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Homework Help: System of ODEs

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Solve the system x'' + αx = 0

    x = [ x1 x2 ]
    α constant

    2. Relevant equations

    x'' + αx = 0 Where x is a vector [x1 x2]

    3. The attempt at a solution

    The solution of x'' + αx = 0 is x = Acosh(√α x) + Bsinh(√α x)

    Differentiating x

    x' = √α*Acosh(√α x) + √α*Bsinh(√α x)
    x'' = α*Acosh(√α x) + α*Bsinh(√α x)

    Plugging this back into the original ODE

    α*Acosh(√α x) + α*Bsinh(√α x) - α(Acosh(√α x) + Bsinh(√α x)) = 0 is satisfied

    I can't get this in Eigenform to solve for the system. What do I do in this case?
  2. jcsd
  3. Mar 13, 2014 #2


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    Science Advisor

    That makes no since at all. x is a two dimensional vector but you have it as both a single function and the independent variable!
    IF we were given a single real function, x, satisfying x''+ αx= 0 THEN the solution would be either x= Asinh(√αt)+ Bcosh(√αt), x= Asin(√αt)+ Bcos(√αt), or x= At+ B, depending upon whether α was positive, negative, or 0 and t is the (unnamed in the equation) independent variable.

    Are you given that α single real number? If so then the general solution would be
    [tex]\begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}= \begin{bmatrix}Asinh(√αt)+ Bcosh(√αt) \\ Csinh(√αt)+ Dcosh(√αt)\end{bmatrix}[/tex]
    if α> 0,
    [tex]\begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}= \begin{bmatrix}Asin(√αt)+ Bcos(√αt) \\ Csin(√αt)+ D cos(√αt)\end{bmatrix}[/tex]
    lf α< 0,
    [tex]\begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}= \begin{bmatrix}At+ B \\ Ct+ D\end{bmatrix}[/tex]
    if α= 0.

    I'm not sure what you mean by "eigenform" but if α is a single number then the differential equation can be written in matrix form as
    [tex]\begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix}'= \begin{bmatrix}α & 0 \\ 0 & α\end{bmatrix}\begin{bmatrix}x_1(t)\\ x_2(t)\end{bmatrix}[/tex]
    which has α as the only eigenvalue.
  4. Mar 13, 2014 #3
    Sorry wasn't clear enough

    Apologies, I should have been clear. This is the system:

    d^2/dx^2(y1) = α(y1 - y2)
    d^2/dx^2(y2) = α(y2 - y1)

    Hence, why your set of solutions won't work. This is why I tried to put it into eigenform, like the harmonic oscillator problem, which is easily done to get the frequencies for an mdof system. This method doesn't work for this type of problem
    Last edited: Mar 13, 2014
  5. Mar 14, 2014 #4


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    Homework Helper

    It is obvious from inspection that if [itex]u = y_1 + y_2[/itex] and [itex]v = y_1 - y_2[/itex] then
    \frac{d^2 u}{dx^2} = 0 \\
    \frac{d^2 v}{dx^2} = 2av
  6. Mar 14, 2014 #5
    Yes but how do you solve it formally? I use the eigenvalue method to get:

    y1 = 0
    y2 = Acosh(sqrt(2a)x) + Bsinh(sqrt(2a)x) but differentiating this twice and subbing this into the second equation and solving for y1 doesn't satisfy the boundary conditions
    Last edited: Mar 14, 2014
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