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Differential equation question

  1. Dec 21, 2012 #1
    What is the logic behind equating differential equations to zero? For example the equation

    Because it can just as easily be written y''-5y'=-6y
    I am interested in the meaning of why if we sum y''+(-5y')+6y equals zero. What is the relationship of its second derivative, first derivative, and 6y?
  2. jcsd
  3. Dec 21, 2012 #2
    I would say it is similar to solving quadratic equations, saying ax^2+bx+c=0. We could move the c over, but after factoring to (x-x_1)(x-x_2)=0, it is easier to read off the roots. Also, sometimes there is a function of t on the right,


    and the business on the left is considered in methods apart and together with f. There's a number of reasons, and at the same time, it's not really a big deal. We can move c over, in ax^2+bx=-c, and we often do if we are completing the square, but it's still usually presented as "=0".

    Try reading some of the methods on techniques for solving the homogeneous second-order constant-coefficient differential equation (f=0). Try also reading about finding a particular solution for the inhomogeneous second-order linear differential equation (they have some techniques that work for linear in general, and some that only work for the constant-coefficient case). Then try to find where they descripe ho to put together the homogeneous solutions and a particular solution to get the full solution set.

    This may give you an idea as to why it's more comfortable to put all the y terms on the left. Also, note that, if anything, it might make more sense to put the y'' on the left, like y''=f(t,y,y'). Compare this with something you may have seen, y'=p(t)y(t)+q(t)=f(t,y).

    In retrospect, one might answer your question by saying we write it this way when our differential equation is a "linear" differential equation, meaning, as opposed to any general differential equation which can be written


    it is of the special form


    That is, the coefficients are functions of t, and there are no terms like sin(y'') or (y')^3. The form we have written it in is in a form ready to use many of the methods of solving. Why on earth do we have such a bizarre definition of linear? I feel this is because differential equations of the above form, obey a very nice property, in that we can add solutions. In other words, the solution space forms a vector space (offset by a particular solution, so I think we call it affine. Think of it as a plane shifted away from the origin.)
    Last edited: Dec 21, 2012
  4. Dec 22, 2012 #3


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    There is no "logic" to it, it's just customary. Whether we write it as y''- 6y'+ 5y= 0 or y''+ 5y= 6y' or y''= 6y'- 5y, we note that the coefficients are constants so that whether we say they add to 0 or y''+ 5y is equal to 6y', etc. they must be the same kind of function so that can cancel or be equal. And, we know that exponentials have the nice property that their derivatives are again exponentials.

    So we "try" a solution of the form [itex]y= e^{rx}[/itex] so that [itex]y'= re^{rx}[/itex] and [itex]y''= r^2e^{rx}[/itex]. Putting those into any of the previous forms gives:
    [itex]y''- 6y'+ 5y= r^2e^{rx}- 6re^{rx}+ 5e^{rx}= e^{rx}(r^2- 6r+ 5)= 0[/itex] or
    [itex]y''+ 5y= e^{rx}(r^2+ 5)= 6y'= 6re^{rx}[/itex] or
    [itex]y''= r^2e^{rx}= 6y'- 5y= (6r- 5)e^{rx}[/itex]

    In any case, since [itex]e^{rx}[/itex] is not 0 for any x, we can cancel it to get
    [itex]r^2- 6r+ 5= 0[/itex] or [itex]r^2+ 5= 6r[/itex] or [itex]r^2= 6r- 5[/itex].
    That is a quadratic equation. In the first form, we can factor it to get [itex](r- 5)(r- 1)[/itex] (in itself, perhaps a good reason for writing it in that form). That tells us that r= 1 or r= 5 so that both [itex]e^x[/itex] and [itex]e^{5x}[/itex] satisfy the differential equation.

    Now, there is some "logic" to the rest- it can be shown that the set of all solutions to a linear nth order differential equation forms an n dimentional vector space. Here that means that since [itex]e^x[/itex] and [itex]e^{5x}[/itex] are independent solutions, any solution to this differential equation can be written in the form [itex]y(x)= C_1e^x+ C_2e^{5x}[/itex]
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