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Why 2nd order differential equation has two solutions

  1. Jan 20, 2016 #1

    goodphy

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    Hello.

    I forgot the reason why 2nd order differential equation has two independent solutions. (Here, source term is zero) Why 3 or 4 independent solutions are not possible?

    Please give me clear answer.
     
  2. jcsd
  3. Jan 20, 2016 #2

    jbriggs444

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    How many independent boundary conditions can there be a 2nd order differential equation? If someone specifies x(0) and x'(0), is x''(0) independent from those?
     
  4. Jan 20, 2016 #3
    You need to be a little careful. The statement as you've stated it is not correct. (This is a common error).

    Consider the equation [itex] \left(y''+y\right) \left(y''-y \right)=0 [/itex]

    You can check that this equation is 2nd order by expanding the left side:
    [itex] \left(y''\right)^2-y^2=0 [/itex]

    The highest derivative of y is 2nd order. However this equation has 4 independent solutions:
    [itex] y = c_1 e^x + c_2 e^{-x} + c_3 cos x + c_4 sin x [/itex]

    You should note that this equation is also nonlinear.

    A correct statement is that a second order linear differential equation of the form [itex] y'' + f(x)y' + g(x) y =0[/itex] has two independent solutions on an interval I if f and g are both continuous on the interval I.

    This theorem can be generalized to higher dimensions.

    A proof of the above statement is given in Tenenbaum and Pollard Lesson 65. In my opinion this is a great reference for differential equations.

    You can probably also find the proof online using some google-fu if you search for general solution of 2nd order (or n-th order) differential equations.
     
  5. Jan 20, 2016 #4

    goodphy

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    I see.

    2nd order linear differential equation without source (homogeneous linear differential equation) can be always written as y'' + P(x)y' + Q(x)y = 0 where ' means differentiation. Full solution is obtained by solving this equation with initial conditions (equation describes evolution of the system from initial conditions). If we know, y(0) and y'(0) then the equation tells y''(0). These initial conditions gives next time values y(t1), y'(t1) and substituting these into the equation tells y''(t1). Next time values are obtained in similar process so full solution is obtained.

    As coefficients of independent solutions are rooms for initial conditions and we need only two conditions to fully describe system evolution thus 2nd order homogeneous differential equation only requires two independent solutions as constituents of general solutions.

    Thanks to remind me of this!
     
  6. Jan 20, 2016 #5

    HallsofIvy

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    Every linear homogeneous second order differential equation has two independent solution because the set of all solutions to an nth order linear homogenous equation is a vector space of dimension n. If you want to include non-linear equations, you will have to specify what you mean by "independent".
     
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