vector differential equation" so you need a function for both components.

    T=As for "the third term which doesn't have a constant", the set of all solutions to a second order linear homogeneous differential equation, here [itex]t^2x_1''- tx_1'+ x_1= 0[/itex], form a two dimensional vector space. The two functions, t and t ln(t) are "basis vectors" for that linear space. Every solution can be written as a linear combination of them: c1t+ c2 t ln(t).

    The set of all solutions to a linear non-homogeneous equation, [itex]t^2x_1''- tx_1+ x_1= 2t[/itex] form a "linear manifold". Where you can think of a linear vector space, geometrically, as a plane containing the origin so a "linear manifold" can be thought of as a line or plane that does NOT contain the origin. What we can do with such a set is imagine as using the plane through the origin parallel to that linear manifold. We can represent a point on that "manifold" as a vector in the "parallel" vector space plus a vector to that plane. The set of all points (here, functions) in that manifold can be represented as a general function in that parallel two dimensional space plus that one vector to the plane.
     
  6. Jul 31, 2015 #5
    Okay, so this problem was the example problem in my engineering math book. I tried to do the example before getting to the actual problems. Now I tried the same technique on the first problem and I checked out my solution to see if it worked and it only half worked.

    The problem I'm having is that when it comes down to combining the x1 and x2 equations, you took a derivative, but is that the only way? I just added them together. I have included my work as attachment. Any help getting me to understand this is much appreciated.
     

    Attached Files:

  7. Aug 1, 2015 #6
    I believe that I've now figured it out. I just used det(a-lamda(I))=0 solve eigen values and then get eigen vectors.
     
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